Revision as of 23:51, 18 November 2023 by Admin (Created page with "'''Solution: A''' The principal repaid in the first payment is 100 – iL. The outstanding principal is <math display = "block"> L – 100 + iL = L + 25 </math> Hence, iL = 125. Also <math display = "block"> \begin{array}{l l}{{L=300(a_{\overline{16}|}-200a_{\overline{8}|})-\frac{300(1- v^{16})-200(1- v^{8})}{i}}}\\ {{125=i L=100+200 v^{8} - 300v^{16}}}\\ {{300 v^{16}-200 v^{8} + 25 = 0}}& \\ {{ v^{8}=\frac{200\pm\sqrt{200^{2}-4(300)(25)}}{600}=\frac{200\pm100}{600}...")
Exercise
ABy Admin
Nov 18'23
Answer
Solution: A
The principal repaid in the first payment is 100 – iL. The outstanding principal is
[[math]]
L – 100 + iL = L + 25
[[/math]]
Hence, iL = 125. Also
[[math]]
\begin{array}{l l}{{L=300(a_{\overline{16}|}-200a_{\overline{8}|})-\frac{300(1- v^{16})-200(1- v^{8})}{i}}}\\ {{125=i L=100+200 v^{8} - 300v^{16}}}\\ {{300 v^{16}-200 v^{8} + 25 = 0}}&
\\ {{ v^{8}=\frac{200\pm\sqrt{200^{2}-4(300)(25)}}{600}=\frac{200\pm100}{600}=0.5.
}}
\end{array}
[[/math]]
The larger of the two values is used due to the value being known to exceed 0.3. The outstanding valance at time eight is the present value of the remaining payments:
[[math]]
300a_{\overline{{{8}}}|}=300\frac{1-0.5}{2^{1/8}-1}=1657.
[[/math]]