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

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A demand (variable rate)mortgage of $137 000.00 is amortized over 20 years by equal monthly payments. After 21 months the original interest rate of 6% p.a was raised to 6.6% p.a. Three years after the mortgage was taken out, it was renewed at the request for the mortgagor for a five-year term at a fixed rate of 7.25% p.a. a)Calculate the mortgage balance after 21 months. b)Compute the size of the new monthly payment at the 6.6% rate of interest. c)Determine the mortgage balance at the end of the five-year term.

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a)p = a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 - 1 = 0.00486755
137000 = PMT a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 137000 = PMT(141.3843090)
$968.99 = PMT
PVn = 968.99 a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 = 968.99(134.5076431)= $130 336.56
b)p = a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 - 1 = 0.005340319
130335.56 = PMT a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 130335.56 = PMT(128.9285035)
$1010.91 = PMT
c)PVn = 1010.91 a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 1010.91(124.0775295)= $125,431.22
p = a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 - 1 = 0.005849741
125431.22 = PMT a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 125431.22 = PMT(118.9350584)
$1054.62 = PMT
PVn = 1054.62 a)p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431)= $130 336.56 b)p = - 1 = 0.005340319 130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)PVn = 1010.91 1010.91(124.0775295)= $125,431.22 p = - 1 = 0.005849741 125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 = 1054.62(97.14104632)= $102 446.89 = 1054.62(97.14104632)= $102 446.89

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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a)n = a)n = = = = 32.5019768 b)PVn = 4570.00 = 4570.00(0.48727778)= $2226.86 Final payment = 2226.86(1.0403)= $2316.60 = a)n = = = = 32.5019768 b)PVn = 4570.00 = 4570.00(0.48727778)= $2226.86 Final payment = 2226.86(1.0403)= $2316.60 = a)n = = = = 32.5019768 b)PVn = 4570.00 = 4570.00(0.48727778)= $2226.86 Final payment = 2226.86(1.0403)= $2316.60 = 32.5019768
b)PVn = 4570.00 a)n = = = = 32.5019768 b)PVn = 4570.00 = 4570.00(0.48727778)= $2226.86 Final payment = 2226.86(1.0403)= $2316.60 = 4570.00(0.48727778)= $2226.86
Final payment = 2226.86(1.0403)= $2316.60

Lisa saved $750 000.00. If she withdraws $12 000.00 at the beginning of every month and interest is 12% compounded monthly, what is the size of the last withdrawal?

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750000 = 12000 750000 = 12000 (1.01) 61.88 = 0.6188 = 1 - 1.01-n ln 0.38118811 = -nln 1.01 (0.9644622765)= -n(-0.0099503309) n = 96.927659 PVn(due)= 12000 (1.01) = 12000(0.9303234212)= $11 135.91 The final payment is $11 135.91 (1.01)
61.88 = 750000 = 12000 (1.01) 61.88 = 0.6188 = 1 - 1.01-n ln 0.38118811 = -nln 1.01 (0.9644622765)= -n(-0.0099503309) n = 96.927659 PVn(due)= 12000 (1.01) = 12000(0.9303234212)= $11 135.91 The final payment is $11 135.91 0.6188 = 1 - 1.01-n
ln 0.38118811 = -nln 1.01
(0.9644622765)= -n(-0.0099503309)
n = 96.927659
PVn(due)= 12000 750000 = 12000 (1.01) 61.88 = 0.6188 = 1 - 1.01-n ln 0.38118811 = -nln 1.01 (0.9644622765)= -n(-0.0099503309) n = 96.927659 PVn(due)= 12000 (1.01) = 12000(0.9303234212)= $11 135.91 The final payment is $11 135.91 (1.01)
= 12000(0.9303234212)= $11 135.91
The final payment is $11 135.91

Andre has saved $152 000.00. If he withdraws $1750.00 at the beginning of every month and interest is 7.5% compounded monthly, what is the size of the last withdrawal?

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Question 4

The Taylors agreed to monthly payments rounded up to the nearest $100.00 on a mortgage of $136 000.00 amortized over 15 years. Interest for the first five years was 8.5% compounded semi-annually. After 30 months, as permitted by the mortgage agreement, the Taylors increased the rounded monthly payment by 10%. a)Determine the mortgage balance at the end of the five-year term. b)If the interest rate remains unchanged over the remaining term, how many more of the increased payments will amortize the mortgage balance? c)How much did the Taylors save by exercising the increase-in-payment option?

