. 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)
Answers
Answer:
Step-by-step explanation:
A.The cost of the house is $ 326,000
The deposit is $ 75,000
The mortgage value is $(326,000 - 75,000) = $ 251,000
=$251,000
The Amortizing period is 360 months (30 years)
rate of interest is 9%.
per month rate of interest is (9/12)= 0.75%.
B. To calculate per month payment :
P ₐ =x∗ r (1−(1+r) ⁻ⁿ )
x: monthly payment
P: principal amount.
r: rate of interest converted to per month equivalent.
n: number of periods.
P = $251,000
r = 9/12=0.75% = 0.0075
n= 30 years = 360 months.
x= 251000∗0.0075 /1−(1.0075) ⁻³⁶⁰
1882.5 / 0.932113 = 2019.59
monthly payment is $ 2019.6
C. After 20 payments the number of months left to be paid is (360-20)=340
We will calculate the outstanding balance of left 340 months when we know family is paying $2019.6 every month @ 9% annually.
Pn = x∗ r (1−(1+r) -n )
x is the monthly payment ,here x = $2019.6
r = rate of interest ,here r= 0.0075
and n = number of months left ,here n = 340
∴
P n =2019.6 0.0075 (1−(1+0.0075) -340 )
2019.6∗ 0.921171 / 0.0075
=2019.6∗122.8228
=248052.92
the balance outstanding after 20 payments is $ 248052.92
D. Principal repaid in the 21st payment:
This can be calculated simply by applying P340 -− P339
Then
P₃₃₉ = 2019.6∗ 1−(1+0.0075) ⁻³³⁹/0.0075
= 2019.6∗ (1-0.0794)/ (0.0075 )
=2019.6∗ 0.9206 / 0.0075
=2019.6∗122.747
= 247899.84
Therefore the principal paid in 21st payment is
P₃₄₀ − P₃₃₉ = 248052.92-247899.84
=153.08
$ 153.08