SOA Exam FM Sample Question 77
Study guide: Exam FM-2
An investor's retirement account pays an annual nominal interest rate of 4.2%, convertible monthly.
On January 1 of year y, the investor's account balance was X. The investor then deposited 100 at the end of every quarter. On May 1 of year (y+10), the account balance was 1.9X.
Determine which of the following is an equation of value that can be used to solve for X.
- A)
\dfrac{1.9X}{(1.0105)^{124/3}} + \displaystyle\sum_{k=1}^{42}\dfrac{100}{(1.0105)^{k-1}} = X - B)
X + \displaystyle\sum_{k=1}^{42}\dfrac{100}{(1.0035)^{3(k-1)}} = \dfrac{1.9X}{(1.0035)^{124}} - C)
X + \displaystyle\sum_{k=1}^{41}\dfrac{100}{(1.0035)^{3k}} = \dfrac{1.9X}{(1.0035)^{124}} - D)
X + \displaystyle\sum_{k=1}^{41}\dfrac{100}{(1.0105)^{k-1}} = \dfrac{1.9X}{(1.0105)^{124/3}} - E)
X + \displaystyle\sum_{k=1}^{42}\dfrac{100}{(1.0105)^{k-1}} = \dfrac{1.9X}{(1.0105)^{124/3}}
Answer this question and see the worked solution