Let's go step by step: ### **Step 1: Fund X (6% convertible monthly)** - **Interest rate:** 6%, convertible monthly. - **Nominal rate:** \( i^{(12)} = 6\% \) - **Monthly rate:** \( i_{monthly} = \frac{0.06}{12} = 0.005 \) - **Number of periods:** \( 12 \text{ years} \times 12 = 144 \) months The accumulation factor for Fund X after 12 years: \[ A_X = P \times (1 + i_{monthly})^{144} \] Given that \( A_X = 2,000 \): \[ 2,000 = P \times (1.005)^{144} \] Let's calculate \( (1.005)^{144} \): First, calculate the exponent: \[ \ln(1.005) = 0.0049875 \] \[ 0.0049875 \times 144 = 0.7182 \] \[ e^{0.7182} = 2.0504 \] So, \[ (1.005)^{144} \approx 2.0504 \] Now, \[ 2,000 = P \times 2.0504 \implies P = \frac{2,000}{2.0504} \approx 975.41 \] --- ### **Step 2: Fund Y (force of interest)** - **Force of interest:** \( \delta_t = \frac{1}{t + 12} \) - **Accumulation over 12 years:** Use the formula for accumulation with a time-varying force of interest: \[ A_Y = P \times \exp\left( \int_0^{12} \delta_t\, dt \right) \] Calculate the integral: \[ \int_0^{12} \frac{1}{t + 12} dt = [\ln(t + 12)]_{0}^{12} = \ln(24) - \ln(12) = \ln\left(\frac{24}{12}\right) = \ln(2) \] So, \[ A_Y = P \times e^{\ln(2)} = P \times 2 \] --- ### **Step 3: Substitute \( P \) from above** Recall \( P \approx 975.41 \) \[ A_Y = 975.41 \times 2 = 1,950.82 \] --- ### **Step 4: Final Answer** The closest possible answer is **1,950**. \[ \boxed{1,950} \] --- **Summary of Steps:** 1. Calculated monthly accumulation for Fund X to get \( P \). 2. Calculated accumulation for Fund Y using the given force of interest. 3. Substituted \( P \) into Fund Y's formula to get \( Z \). 4. Rounded and matched to the closest answer. **Final Answer:** \[ \boxed{1,950} \]