Free SOA Exam FM (Financial Mathematics) Loans Practice Questions

Loans on SOA Exam FM cover amortization schedules, outstanding balances by the prospective and retrospective methods, the interest and principal split of each payment, drop and balloon payments, and refinancing. The SOA syllabus allocates 15-25% of Exam FM to loans (SOA).

170 questions 63 easy 67 medium 40 hard 2026 syllabus

Sample Questions

Question 1 Easy
Which of the following best describes the term of a loan?
Solution
E is correct.

The term of a loan is the length of time from when the loan is originated to when the final payment is made and the loan is fully retired. For example, a 30-year mortgage has a term of 30 years.
Question 2 Medium
A $28,000 loan at 6% effective annual interest is repaid with 8 level annual payments. Determine the total interest paid in the first 3 payments.
Solution
C is correct.

Annual payment:
a8‾∣0.06=1−(1.06)−80.06=6.20979a_{\overline{8}|0.06} = \frac{1-(1.06)^{-8}}{0.06} = 6.20979
P=28,0006.20979=4,509.17P = \frac{28{,}000}{6.20979} = 4{,}509.17

Compute the interest portion of each of the first three payments (interest = rate ×\times outstanding balance):
I1=0.06×28,000=1,680.00I_1 = 0.06 \times 28{,}000 = 1{,}680.00
B1=28,000−(4,509.17−1,680.00)=25,170.83B_1 = 28{,}000 - (4{,}509.17 - 1{,}680.00) = 25{,}170.83
I2=0.06×25,170.83=1,510.25I_2 = 0.06 \times 25{,}170.83 = 1{,}510.25
B2=25,170.83−(4,509.17−1,510.25)=22,171.91B_2 = 25{,}170.83 - (4{,}509.17 - 1{,}510.25) = 22{,}171.91
I3=0.06×22,171.91=1,330.31I_3 = 0.06 \times 22{,}171.91 = 1{,}330.31

Total interest in the first three payments:
1,680.00+1,510.25+1,330.31=4,520.56≈4,5211{,}680.00 + 1{,}510.25 + 1{,}330.31 = 4{,}520.56 \approx 4{,}521
Question 3 Hard
A loan of $80,000 is repaid with level annual payments at the end of each year for 25 years at an annual effective interest rate of 6%. Determine the interest portion of the 10th payment.
Solution
D is correct.

The annual payment is:

PMT=80,000a25‾∣at i=0.06PMT = \frac{80{,}000}{a_{\overline{25}|}} \quad\text{at } i = 0.06

v=1/1.06v = 1/1.06, v25=0.232999v^{25} = 0.232999.

a25‾∣=1−0.2329990.06=0.7670010.06=12.783356a_{\overline{25}|} = \frac{1 - 0.232999}{0.06} = \frac{0.767001}{0.06} = 12.783356

PMT=80,00012.783356=6,258.14PMT = \frac{80{,}000}{12.783356} = 6{,}258.14

The interest portion of the 10th payment equals i×OB9i \times OB_9, where OB9OB_9 is the outstanding balance after 9 payments:

OB9=PMT⋅a16‾∣OB_9 = PMT \cdot a_{\overline{16}|}

v16=0.393646v^{16} = 0.393646

a16‾∣=1−0.3936460.06=0.6063540.06=10.105895a_{\overline{16}|} = \frac{1 - 0.393646}{0.06} = \frac{0.606354}{0.06} = 10.105895

OB9=6,258.14×10.105895=63,244.08OB_9 = 6{,}258.14 \times 10.105895 = 63{,}244.08

I10=0.06×63,244.08=3,794.64≈3,795I_{10} = 0.06 \times 63{,}244.08 = 3{,}794.64 \approx 3{,}795
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