Revision as of 22:56, 17 November 2023 by Admin (Created page with "'''Solution: E''' The accumulated values for Funds <math>X</math> and <math>Y</math> are <math>1000\left(1+\frac{k}{2}\right)^{10}</math> and <math>921.90\left(1-\frac{k}{2}\right)^{-10}</math> respectively. Equating them and solving for <math>k</math> : <math display = "block"> \begin{aligned} & 1000\left(1+\frac{k}{2}\right)^{10}=921.90\left(1-\frac{k}{2}\right)^{-10} \\ & 0.9219=\left[\left(1+\frac{k}{2}\right)\left(1-\frac{k}{2}\right)\right]^{10}=\left(1-\frac{k^2...")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: E
The accumulated values for Funds [math]X[/math] and [math]Y[/math] are [math]1000\left(1+\frac{k}{2}\right)^{10}[/math] and [math]921.90\left(1-\frac{k}{2}\right)^{-10}[/math] respectively. Equating them and solving for [math]k[/math] :
[[math]]
\begin{aligned}
& 1000\left(1+\frac{k}{2}\right)^{10}=921.90\left(1-\frac{k}{2}\right)^{-10} \\
& 0.9219=\left[\left(1+\frac{k}{2}\right)\left(1-\frac{k}{2}\right)\right]^{10}=\left(1-\frac{k^2}{4}\right)^{10} \\
& 1-\frac{k^2}{4}=0.9919 \\
& k^2=0.0324 \\
& k=0.18
\end{aligned}
[[/math]]
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