Question

QUESTION 2: LOAN AMORTIZATION [35 MARKS]

I. A family buys a house worth $326,000. They pay $75,000 deposit and take a mortgage for the balance at J12=9% p.a. to be amortized over 30 years with monthly payments.

A. Find the value of the mortgage on their house? (1 mark)
B. Find the value of the monthly payment? (3 marks)
C. Find the loan outstanding after making 20 payments? (4 marks)
D. Find the principal repaid in the 21st payment? (5 marks)

II. Fill out the loan amortization schedule provided in the solution template for the first 5 loan payments. What do you notice about the composition of the payment amount? (6 marks)

III. Suppose that after making 50 payments, the interest rate changes to J2=9% p.a.:

A. Convert the interest rate J2=9% to J12 equivalent (2 marks
B. Assuming that the family seeks to accept the change in interest rates, what would be their new payment based on the new interest rate? (5 marks)
C. Assuming that the family seeks to continue their initial monthly payment calculated in part I, how many full payments would be required to pay off the loan and what would be the final concluding smaller payment one period later? (9 marks)

Answer 1

Answer:

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I. A family buys a house worth $326,000. They pay $75,000 deposit and take a mortgage for the balance at J12=9% p.a. to be amortized over 30 years with monthly payments.

A. Find the value of the mortgage on their house? (1 mark)

B. Find the value of the monthly payment? (3 marks)

C. Find the loan outstanding after making 20 payments? (4 marks)

D. Find the principal repaid in the 21st payment? (5 marks)

II. Fill out the loan amortization schedule provided in the solution template for the first 5 loan payments. What do you notice about the composition of the payment amount? (6 marks)

III. Suppose that after making 50 payments, the interest rate changes to J2=9% p.a.:

A. Convert the interest rate J2=9% to J12 equivalent (2 marks

B. Assuming that the family seeks to accept the change in interest rates, what would be their new payment based on the new interest rate? (5 marks)

C. Assuming that the family seeks to continue their initial monthly payment calculated in part I, how many full payments would be required to pay off the loan and what would be the final concluding smaller payment one period later? (9

Answer 2

Answer:

${\fbox{\boxed {\huge{\rm{\red{Answer}}}}}}$

Given:

• Cost of the house $ 326,000
• Initial deposit : $ 75,000
• mortgage value : $(326,000 - 75,000) = $ 251,000

${\underline{\underline{\large{\tt{Answer:A:251,000}}}}}$

=> Now Amortizing period = 360 months (30 years)

=> rate of interest is 9%.

=> and per month rate of interest is (9/12)= 0.75%.

${\bold{\large{(B.)}}}$

=> To calculate per month payment we will use the following formula :

${:{\implies{\rm{Pn = x \times \frac{( 1 - {1 + r})^{ - n} }{r}}}}}$

${\longmapsto{\rm{x: monthly \: payment}}}\\</p><p></p><p> {\longmapsto{\rm{P: principal\: amount.}}} \\ </p><p></p><p>{\longmapsto{\rm{ r: rate \: of\: interest \: converted \: to }}} \\{\rm{\:\:\:\:\:\:\:\:per month \: equivalent}}. \\ </p><p></p><p>{\longmapsto{\rm{n: number \: of \: periods.}}}$

Now here

• P = $251,000

• r = 9/12=0.75% = 0.0075

• n= 30 years = 360 months.

${\implies{\rm{251000 = (x) \frac{1 - ( {1 + 0.0075)}^{ - 360} }{0.0075}}}}</p><p>$

${\rm{or \: x = \frac{251000 \times0.0075 }{1 - {(1.0075}^{-360}}}}$

${:{\implies{\rm{ \frac{1882.5}{0.932113}}}}}$

${\boxed{\LARGE{:{\implies{\rm{2019.59}}}}}}$

${\bold{Answer:B\:monthly\:payment\:is\: \ 2019.6}}$

${\bold{\large{</strong><strong>(</strong><strong>C</strong><strong>.</strong><strong>)</strong><strong>}}}$

=> After making 20 payments ,number of months left to be paid is (360-20)=340

=> so we need to calculate the outstanding balance of left 340 months when we know family is paying $2019.6 every month @ 9% annually.

So,

${:{ \implies{\rm{P_n=2019.6 \frac{1 - {1 + 0.0075}^{ - 340} }{0.0075}}}}}</p><p>$

${:{\implies{\rm{2019.6 \times \frac{0.921171}{0.0075}}}}}</p><p>$

${:{\implies{\rm{2019.6 \times122.8228}}}}$

${\boxed{\LARGE{:{\implies{\rm{ 248052.92}}}}}}$

${\bold{Answer : C\: the \:balance\: outstanding \:after}} \\ {\bold{making\: 20\: payments \: is \: \ 248052.92}}</p><p></p><p></p><p>$

${\bold{\large{(D.)}}}$

${\Rightarrow{\bold {\tt{Principal \: repaid \: in \: the \: 21st \: payment:}}}}$

=>This can be calculated simply by applying

${\Rightarrow{\rm{P_{340} - P_{339}}}}</p><p>$

Now

${:{\implies{\rm{ P_ {339}= 2019.6 \times \frac{1 - ( {1 + 0.0075)}^{ - 339} }{0.0075}}}}}$

${:{\implies{\rm{ 2019.6 \times \frac{(1 - 0.0794)}{0.0075}}}}}$

${:{\implies{\rm{ 2019.6 \times \frac{ (0.9206)}{0.0075}}}}}$

${:{\implies{\rm{2019.6 \times 122.747 }}}}$

${\LARGE{\blue{\boxed{:{\implies{\rm{247899.84 }}}}}}}$

so principal paid in 21st payment is

$=> {\rm{P_{340}-P_{339}= 248052.92-247899.84= 153.08}}$

${\huge{\pink{\boxed{\rm{Answer : D: 153.08}}}}}$

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