SOA Exam FM Sample Question 311
Study guide: Exam FM-2
A credit card company charges an annual effective interest rate of 16.8%. Interest accumulates from the date of purchase.
A borrower's credit card balance was X at the beginning of month 1. Starting with month 1, the borrower purchased satellite internet service, resulting in a 79.99 charge on the credit card at the middle of each month.
The borrower paid 250 at the end of each month. Immediately after the payment in month 15, the balance was 3000.
Determine which of the following is an equation of value that can be used to solve for X.
- A)
X+\displaystyle\sum_{n=1}^{15}\frac{79.99}{(1.014)^{n-0.5}}=\frac{3000}{(1.014)^{15}}+\sum_{n=1}^{15}\frac{250}{(1.014)^n} - B)
X+\displaystyle\sum_{n=1}^{16}\frac{79.99}{(1.168)^{\frac{n+0.5}{12}}}=\frac{3000}{(1.168)^{\frac{5}{4}}}+\sum_{n=1}^{16}\frac{250}{(1.168)^{\frac{n}{12}}} - C)
X+\displaystyle\sum_{n=1}^{16}\frac{79.99}{(1.168)^{\frac{n-0.5}{12}}}=\frac{3000}{(1.168)^{\frac{4}{3}}}+\sum_{n=1}^{16}\frac{250}{(1.168)^{\frac{n}{12}}} - D)
X+\displaystyle\sum_{n=1}^{15}\frac{79.99}{(1.168)^{\frac{n+0.5}{12}}}=\frac{3000}{(1.168)^{\frac{5}{4}}}+\sum_{n=1}^{15}\frac{250}{(1.168)^{\frac{n}{12}}} - E)
X+\displaystyle\sum_{n=1}^{15}\frac{79.99}{(1.168)^{\frac{n-0.5}{12}}}=\frac{3000}{(1.168)^{\frac{5}{4}}}+\sum_{n=1}^{15}\frac{250}{(1.168)^{\frac{n}{12}}}
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