Answer to Question #119697 in Financial Math for Piya

Question

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)


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

I.326 000-75 000=2 51 000

A.

$2019.60\times360=727056$

interests=727 056-251 000=476056

B.

$251 000=A\times\frac{1-(1.0075)^{-360}}{0.0075}251000$

A=2019.60

C.

$727056-2019.60\times20=686664727056−2019.60×20=686664$

D.

$2019.60*21=42411.602019.60∗21=42411.60$

As part of the amount of payments, the principal amount and interest for using the loan

III.

A. $J1=(1+\frac{j2}{2})^2-1=(1+0.045)^2-1=0.092025$

$i=q\times[(1+\frac{r}{m})^{m/q}-1]$

$i=12\times[(1+\frac{0.092025.}{1})^{1/12}-1]$

i=0.088357 8.8357%

B.

242 720 - The remainder

$242720=A\times\frac{1-(1.007363)^{-310}}{0.007363}$

A=1 992.12

C.

360 full payments required to repay the loan

727 056-725 036.4=2019.6