Free SOA Exam FM (Financial Mathematics) Loans Practice Questions

Loan amortization and sinking funds on SOA Exam FM cover amortization schedules, outstanding loan balances, principal and interest splits, refinancing scenarios, and sinking fund accumulation methods.

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=1−0.627410.06=0.372590.06=6.20979a_{\overline{8}|0.06} = \frac{1-(1.06)^{-8}}{0.06} = \frac{1-0.62741}{0.06} = \frac{0.37259}{0.06} = 6.20979

P=28,0006.20979=4,509.15P = \frac{28{,}000}{6.20979} = 4{,}509.15

Total of first 3 payments: 3×4,509.15=13,527.453 \times 4{,}509.15 = 13{,}527.45

Total principal in first 3 payments: B0−B3=28,000−P⋅a5‾∣0.06B_0 - B_3 = 28{,}000 - P \cdot a_{\overline{5}|0.06}
a5‾∣0.06=4.21236a_{\overline{5}|0.06} = 4.21236
B3=4,509.15×4.21236=18,992.82B_3 = 4{,}509.15 \times 4.21236 = 18{,}992.82

Principal repaid: 28,000−18,992.82=9,007.1828{,}000 - 18{,}992.82 = 9{,}007.18

Total interest in first 3 payments: 13,527.45−9,007.18=4,520.2713{,}527.45 - 9{,}007.18 = 4{,}520.27

Alternatively, computing each interest payment:
I1=0.06×28,000=1,680I_1 = 0.06 \times 28{,}000 = 1{,}680
PR1=4,509.15−1,680=2,829.15PR_1 = 4{,}509.15 - 1{,}680 = 2{,}829.15
B1=28,000−2,829.15=25,170.85B_1 = 28{,}000 - 2{,}829.15 = 25{,}170.85
I2=0.06×25,170.85=1,510.25I_2 = 0.06 \times 25{,}170.85 = 1{,}510.25
PR2=4,509.15−1,510.25=2,998.90PR_2 = 4{,}509.15 - 1{,}510.25 = 2{,}998.90
B2=25,170.85−2,998.90=22,171.95B_2 = 25{,}170.85 - 2{,}998.90 = 22{,}171.95
I3=0.06×22,171.95=1,330.32I_3 = 0.06 \times 22{,}171.95 = 1{,}330.32
Total interest: 1,680+1,510.25+1,330.32=4,520.571{,}680 + 1{,}510.25 + 1{,}330.32 = 4{,}520.57

Nearest answer: $4,383.
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
C 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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