Revision as of 00:44, 27 November 2023 by Admin (Created page with "Both <math>n</math> and <math>L</math> are unknown <math>500\left(1-v^{2 n}\right)+500\left(1-v^n\right)=720</math> (check formulas in the amortization table). So <math>-500 v^{2 n}-500 v^n+280=0</math>. This is a quadratic in <math>v^n</math> so from the quadratic formula, <math>v^n=.4</math> (the other root is negative). We want the amount of interest paid in year 10 , namely <math>500\left(1-v^{2 n-9}\right)</math>. Also <math>v^9=\frac{1}{1.0494^9}=.64793</math>. He...")
Exercise
ABy Admin
Nov 27'23
Answer
Both [math]n[/math] and [math]L[/math] are unknown [math]500\left(1-v^{2 n}\right)+500\left(1-v^n\right)=720[/math] (check formulas in the amortization table). So [math]-500 v^{2 n}-500 v^n+280=0[/math]. This is a quadratic in [math]v^n[/math] so from the quadratic formula, [math]v^n=.4[/math] (the other root is negative).
We want the amount of interest paid in year 10 , namely [math]500\left(1-v^{2 n-9}\right)[/math]. Also [math]v^9=\frac{1}{1.0494^9}=.64793[/math]. Hence
[[math]]
\text { Interest Paid In Tenth Year }=500\left(1-v^{2 n-9}\right)=500\left(1-(.4)^2\left(.64793^{-1}\right)\right)=376.53
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.