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.

Solution: C

[[math]]A_X(t)=1 e^{\int_0^t \delta_r d r}=e^{.005 t^2+.1 t} \cdot 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.