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Exercise


ABy Admin
Nov 18'23

Answer

Solution: A

[[math]] \begin{align*} 167.50 &=10a_{\overline{5}|9.2\%}+10(1.092)^{-5}\sum_{t=1}^{c}\left[\frac{(1+k)}{1\cdot092}\right]^{t} \\ 167.50 &=38.6955+6.44001{\frac{(1+k)/1.092}{1-(1+k)/1.092}} \\ (167.50-38.6955)[1-(1+k)/1.092] &= 6.4400](1+k)/1.092 \\ 128.8045 &= 135.24451(1+k)/1.092 \\ 1+k &= 0.0400 \\ k &= 0.0400 \Rightarrow K= 4.0\% \end{align*} [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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