Revision as of 23:14, 26 November 2023 by Admin (Created page with "'''Solution: A''' <math>\begin{aligned} & 55 a_{\overline{20} \mid}=30 a_{\overline{10} \mid}+60 a_{\overline{10} \mid} v^{10}+90 a_{\overline{10} \mid} v^{20} \text { so } \\ & 55\left(1-v^{20}\right)=55\left(1-v^{10}\right)\left(1+v^{10}\right)=30\left(1-v^{10}\right)+60\left(1-v^{10}\right) v^{10}+90\left(1-v^{10}\right) v^{20} \text {. But } v^{10} \neq 1 \text { so } \\ & 55\left(1+v^{10}\right)=30+60 v^{10}+90 v^{20} \text { so } 11(1+x)=6+12 x+18 x^2 \text { wher...")
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Exercise


Nov 26'23

Answer

Solution: A

[math]\begin{aligned} & 55 a_{\overline{20} \mid}=30 a_{\overline{10} \mid}+60 a_{\overline{10} \mid} v^{10}+90 a_{\overline{10} \mid} v^{20} \text { so } \\ & 55\left(1-v^{20}\right)=55\left(1-v^{10}\right)\left(1+v^{10}\right)=30\left(1-v^{10}\right)+60\left(1-v^{10}\right) v^{10}+90\left(1-v^{10}\right) v^{20} \text {. But } v^{10} \neq 1 \text { so } \\ & 55\left(1+v^{10}\right)=30+60 v^{10}+90 v^{20} \text { so } 11(1+x)=6+12 x+18 x^2 \text { where } x=v^{10} \text {. Thus } 18 x^2+x-5=0 \text {. So } \\ & v^{10}=x=\frac{-1 \pm \sqrt{1+4(18)(5)}}{36}=18 / 36=.5 \text { so } i=2^{.1}-1=.072 \text {. } \\ & \text { Thus } P V=55 a_{\overline{20} \mid}=55 \frac{1-v^{20}}{i}=55 \frac{1-.5^2}{.072}=574.725 \\ & \end{aligned}[/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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