Revision as of 22:44, 17 November 2023 by Admin (Created page with "'''Solution: A''' The monthly interest rate is 0.072/12 = 0.006. 6500 five years from today has value 6500(1.006)<sup>-60</sup> = 4359.77. The equation of value is <math display = "block"> 4539.77=1700(1.006)^{-n}+3400(1.006)^{-2n}. </math> Let <math>x = 1.006^{-n}</math>. Then, solve the quadratic equation <math display = "block"> \begin{align*} 3400x^{2}+1700x-4539.77=0 \\ x={\frac{-1700+{\sqrt{1700^{2}-4(34000)(-4539.77)}}}{2(3400)}}=0.93225. \end{align*} </mat...")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: A
The monthly interest rate is 0.072/12 = 0.006. 6500 five years from today has value 6500(1.006)-60 = 4359.77. The equation of value is
[[math]]
4539.77=1700(1.006)^{-n}+3400(1.006)^{-2n}.
[[/math]]
Let [math]x = 1.006^{-n}[/math]. Then, solve the quadratic equation
[[math]]
\begin{align*}
3400x^{2}+1700x-4539.77=0
\\
x={\frac{-1700+{\sqrt{1700^{2}-4(34000)(-4539.77)}}}{2(3400)}}=0.93225.
\end{align*}
[[/math]]
Then
[[math]]
1.006^{-\,n}=0.9325\Rightarrow-n\ln(1.006)=\ln(0.93225)\Rightarrow n=11.73.
[[/math]]
To ensure there is 6500 in five years, the deposits must be made earlier and thus the maximum integral value is 11.
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.