Revision as of 21:07, 29 November 2023 by Admin (Created page with "'''Solution: B''' <math display="block"> \begin{aligned} & \begin{array}{llllllll} 0 & 12 i & 12 i & \ldots & 12 i & 24 i & 24 i \ldots 24 i & \text { (Interest Payment) } \end{array} \\ & 1-1-|-\ldots-|-|-| \ldots \mid \\ & \begin{array}{lllllllll} 0 & 1 & 2 & \ldots & 10 & 11 & 12 & \ldots & 20 \end{array} \\ & \text { (Time) } \\ & 64=12 i s_{\overline{20} \mid .75 i}+12 i s_{\overline{10} \mid .75 i}=12 i \frac{(1+.75 i)^{20}-1}{.75 i}+12 i \frac{(1+.75 i)^{10}-1}{...")
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Exercise


Nov 29'23

Answer

Solution: B

[[math]] \begin{aligned} & \begin{array}{llllllll} 0 & 12 i & 12 i & \ldots & 12 i & 24 i & 24 i \ldots 24 i & \text { (Interest Payment) } \end{array} \\ & 1-1-|-\ldots-|-|-| \ldots \mid \\ & \begin{array}{lllllllll} 0 & 1 & 2 & \ldots & 10 & 11 & 12 & \ldots & 20 \end{array} \\ & \text { (Time) } \\ & 64=12 i s_{\overline{20} \mid .75 i}+12 i s_{\overline{10} \mid .75 i}=12 i \frac{(1+.75 i)^{20}-1}{.75 i}+12 i \frac{(1+.75 i)^{10}-1}{.75 i}=16 x^2+16 x-32 \\ & \end{aligned} [[/math]]

where [math]x=(1+.75 i)^{10}[/math]. So [math]16 x^2+16 x-96=0[/math] so [math]x^2+x-6=0[/math] so [math](x+3)(x-2)=0[/math] so [math]x=2[/math]. Then [math](1+.75 i)^{10}=2[/math] so [math]i=.095698[/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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