Revision as of 14:10, 26 November 2023 by Admin (Created page with "In Fund X money accumulates at force of interest <math>\delta_t=.01 t+.10</math>, for <math>0 < t< 20</math>. In Fund Y money accumulates at annual effective rate <math>i</math>. An amount of $1 is invested in each fund, and the accumulated values are the same at the end of 20 years. Find the value in Fund Y at the end of 1.5 years. <ul class="mw-excansopts"> <li>1.3</li> <li>1.32</li> <li>1.35</li> <li>1.37</li> <li>1.4</li> </ul> '''References''' {{cite web |url=htt...")
ABy Admin
Nov 26'23
Exercise
In Fund X money accumulates at force of interest [math]\delta_t=.01 t+.10[/math], for [math]0 \lt t\lt 20[/math]. In Fund Y money accumulates at annual effective rate [math]i[/math]. An amount of $1 is invested in each fund, and the accumulated values are the same at the end of 20 years. Find the value in Fund Y at the end of 1.5 years.
- 1.3
- 1.32
- 1.35
- 1.37
- 1.4
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
[[math]]A_X(t)=1 e^{\int_0^t \delta_r d r}=e^{.005 t^2+.1 t}, \, A_Y(t)=(1+i)^t[[/math]]
Thus [math]e^{.005(20)^2+.1(20)}=e^4=[/math] set [math]=(1+i)^{20}[/math]. We want to find
[[math]]
(1+i)^{1.5}=\left((1+i)^{20}\right)^{1.5 / 20}=\left(e^4\right)^{1.5 / 20}=e^{.3}=1.34986
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.