SOA Exam FM Sample Question 162
Study guide: Exam FM-2
Trish had a loan with a balance of 4000 at the beginning of month 1. Starting with month 1, and every month thereafter, she made a payment of X in the middle of the month. At the beginning of month 4, and every 6 months thereafter, she borrowed an additional 800.
Trish's loan balance became 4000 again at the end of month 36.
The annual nominal interest rate for the loan is 26.4%, convertible quarterly.
Determine which of the following is an equation of value that can be used to solve for X.
- A)
\displaystyle\sum_{n=1}^{6}\frac{800}{\left(1+\frac{0.2640}{4}\right)^{2n-1}}=\sum_{n=1}^{36}\frac{X}{\left(1+\frac{0.2640}{4}\right)^{\frac{n-0.5}{3}}} - B)
\displaystyle\sum_{n=1}^{6}\frac{800}{\left(1+\frac{0.2640}{12}\right)^{6n-3}}=\sum_{n=1}^{36}\frac{X}{\left(1+\frac{0.2640}{12}\right)^{n-0.5}} - C)
\displaystyle 4000+\sum_{n=1}^{6}\frac{800}{\left(1+\frac{0.2640}{4}\right)^{\frac{6n-2}{3}}}=\frac{4000}{\left(1+\frac{0.2640}{4}\right)^{12}}+\sum_{n=1}^{36}\frac{X}{\left(1+\frac{0.2640}{4}\right)^{\frac{n}{3}}} - D)
\displaystyle 4000+\sum_{n=1}^{6}\frac{800}{\left(1+\frac{0.2640}{12}\right)^{6n-3}}=\frac{4000}{\left(1+\frac{0.2640}{12}\right)^{36}}+\sum_{n=1}^{36}\frac{X}{\left(1+\frac{0.2640}{12}\right)^{n-0.5}} - E)
\displaystyle 4000+\sum_{n=1}^{6}\frac{800}{\left(1+\frac{0.2640}{4}\right)^{2n-1}}=\frac{4000}{\left(1+\frac{0.2640}{4}\right)^{12}}+\sum_{n=1}^{36}\frac{X}{\left(1+\frac{0.2640}{4}\right)^{\frac{n-0.5}{3}}}
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