Revision as of 19:58, 26 November 2023 by Admin (Created page with "'''Solution: A''' There are 10 payments at times <math>0,4,8, \ldots, 36</math>. Let <math>j</math> be the interest rate for a 4 year period and let <math>i</math> be the annual rate. Then <math>(1+j)=(1+i)^4</math>. Thus <math>100 \ddot{s}_{\overline{10} \mid j}=5(100) \ddot{s}_{\overline{5} \mid j}</math>. So <math display = "block">100 \frac{(1+j)^{10}-1}{j}(1+j)=5(100) \frac{(1+j)^5-1}{j}(1+j)</math>. Simplifying, using <math>(1+j)^{10}-1=\left((1+j)^5-1\right)\left...")
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Exercise


Nov 26'23

Answer

Solution: A

There are 10 payments at times [math]0,4,8, \ldots, 36[/math]. Let [math]j[/math] be the interest rate for a 4 year period and let [math]i[/math] be the annual rate. Then [math](1+j)=(1+i)^4[/math]. Thus [math]100 \ddot{s}_{\overline{10} \mid j}=5(100) \ddot{s}_{\overline{5} \mid j}[/math]. So

[[math]]100 \frac{(1+j)^{10}-1}{j}(1+j)=5(100) \frac{(1+j)^5-1}{j}(1+j)[[/math]]

. Simplifying, using [math](1+j)^{10}-1=\left((1+j)^5-1\right)\left((1+j)^5+1\right)[/math] gives [math]\left((1+j)^5+1\right)=5[/math] so [math](1+j)^5=4[/math] so

[[math]]X=100 \frac{(1+j)^{10}-1}{j}(1+j)=100 \frac{4^2-1}{4^{1 / 5}-1} 4^{1 / 5}=6194.72[[/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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