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Question 5

A mortgage balance of $137 960.70 is to be repaid over a eleven-year term by equal monthly payments at 8.1% compounded semi-annually. At the request of the mortgagor, the monthly payments were set at $1140.00. a)How many payments will the mortgagor have to make? b)What is the size of last payment? c)Determine the difference between the total actual amount paid and the total amount required to amortize the mortgage by the contractual monthly payments.

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Question 6

Rebecca borrowed $20 400 from her local credit union to pay for the upgrades in her kitchen. She wants to pay the loan with monthly payments over the period of 3 years. What is the interest included in the tenth payment, if the rate charged by the credit union is 6% monthly?

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Question 7

Rebecca bought a house in Oshawa and took a $259 000 mortgage amortized by monthly payments over 25 years that is renewable after 5 years. If interest is 4.4% compounded semi-annually, what is the outstanding balance at the end of the five year term? If the mortgage is renewed for a further 5 year term at 3.7% compounded semi-annually, what is the balance after the second 5 year term?

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Question 8

Mr. Lamb borrowed $8321.00 at 11.12% compounded monthly. He agreed to repay the loan in equal monthly payments over five years. a)What is the size of the monthly payment? b)How much of the 24th payment is interest? c)What is the principal repaid in the 37th payment? d)Prepare a partial amortization schedule showing details of the first three payments, the last three payments, and totals.

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Question 9

Cody borrowed $9100.00 from his credit union. He agreed to repay the loan by making equal monthly payments for 4 years. Interest is 9% compounded monthly. a)What is the size of the monthly payments? b)How much will the loan cost him? c)How much will Cody owe after 16 months? d)How much interest will he pay in his 26th payment? e)How much of the principal will be repaid in the 18th payment? f)Prepare a partial amortization schedule showing details of the first three payments, Payments 24, 25, 26, the last three payments, and totals.

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Question 10

Shiva transferred his mortgage of $145 981 from CIBC to BMO for the remaining amortization period at 3.59% compounded semi-annually. He has agreed to pay rounded up payment of $2000 every month. How long will it take Shiva to pay-off his mortgage?

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Question 11

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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Question 12

How much principal is repaid in the 74th payment interval on a $142 300 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 7.44% compounded monthly.

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Question 13

Big Printing Inc. is repaying a loan of $37 300.00 by making payments of $2300.00 at the end of every three months. If interest is 3.5% compounded quarterly, construct an amortization schedule showing the total paid and the total cost of the loan. What is the total interest paid and cost of the loan?

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Question 14

How much interest is paid in total on a 1-year loan for $5000? The interest rate is 12 % compounded monthly and the payments are monthly.

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Question 15

Big Printing Inc. is repaying a loan of $37 300.00 by making payments of $2300.00 at the end of every three months. If interest is 3.5% compounded monthly, construct an amortization schedule showing the total paid and the total cost of the loan. What is the total interest paid and cost of the loan?

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Question 16

A loan of $19 000.00 is repaid by quarterly payments of $900.00 each at 8% compounded quarterly. What is the principal repaid by the 21st payment?

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Question 17

John makes a monthly mortgage payment of $1940 at the end of every month. On October 31, his loan balance was $145 000 after making a payment of $1940. His financial institute allows him to make one additional payment equal to or less than his monthly payment, once a month. Taking advantage of that, John made a payment of $1500 on November 15 and $1900 on December 20. What is his year-end balance after making monthly payment on December 31? The effective monthly fixed rate is 0.306%.

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Question 18

A $30 000.00 mortgage is amortized by monthly payments over twenty years and is renewable after five years. a)If the interest rate is 8.5% compounded semi-annually, calculate the outstanding balance at the end of the five-year term. b)If the mortgage is renewed for a further three-year term at 8% compounded semi-annually, calculate the size of the new monthly payment. c)Calculate the payout figure at the end of the three-year term.

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Question 19

Catrina receives payments of $1120.00 at the beginning of each month from a pension fund of $79 500.00. Interest earned by the fund is 11.36% compounded monthly. a)What is the number of payments Catrina will receive? b)What is the size of the final payment?

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