Exercise


ABy Admin
Nov 27'23

Answer

Solution: E

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.

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