Free SOA Exam FM (Financial Mathematics) Time Value of Money Practice Questions

The present value is:

$$PV = FV \cdot v^n = \frac{FV}{(1+i)^n} = \frac{50{,}000}{(1.03)^{12}}$$

Compute $(1.03)^{12}$ by repeated squaring:

$$(1.03)^2 = 1.0609$$

$$(1.03)^4 = (1.0609)^2 = 1.12550881$$

$$(1.03)^8 = (1.12550881)^2 = 1.26677008$$

$$(1.03)^{12} = (1.03)^8 \times (1.03)^4 = 1.26677008 \times 1.12550881 = 1.42576089$$

$PV = \frac{50{,}000}{1.42576089} = 35{,}069$

The present value is \$35,069.

Step 1: Find the effective annual interest rate.

$$1 + i = \left(1 + \frac{0.06}{2}\right)^2 = (1.03)^2 = 1.0609$$

Step 2: Find the force of interest.

$$\delta = \ln(1 + i) = \ln(1.0609)$$

Alternatively: $\delta = 2\ln(1.03) = 2 \times 0.029559 = 0.059118$.

So $\delta \approx 0.05912$.

Common errors: (B) uses $\delta = \ln(1.06) = 0.05827$. (B) arises from a rounding error in the computation; (C) assumes $\delta$ equals the nominal rate; (D) uses $\delta = \ln(1.0609)$ but rounds up; (E) uses $\delta = i = 0.0609$ and confuses the effective rate with the force.

Convert $i^{(4)}$ to the force of interest $\delta$.

$$\delta = 4 \ln\left(1 + \frac{i^{(4)}}{4}\right) = 4 \ln(1.025) = 4 \times 0.024693 = 0.098771$$

Convert $\delta$ to $d^{(6)}$:
$$d^{(6)} = 6\left(1 - e^{-\delta/6}\right) = 6\left(1 - e^{-0.016462}\right)$$

$$e^{-0.016462} = 0.98367$$
$$d^{(6)} = 6 \times 0.01633 = 0.09796 \approx 9.80\%$$

Choice C is correct.
Choice A is incorrect because it uses an incorrect approximation for the exponential.
Choice B is incorrect because it confuses $d^{(6)}$ with the nominal interest $i^{(6)}$.
Choice D is incorrect because it fails to convert from nominal interest to nominal discount.
Choice E is incorrect because it uses $i^{(6)}$ directly without the discount conversion.