A bank agrees to supply you a house mortgage on a home valued at $2 000 000, if you pay %20 deposit and repay the balance over 30 years.

a) How much is your monthly repayment at a fixed rate of interest 3% compounded monthly?

b) Find the principal repaid and the interest paid by the end of 5 years period (i.e. after 60 payments).

Guest Aug 25, 2019

#1**0 **

20% down payment on $2,000,000.00 = $2,000,000 x 0.20 =$400,000.00

$2,000,000 - $400,000 =$1,600,000 - amount of mortgage.

Here is an amortization schedule which has All the numbers you need in it. Just press "monthly schedule" and you will see all the payments taken out. Go down to 60th payment, or year end 5, and you will see the balance of the mortgage after 60 payments.

Subtract the remaining balance from $1,600,000 and you should get: $177,498.19 which is the principal amount paid in 60 months or 5 years.

Then, take the monthly payment at the top =$6,745.66 x 60 months =$404,739.60 which is total principal + interest paid over 5 years or 60 months.

$404,739.60 - $177,498.19(principal paid above) =$227,241.41 - which is the interest paid on the mortgage over that period of 5 years or 60 months.

Here is your amortization schedule: **https://www.calculator.net/amortization-calculator.html?cloanamount=1600%2C000.00&cloanterm=30&cinterestrate=3.00&printit=0&x=44&y=17**

**Note: If you don't understand something, just let us know here.**

Guest Aug 25, 2019

edited by
Guest
Aug 25, 2019

edited by Guest Aug 25, 2019

edited by Guest Aug 25, 2019

edited by Guest Aug 25, 2019

edited by Guest Aug 25, 2019

#2**0 **

\(\text{Here's a bit of a derivation of the monthly payment}\\ \text{ignore months and years for a moment and just let $r$ be the periodic rate}\\ \text{$n$ will be the number of periods. Let $\alpha = (1+r)$}\\ \text{every period we take the principal, apply interest, and subtract a payment}\\ p(k+1) = \alpha p(k) -pmt,~p(0) = loan\\ \text{If you work through this you find that}\\ p(n) = \alpha^n loan - pmt \sum \limits_{k=0}^{n-1} \alpha^k = \alpha^n loan - pmt\dfrac{\alpha^n-1}{\alpha -1}\)

\(\text{We want $p(n)=0$ and we need to solve for $pmt$}\\ \alpha^n loan = pmt \dfrac{\alpha^n -1}{\alpha-1}\\ loan = pmt \dfrac{1-\alpha^{-n}}{\alpha -1}\\ pmt =loan \dfrac{(\alpha-1)}{1-\alpha^{-n}} = loan \dfrac{r}{1-(1+r)^{-n}}\)

\(\text{Here we plug in $loan=1600000,~n=360,~r=\dfrac{0.03}{12}=0.0025$}\\ pmt = \$1600000\dfrac{0.0025}{1-(1.0025)^{-360}} = \$6745.66\)

.Rom Aug 25, 2019