True or false. A mortgage bond of R800k with 10.5% interest over 20 years is calculated at R800k/1000 x 09.98 = R 7984.00/ month over the 20 years

Question

Calculate monthly interest on a loan of R800000 with 10.5% interest over 20 years

Answer 1

Let's solve this problem algebraically, then we can substitute actual values to see if the monthly payment is correct.

Let m be the monthly payment and r% the annual percentage interest rate. We start with the principal P₀ (the initial amount borrowed). The interest rate is per annum so it needs to be calculated as a monthly rate=r/1200 as a fraction.

After the first month before the monthly payment the initial amount grows to P₀(1+r/1200) due to interest. Let g (growth factor)=1+r/1200. After the monthly payment this is reduced to P₁=P₀g-m, and this becomes the principal for the second month, P₂=P₁g-m=(P₀g-m)g-m=P₀g²-mg-m, P₃=P₂g-m.

So P₃=(P₀g²-mg-m)g-m=P₀g³-mg²-mg-m=P₀g³-m(1+g+g²). So there's a pattern.

In general, P_n=P_n-1(1+r/1200)-m is the recursive equation for the principal after n months.

N years=12N months. So n=12N and P_n=0 because it's an N-year mortgage so there will be nothing to pay after N years.

So P_12N=P₀g^12N-m(1+g+g²+...+g^12N-1).

1+g+g²+...+g^12N-1 is a geometric sequence and it sums to (g^12N​​​​​​​-1)/(g-1).

Therefore, since P_12N=0, m(g^12N​​​​​​​-1)/(g-1)=P₀g^12N. From this equation m can be calculated given P₀, r and n. P₀=800000; N=20, so n=240; r=10.5/1200, so g=1+10.5/1200=1.00875.

g²⁴⁰=8.09192 approx; g-1=0.00875.

Therefore m=800000×8.09192/(7.09192/0.00875)=R7987.04. Compare with R7984.00. The discrepancy is probably due to rounding errors, so the given amount is 99.96% correct.