On January 1, 2010, Amil has the following options for repaying a loan.

1. Sixty monthly payments of 100 beginning February 1, 2010.
2. A single payment of 6000 at the end of K months.

Interest is at a nominal rate of 12% compounded monthly. The two options have the same present value.

Determine K.

• 29
• 29.5
• 30
• 30.5
• 31

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

Solution: A

[[math]] \begin{aligned} & 100 a_{\overline{60} \mid .01}=6000 v^K \text { so } \\ & 100 \frac{1-1.01^{-60}}{.01}=6000(1.01)^{-K} \text { so } \\ & K=-\left(\ln (10 / 6)+\ln \left(1-1.01^{-60}\right)\right) / \ln (1.01)=29.01227 .\end{aligned} [[/math]]

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.