Revision as of 22:07, 17 November 2023 by Admin (Created page with "'''Solution: E''' The accumulation function is <math display = "block"> a(t)=\exp\biggl[\int_{0}^{t}(s^{2}/100)d s\biggr]=\exp(t^{3}/300). </math> The accumulated value of 100 at time 3 is <math>100 \exp(3^3 / 300) = 109.41743.</math> The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6 minus the accumulated value at time 3. Thus <math display = "block"> \begin{array}{l}{{\left(109.41743+X\right)[a(6)/a(3)-1]=X}}\\ {{\left(109....")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: E
The accumulation function is
[[math]]
a(t)=\exp\biggl[\int_{0}^{t}(s^{2}/100)d s\biggr]=\exp(t^{3}/300).
[[/math]]
The accumulated value of 100 at time 3 is [math]100 \exp(3^3 / 300) = 109.41743.[/math]
The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6 minus the accumulated value at time 3. Thus
[[math]]
\begin{array}{l}{{\left(109.41743+X\right)[a(6)/a(3)-1]=X}}\\ {{\left(109.41743+X\right)(2.0544332/1..0941743-1)=X}}\\ {{\left(109.41743+X\right)(3.877613=X}}\\ {{X=784.61.}}\end{array}
[[/math]]
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