Revision as of 00:29, 27 November 2023 by Admin (Created page with "On January 1, 2010, Amil has the following options for repaying a loan. # Sixty monthly payments of 100 beginning February 1, 2010. # 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. <ul class="mw-excansopts"><li>29</li><li>29.5</li><li>30</li><li>30.5</li><li>31</li></ul> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtest...")
ABy Admin
Nov 27'23
Exercise
On January 1, 2010, Amil has the following options for repaying a loan.
- Sixty monthly payments of 100 beginning February 1, 2010.
- 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.
ABy Admin
Nov 27'23
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.