A couple decides to save money for their child's first year college tuition. The parents will deposit 1700 n months from today and another 3400 2n months from today. All deposits earn interest at a nominal annual rate of 7.2%, compounded monthly.

Calculate the maximum integral value of n such that the parents will have accumulated at least 6500 five years from today.

• 11
• 12
• 18
• 24
• 25

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

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.