Outstanding loan balance

Question

A loan of ${$1000}$ is being repaid with annual payments over 10 years. The size of the payment in the first five years is ${$ k}$. It is found that the payments in the last five years are five times the payments in the first five years. Let the annual rate $r=0.08$, calculate

(i) The value of $K$

(ii)The outstanding loan balance after making the $4th$ payment

(iii) The outstanding loan balance after making the $5th$, hence the principal amortised in the $5th$ payment.

$$ \begin{array}{c|c|c} t=0 & ...t=5 & t=6 ... t=10 \\ \hline \\ K... &K & (5K)... \\ \end{array} $$

(i) In order to find $K$:

$L=K a_{5|0.08}+(5K)(a_{5|0.08}).(1+i)^{-5}$

$1000=K[\frac{1-1.08^{-5}}{0.08}]+5K[\frac{1-1.08^{-5}}{0.08}](1.08)^{-5} \rightarrow K=56.88$

(ii)Using the retrospective method to find $B_4^r$

$B^r_4=1000(1.08)^4-56.88[\frac{1.08^4-1}{0.08}]=1104.18$

(iii)I have tried some different thing. I doubt it works.

$B^5_r=1104.18-56.88[\frac{1.08-1}{0.08}]$=1047.30

My answers are absurd. I get to play more than the loan itself.

Answer 1

1. In order to find $K$ $$ L=K\,a_{\overline{5}|i}+5K\,v^5\,a_{\overline{5}|i}=K\,a_{\overline{5}|i}(1+5v^5) $$ and then $$ K=\frac{L}{a_{\overline{5}|i}(1+5v^5)}=56.88422\approx 56.88 $$
2. The outstanding balance at $t=4$ $$ B_4=K\,a_{\overline{5-4}|i}+5K\,v^5\,a_{\overline{5}|i}=Kv+5K\,v^5\,a_{\overline{5}|i}= 825.55 $$
3. The outstanding balance at $t=5$ $$ B_5=5K\,v^5\,a_{\overline{5}|i}= 772.88 $$

 t    Paymet   Principal   Interest     Debt
 0                                    1,000.00 
 1    56.88       38.71      18.17      961.29 
 2    56.88       41.81      15.07      919.47 
 3    56.88       45.16      11.73      874.32 
 4    56.88       48.77       8.12      825.55 
 5    56.88       52.67       4.21      772.88 
 6   284.42      131.74     152.68      641.14 
 7   284.42      142.28     142.14      498.85 
 8   284.42      153.66     130.76      345.19 
 9   284.42      165.96     118.46      179.23 
10   284.42      179.23     105.19        0.00