Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Present Value Random Variables for Long-Term Insurance Coverages Practice Questions

Present value random variables on SOA Exam FAM test insurance and annuity present values, variance of future loss, and the mathematical relationship between whole life insurance and whole life annuity functions.

157 questions 87 easy 43 medium 27 hard 2026 syllabus

Sample Questions

Question 1 Easy
Which of the following correctly describes Ax:n‾∣A_{x:\overline{n}|}?
Solution
A is correct.

No bar means discrete. No superscript 1 means the benefit pays on the first of death or survival to nn. This is an endowment insurance: Ax:n‾∣=Ax:n‾∣1+nExA_{x:\overline{n}|} = A^{1}_{x:\overline{n}|} + {}_{n}E_x.
Question 2 Medium
You are given Ax=0.35A_x = 0.35, Ax:10‾∣1=0.05A^{1}_{x:\overline{10}|} = 0.05, and 10Ex=0.45{}_{10}E_x = 0.45. Calculate Ax+10A_{x+10}.
Solution
E is correct.

The deferred insurance is:

10∣Ax=Ax−Ax:10‾∣1=0.35−0.05=0.30{}_{10|}A_x = A_x - A^{1}_{x:\overline{10}|} = 0.35 - 0.05 = 0.30

Also 10∣Ax=10Ex⋅Ax+10{}_{10|}A_x = {}_{10}E_x \cdot A_{x+10}, so:

Ax+10=10∣Ax10Ex=0.300.45=23=0.667A_{x+10} = \frac{{}_{10|}A_x}{{}_{10}E_x} = \frac{0.30}{0.45} = \frac{2}{3} = 0.667
Question 3 Hard
You are given: Aˉx:n‾∣=0.60\bar{A}_{x:\overline{n}|} = 0.60, nEx=0.35{}_{n}E_x = 0.35, and Aˉx+n=0.50\bar{A}_{x+n} = 0.50. Calculate Aˉx\bar{A}_x.
Solution
C is correct.

First: Aˉx:n‾∣1=Aˉx:n‾∣−nEx=0.60−0.35=0.25\bar{A}^{1}_{x:\overline{n}|} = \bar{A}_{x:\overline{n}|} - {}_{n}E_x = 0.60 - 0.35 = 0.25.

The deferred insurance: n∣Aˉx=nEx⋅Aˉx+n=0.35×0.50=0.175{}_{n|}\bar{A}_x = {}_{n}E_x \cdot \bar{A}_{x+n} = 0.35 \times 0.50 = 0.175.

The whole life insurance: Aˉx=Aˉx:n‾∣1+n∣Aˉx=0.25+0.175=0.425\bar{A}_x = \bar{A}^{1}_{x:\overline{n}|} + {}_{n|}\bar{A}_x = 0.25 + 0.175 = 0.425.
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