Revision as of 18:58, 19 November 2023 by Admin (Created page with "'''Solution: B''' The two equations are: <math display = "block"> \begin{array}{l}{{P=(10,000r)a_{\overline{50}|0.04}+9,000(1.04)^{-5}=44,518.22r+7,397.34}}\\ {{1.2P=[10,000(r+0.01)]a_{\overline{50}|0.04}+11,000(1.04)^{-5}=44,518.22r+9,486.38}}\end{array} </math> Subtracting the first equation from the second gives 0.2P = 2089.04 for P = 10,445.20. Inserting this in the first equation gives r = (10,445.20 – 7,397.34)/44,518.22 = 0.0685. {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 19'23
Answer
Solution: B
The two equations are:
[[math]]
\begin{array}{l}{{P=(10,000r)a_{\overline{50}|0.04}+9,000(1.04)^{-5}=44,518.22r+7,397.34}}\\ {{1.2P=[10,000(r+0.01)]a_{\overline{50}|0.04}+11,000(1.04)^{-5}=44,518.22r+9,486.38}}\end{array}
[[/math]]
Subtracting the first equation from the second gives 0.2P = 2089.04 for P = 10,445.20. Inserting this in the first equation gives r = (10,445.20 – 7,397.34)/44,518.22 = 0.0685.
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.