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

Work with present value random variables for long-term coverages on Exam FAM. Questions test insurance and annuity present values, variance calculations, and the relationship between life insurance and annuity functions.

162 Questions
87 Easy
48 Medium
27 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
Which of the following correctly describes Ax:nA_{x:\overline{n}|}?
Solution
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:n1+nExA_{x:\overline{n}|} = A^{1}_{x:\overline{n}|} + {}_{n}E_x.

The answer is (E).
Question 2 Medium
Under constant forces μ\mu and δ\delta, which expression equals Var(vTx)\text{Var}(v^{T_x})?
Solution
Var(Z)=μμ+2δμ2(μ+δ)2=μ[(μ+δ)2μ(μ+2δ)](μ+δ)2(μ+2δ)=μδ2(μ+δ)2(μ+2δ)\text{Var}(Z) = \frac{\mu}{\mu+2\delta} - \frac{\mu^2}{(\mu+\delta)^2} = \frac{\mu[(\mu+\delta)^2 - \mu(\mu+2\delta)]}{(\mu+\delta)^2(\mu+2\delta)} = \frac{\mu\delta^2}{(\mu+\delta)^2(\mu+2\delta)}

The answer is (C).
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
First: Aˉx:n1=Aˉx:nnEx=0.600.35=0.25\bar{A}^{1}_{x:\overline{n}|} = \bar{A}_{x:\overline{n}|} - {}_{n}E_x = 0.60 - 0.35 = 0.25.\n\nThe deferred insurance: nAˉx=nExAˉx+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.\n\nThe whole life insurance: Aˉx=Aˉx:n1+nAˉx=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.\n\n(A) 0.325 is wrong — perhaps uses Aˉx+n=0.30\bar{A}_{x+n} = 0.30.\n(B) 0.375 is wrong — perhaps ignores the term insurance.\n(D) 0.475 is wrong — perhaps adds nEx{}_{n}E_x instead of nExAˉx+n{}_{n}E_x \cdot \bar{A}_{x+n}.\n(E) 0.525 is wrong — perhaps adds the endowment and deferred directly.
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