Revision as of 12:22, 19 November 2023 by Admin (Created page with "'''Solution: E''' Let L = the loan amount. Note that <math></math>. The equation of value is <math display = "block"> P\cdot a_{\overline{{{k}}}|\;i}=L=120\cdot a_{\overline{5k}|j} </math> so that <math display = "block"> \begin{aligned} P=\frac{120a_{\overline{{{5k}}}\mid j}}{a_{\overline{{{k}}}\mid i}} \\ =120\frac{1-(1+j)^{-5k}}{j}\frac{i}{1-(1+i)^{-k}} \\ =120\frac{1-\left(1+i\right)^{-k}}{j}\frac{i}{1-\left(1+i\right)^{-k}} \\ =120 \frac{i}{j} \\ =120\frac{\lef...")
Exercise
ABy Admin
Nov 19'23
Answer
Solution: E
Let L = the loan amount. Note that [math][/math]. The equation of value is
[[math]]
P\cdot a_{\overline{{{k}}}|\;i}=L=120\cdot a_{\overline{5k}|j}
[[/math]]
so that
[[math]]
\begin{aligned}
P=\frac{120a_{\overline{{{5k}}}\mid j}}{a_{\overline{{{k}}}\mid i}} \\
=120\frac{1-(1+j)^{-5k}}{j}\frac{i}{1-(1+i)^{-k}} \\
=120\frac{1-\left(1+i\right)^{-k}}{j}\frac{i}{1-\left(1+i\right)^{-k}} \\
=120 \frac{i}{j} \\
=120\frac{\left(1+j\right)^{5}-1}{j}
\end{aligned}
[[/math]]
Next, using the fact that 0 < j < 0.04, we get
[[math]]5\lt{\frac{\left(1+j\right)^{5}-1}{j}}\lt5.41\,633[[/math]]
by plugging in a small value like 0.000000001 and 0.04 resulting in P equaling more than 600 but less than 650.