SOA Exam FM Sample Question 233
Study guide: Exam FM-2
An investor's new savings account earns an annual effective interest rate of 3% for each of the first ten years and an annual effective interest rate of 2% for each year thereafter.
The investor deposits an amount X at the beginning of each year, starting with year 1, so that the account balance just after the deposit in the beginning of year 26 is 100,000.
Determine which of the following is an equation of value that can be used to solve for X.
- A)
\dfrac{100{,}000}{(1.03)^{10}(1.02)^{15}}=X\displaystyle\sum_{k=1}^{11}\frac{1}{(1.03)^{k-1}}+X\sum_{k=12}^{26}\frac{1}{(1.02)^{k-1}} - B)
\dfrac{100{,}000}{(1.03)^{10}(1.02)^{16}}=X\displaystyle\sum_{k=1}^{10}\frac{1}{(1.03)^k}+X\sum_{k=11}^{26}\frac{1}{(1.02)^k} - C)
\dfrac{100{,}000}{(1.03)^{10}(1.02)^{16}}=X\displaystyle\sum_{k=1}^{11}\frac{1}{(1.03)^{k-1}}+X\sum_{k=12}^{26}\frac{1}{(1.03)^{10}(1.02)^{k-11}} - D)
\dfrac{100{,}000}{(1.03)^{10}(1.02)^{15}}=X\displaystyle\sum_{k=1}^{11}\frac{1}{(1.03)^{k-1}}+X\sum_{k=12}^{26}\frac{1}{(1.03)^{10}(1.02)^{k-11}} - E)
\dfrac{100{,}000}{(1.03)^{10}(1.02)^{16}}=X\displaystyle\sum_{k=1}^{10}\frac{1}{(1.03)^k}+X\sum_{k=11}^{26}\frac{1}{(1.03)^{10}(1.02)^{k-10}}
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