Revision as of 17:47, 26 November 2023 by Admin (Created page with "'''Solution: D''' <math display = "block"> \begin{aligned} & 1000(1+i / 2)^{10}=X .(1) \\ & 1980=1000(1+i / 2)^{14}(1+i / 2)^{14}=1000(1+i / 2)^{28}(2) \\ & \text { From }(2), 1.98=(1+i / 2)^{28} \text { so }(1+i / 2)^{10}=1.98^{10 / 28} \\ & \text { Thus } X=1000(1+i / 2)^{10}=1000 * 1.98^{10 / 28}=1276.30\end{aligned} </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor.ca | t...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: D
[[math]]
\begin{aligned} & 1000(1+i / 2)^{10}=X .(1) \\ & 1980=1000(1+i / 2)^{14}(1+i / 2)^{14}=1000(1+i / 2)^{28}(2) \\ & \text { From }(2), 1.98=(1+i / 2)^{28} \text { so }(1+i / 2)^{10}=1.98^{10 / 28} \\ & \text { Thus } X=1000(1+i / 2)^{10}=1000 * 1.98^{10 / 28}=1276.30\end{aligned}
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.