Revision as of 00:06, 27 November 2023 by Admin (Created page with "'''Solution: A''' <math>1000=P a_{\overline{10} \mid .10}=P \frac{1-1.1^{-10}}{.10}</math> so <math>P=\frac{.10(1000)}{1-1.1^{-10}}=162.7454</math>. Amount in sinking fund after payment of loan is <math display="block"> (P-100) s_{\overline{10} \mid .14}-1000=62.7454 \frac{1.14^{10}-1}{.14}-1000=213.3263 . </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor.ca | title = Unive...")
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Exercise


ABy Admin
Nov 27'23

Answer

Solution: A

[math]1000=P a_{\overline{10} \mid .10}=P \frac{1-1.1^{-10}}{.10}[/math] so [math]P=\frac{.10(1000)}{1-1.1^{-10}}=162.7454[/math]. Amount in sinking fund after payment of loan is

[[math]] (P-100) s_{\overline{10} \mid .14}-1000=62.7454 \frac{1.14^{10}-1}{.14}-1000=213.3263 . [[/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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