Investment Choice 4F6294

1. **State the problem:** We want to find the accumulated value of two different monthly deposit investments with compound interest. 2. **Formula used:** The future value of an ordinary annuity (deposits at the end of each period) compounded monthly is given by: $$ A = P \times \frac{(1 + r)^n - 1}{r} $$ where: - $A$ is the accumulated amount - $P$ is the monthly deposit - $r$ is the monthly interest rate (annual rate divided by 12) - $n$ is the total number of deposits (months) 3. **Calculate for the first investment:** - $P = 1500$ - Annual interest rate = 6%, so monthly rate $r = \frac{6}{100 \times 12} = 0.005$ - Time = 2 years, so $n = 2 \times 12 = 24$ Calculate: $$ A_1 = 1500 \times \frac{(1 + 0.005)^{24} - 1}{0.005} $$ Calculate $(1 + 0.005)^{24}$: $$ (1.005)^{24} \approx 1.12749 $$ So: $$ A_1 = 1500 \times \frac{1.12749 - 1}{0.005} = 1500 \times \frac{0.12749}{0.005} $$ Simplify fraction: $$ \frac{0.12749}{0.005} = 25.498 $$ Therefore: $$ A_1 = 1500 \times 25.498 = 38247 $$ 4. **Calculate for the second investment:** - $P = 2500$ - Annual interest rate = 4%, so monthly rate $r = \frac{4}{100 \times 12} = 0.0033333$ - Time = 1.5 years, so $n = 1.5 \times 12 = 18$ Calculate: $$ A_2 = 2500 \times \frac{(1 + 0.0033333)^{18} - 1}{0.0033333} $$ Calculate $(1 + 0.0033333)^{18}$: $$ (1.0033333)^{18} \approx 1.06168 $$ So: $$ A_2 = 2500 \times \frac{1.06168 - 1}{0.0033333} = 2500 \times \frac{0.06168}{0.0033333} $$ Simplify fraction: $$ \frac{0.06168}{0.0033333} = 18.504 $$ Therefore: $$ A_2 = 2500 \times 18.504 = 46260 $$ 5. **Conclusion:** - The accumulated value of the first investment after 2 years is approximately $38247$. - The accumulated value of the second investment after 1.5 years is approximately $46260$. So, based on accumulated value alone, the second investment yields a higher amount despite the lower interest rate because of the larger monthly deposit. **Final answer:** - First investment accumulated value: $38247$ - Second investment accumulated value: $46260$