Free SOA Exam FM (Financial Mathematics) Bonds Practice Questions

Bond pricing and yield calculations on SOA Exam FM test par, premium, and discount bonds, callable bonds, yield curves, and Macaulay and modified duration. Bond valuation is one of the most heavily weighted topics on the exam.

224 questions 87 easy 76 medium 61 hard 2026 syllabus

Sample Questions

Question 1 Easy
A bond is said to be "callable" if:
Solution
B is correct.

A callable bond gives the issuer (not the bondholder) the right to redeem the bond before its stated maturity date, typically at a specified call price.
Question 2 Medium
Which of the following correctly describes a premium bond's amortization schedule?
Solution
D is correct.

For a premium bond (P>CP > C), we have Fr>CjFr > Cj. The book value starts above CC and decreases each period toward CC. The amortization of premium in period tt is PAt=Fr−j⋅Bt−1PA_t = Fr - j \cdot B_{t-1}. Since Bt−1B_{t-1} decreases over time, j⋅Bt−1j \cdot B_{t-1} decreases, making PAtPA_t larger. So the amortization of premium increases each period. Book value decreases and amortization amounts increase.
Question 3 Hard
A 1,000 par value 20-year bond with 6% annual coupons is purchased at par. Coupons are reinvested at 4%. Calculate the annual realized yield over the 20-year period.
Solution
E is correct.

The bond is bought at par, so P=1,000P = 1{,}000 and each annual coupon is 0.06⋅1,000=600.06 \cdot 1{,}000 = 60.

Coupons accumulate at the reinvestment rate of 4%:
s20‾∣0.04=(1.04)20−10.04=2.191123−10.04=29.778079.s_{\overline{20}|0.04} = \frac{(1.04)^{20} - 1}{0.04} = \frac{2.191123 - 1}{0.04} = 29.778079.

Accumulated value at maturity (reinvested coupons plus redemption of par):
AV=60(29.778079)+1,000=1,786.68+1,000=2,786.68.AV = 60(29.778079) + 1{,}000 = 1{,}786.68 + 1{,}000 = 2{,}786.68.

The realized annual yield jj satisfies P(1+j)20=AVP(1+j)^{20} = AV:
j=(2,786.681,000)1/20−1=(2.78668)0.05−1.j = \left(\frac{2{,}786.68}{1{,}000}\right)^{1/20} - 1 = (2.78668)^{0.05} - 1.

Using ln⁡(2.78668)=1.02485\ln(2.78668) = 1.02485, we get 1.0248520=0.051243\frac{1.02485}{20} = 0.051243 and e0.051243=1.05258e^{0.051243} = 1.05258.

j=0.0526=5.26%.j = 0.0526 = 5.26\%.
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