Answer to Question #119811 in Financial Math for Sela

Question

. 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

Answer 1

A. P=$326,000-$75,000=$251,000 is the loan amount, loan's principal.

В. A = K · P is the value of the monthly payment

$K=\frac{r\cdot(1+r)^{n}}{(1+r)^n-1}$ - annual factor

$r= \frac{9}{12}=0.75$% = 0.0075 is the rate of interest expressed as a fraction; 9% - annual rate.

$n=12\cdot30=360$ is the number of payments; for monthly payments over 30 years, 12 months x 30 years = 360 payments.

$K=\frac{0.0075\cdot(1+0.0075)^{360}}{(1+0.0075)^{360}-1}$ =0,00804622616

A=0,00804622616*$251,000=$2 019.60276853$\approx$ $2019.61 is the value of the monthly payment

360*A=$727057 is the total mortgage value

С. Amount owed at end of month N:

$P_{N}=(1+r)^{N}P-{\frac {(1+r)^{N}-1}{r}}A$

the loan outstanding after making 20 payments:

$P_{20}=(1+0.0075)^{20}\cdot251000-\frac{(1+0.0075)^{20}-1}{0.0075}\cdot2019.61=248 053,338967\approx$ $248,053.34

D. The principal repaid in the 21st payment:

$P-P_{21}=251000-(1+0.0075)^{21}\cdot251000-\frac{(1+0.0075)^{21}-1}{0.0075}\cdot2019.61=3 105,87099124\approx 3105.87$

Answer:

А. $251,000 =-the loan's principal

$727057 = the total mortgage value

B. $2019.61 =the value of the monthly payment

C. $248,053.34 = the loan outstanding after making 20 payments

D. $3105.87 = the principal repaid in the 21st payment.