Revision as of 20:41, 26 November 2023 by Admin (Created page with "For a given positive integer <math>n</math>, a rate of interest <math>i</math> can be found such that <math>4s_{\overline{2n}|} = 9s_{\overline{n}|}</math>. Express in terms of <math>n</math> how long it will take for money to double at this rate of interest. <ul class="mw-excansopts"> <li>3n</li> <li>3.05n</li> <li>3.11n</li> <li>3.15n</li> <li>3.22n</li> </ul> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=...")
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ABy Admin
Nov 26'23

Exercise

For a given positive integer [math]n[/math], a rate of interest [math]i[/math] can be found such that [math]4s_{\overline{2n}|} = 9s_{\overline{n}|}[/math]. Express in terms of [math]n[/math] how long it will take for money to double at this rate of interest.

  • 3n
  • 3.05n
  • 3.11n
  • 3.15n
  • 3.22n

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

ABy Admin
Nov 26'23

Solution: C

We need [math]t[/math] such that [math](1+i)^t=2[/math] so [math]t=\ln 2 /(\ln (1+i))[/math]. But [math]4 s_{\overline{2 n} \mid}=9 s_{\bar{n} \mid}[/math] so [math]4 \frac{(1+i)^{2 n}-1}{i}=[/math] [math]9 \frac{(1+i)^n-1}{i}[/math] so [math]4 \frac{\left((1+i)^n-1\right)\left((1+i)^n+1\right)}{i}=9 \frac{(1+i)^n-1}{i}[/math] so [math](1+i)^n+1=9 / 4[/math] so [math](1+i)^n=5 / 4=1.25[/math]. Thus [math]n \ln (1+i)=\ln 1.25[/math]. Finally [math]t=\ln 2 /(\ln (1+i))=n \ln 2 / \ln 1.25=3.106 n[/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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