Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Premium and Policy Value Calculation for Long-Term Insurance Coverages Practice Questions

Premium and policy value calculations on SOA Exam FAM cover net and gross premium computation, benefit reserves (prospective and retrospective), Thiele's differential equation, and profit testing.

190 Questions
93 Easy
72 Medium
25 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
Which expression equals P(Ax)P(A_x)?
Solution
E is correct.

P=(1−da¨x)/a¨x=1/a¨x−dP=(1-d\ddot{a}_x)/\ddot{a}_x=1/\ddot{a}_x-d.
Question 2 Medium
For a fully discrete whole life insurance of 1 with 10V=0.18{}_{10}V=0.18, P=0.02P=0.02, qx+10=0.015q_{x+10}=0.015, i=0.06i=0.06, calculate 11V{}_{11}V.
Solution
D is correct.

(0.18+0.02)(1.06)=0.015+0.985⋅11V(0.18+0.02)(1.06)=0.015+0.985\cdot{}_{11}V. 0.212=0.015+0.985⋅11V0.212=0.015+0.985\cdot{}_{11}V. 11V=0.197/0.985=0.200{}_{11}V=0.197/0.985=0.200.
Question 3 Hard
For a fully discrete whole life insurance of 1 on (x), you are given:

- Ax=0.25A_x = 0.25, i=0.06i = 0.06

If mortality increases so that A'_x = 0.35, calculate the change in the net premium PxP_x.
Solution
D is correct.

d=0.06/1.06=3/53d = 0.06/1.06 = 3/53.

Original: a¨x=0.753/53=0.75×533=39.753=13.25\ddot{a}_x = \frac{0.75}{3/53} = \frac{0.75 \times 53}{3} = \frac{39.75}{3} = 13.25
Px=0.2513.25=0.01887P_x = \frac{0.25}{13.25} = 0.01887

New: a¨x′=0.653/53=0.65×533=34.453=11.483\ddot{a}'_x = \frac{0.65}{3/53} = \frac{0.65 \times 53}{3} = \frac{34.45}{3} = 11.483
Px′=0.3511.483=0.03048P'_x = \frac{0.35}{11.483} = 0.03048

Change =0.03048−0.01887=0.01161= 0.03048 - 0.01887 = 0.01161.

0.011610.01161, closest to (C) 0.010000.01000.
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