SOCIETY OF ACTUARIES
EXAM FM
FINANCIAL MATHEMATICS
Practice Question Bank
Time Value of Money • Annuities • Loans • Bonds
Duration • Convexity • Immunization • Swaps • Yield Curves
Compiled from study materials, sample questions, and past examination patterns
in accordance with the current SOA Exam FM syllabus.
150 Unique Multiple-Choice Questions with Step-by-Step Solutions
,Contents
1. Time Value of Money
2. Annuities and Non-Contingent Cash Flows
3. Loans and Amortization
4. Bonds
5. Duration, Convexity, Immunization, Swaps and Yield Curves
1. Time Value of Money
Q1. An investor places 4,000 into a savings account that credits interest at a nominal annual rate of 6
percent convertible monthly. The investor leaves the money undisturbed for exactly five years. Calculate
the accumulated value of the account at the end of the five-year period, rounded to the nearest dollar.
A. 5,348
B. 5,395
C. 5,442
D. 5,489
E. 5,536
Answer: B
Rationale: Monthly effective rate equals 0.06/12 = 0.005 and the number of compounding periods is 60.
Accumulated value = 4000 × (1.005)^60 ≈ 4000 × 1.34885 = 5,395.40, which rounds to 5,395.
SOA Exam FM Practice Bank | Page 2
,Q2. The force of interest earned by a fund is constant and equal to 0.04. A single payment of 7,500 is
scheduled to be received six years from today. Determine the present value of that payment under the
stated force of interest, rounded to the nearest dollar.
A. 5,900
B. 5,950
C. 6,000
D. 6,050
E. 6,100
Answer: A
Rationale: Under a constant force of interest the discount factor is e^(−δt). Present value = 7500 × e^(−0.04
× 6) = 7500 × e^(−0.24) ≈ 7500 × 0.786627 = 5,899.70, which rounds to 5,900.
Q3. An account credits interest according to a simple-interest accumulation function at an annual rate of 8
percent. A payment of 12,000 is due four years from now. Calculate the present value of this future
payment under the simple-interest law, rounded to the nearest dollar.
A. 8,571
B. 9,000
C. 9,091
D. 9,600
E. 10,000
Answer: C
Rationale: The accumulation function is a(t) = 1 + 0.08t. Therefore the present value is 12000 / (1 + 0.08 × 4)
= .32 ≈ 9,090.91, which rounds to 9,091.
SOA Exam FM Practice Bank | Page 3
, Q4. The annual effective rate of discount applicable to a financial transaction is 5 percent. Convert this
discount rate into the equivalent annual effective rate of interest and express the result as a percentage
rounded to two decimal places.
A. 5.00%
B. 5.13%
C. 5.26%
D. 5.41%
E. 5.56%
Answer: C
Rationale: The relationship between effective interest and effective discount is i = d / (1 − d). Substituting d
= 0.05 yields i = 0..95 ≈ 0.0526316, or 5.26 percent.
Q5. A fund grows according to the accumulation function a(t) = exp(0.03t + 0.001t²) for t ≥ 0. An actuary
needs the instantaneous rate of growth at a specific moment. Calculate the force of interest that prevails
at time t = 4.
A. 0.034
B. 0.038
C. 0.042
D. 0.046
E. 0.050
Answer: B
Rationale: The force of interest is the derivative of the natural logarithm of the accumulation function. ln
a(t) = 0.03t + 0.001t², so δ_t = 0.03 + 0.002t. At t = 4 the force equals 0.03 + 0.008 = 0.038.
SOA Exam FM Practice Bank | Page 4
EXAM FM
FINANCIAL MATHEMATICS
Practice Question Bank
Time Value of Money • Annuities • Loans • Bonds
Duration • Convexity • Immunization • Swaps • Yield Curves
Compiled from study materials, sample questions, and past examination patterns
in accordance with the current SOA Exam FM syllabus.
150 Unique Multiple-Choice Questions with Step-by-Step Solutions
,Contents
1. Time Value of Money
2. Annuities and Non-Contingent Cash Flows
3. Loans and Amortization
4. Bonds
5. Duration, Convexity, Immunization, Swaps and Yield Curves
1. Time Value of Money
Q1. An investor places 4,000 into a savings account that credits interest at a nominal annual rate of 6
percent convertible monthly. The investor leaves the money undisturbed for exactly five years. Calculate
the accumulated value of the account at the end of the five-year period, rounded to the nearest dollar.
A. 5,348
B. 5,395
C. 5,442
D. 5,489
E. 5,536
Answer: B
Rationale: Monthly effective rate equals 0.06/12 = 0.005 and the number of compounding periods is 60.
Accumulated value = 4000 × (1.005)^60 ≈ 4000 × 1.34885 = 5,395.40, which rounds to 5,395.
SOA Exam FM Practice Bank | Page 2
,Q2. The force of interest earned by a fund is constant and equal to 0.04. A single payment of 7,500 is
scheduled to be received six years from today. Determine the present value of that payment under the
stated force of interest, rounded to the nearest dollar.
A. 5,900
B. 5,950
C. 6,000
D. 6,050
E. 6,100
Answer: A
Rationale: Under a constant force of interest the discount factor is e^(−δt). Present value = 7500 × e^(−0.04
× 6) = 7500 × e^(−0.24) ≈ 7500 × 0.786627 = 5,899.70, which rounds to 5,900.
Q3. An account credits interest according to a simple-interest accumulation function at an annual rate of 8
percent. A payment of 12,000 is due four years from now. Calculate the present value of this future
payment under the simple-interest law, rounded to the nearest dollar.
A. 8,571
B. 9,000
C. 9,091
D. 9,600
E. 10,000
Answer: C
Rationale: The accumulation function is a(t) = 1 + 0.08t. Therefore the present value is 12000 / (1 + 0.08 × 4)
= .32 ≈ 9,090.91, which rounds to 9,091.
SOA Exam FM Practice Bank | Page 3
, Q4. The annual effective rate of discount applicable to a financial transaction is 5 percent. Convert this
discount rate into the equivalent annual effective rate of interest and express the result as a percentage
rounded to two decimal places.
A. 5.00%
B. 5.13%
C. 5.26%
D. 5.41%
E. 5.56%
Answer: C
Rationale: The relationship between effective interest and effective discount is i = d / (1 − d). Substituting d
= 0.05 yields i = 0..95 ≈ 0.0526316, or 5.26 percent.
Q5. A fund grows according to the accumulation function a(t) = exp(0.03t + 0.001t²) for t ≥ 0. An actuary
needs the instantaneous rate of growth at a specific moment. Calculate the force of interest that prevails
at time t = 4.
A. 0.034
B. 0.038
C. 0.042
D. 0.046
E. 0.050
Answer: B
Rationale: The force of interest is the derivative of the natural logarithm of the accumulation function. ln
a(t) = 0.03t + 0.001t², so δ_t = 0.03 + 0.002t. At t = 4 the force equals 0.03 + 0.008 = 0.038.
SOA Exam FM Practice Bank | Page 4
