To compare the two investments from Person A and Person B, we can calculate the future value of their respective deposits using the compound interest formula:

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Where:

• $A$ is the amount of money accumulated after $t$ years, including interest.
• $P$ is the principal amount (the initial deposit).
• $r$ is the annual interest rate (as a decimal).
• $n$ is the number of times that interest is compounded per year.
• $t$ is the number of years.

Let's break this down for both persons:

For Person A:

• $P_A = 2700$ (the principal deposit)
• $r_A = 0.04$ (4% annual interest)
• $n_A = 1$ (compounded once a year)

So the future value formula for Person A becomes: $A_A = 2700 \left(1 + \frac{0.04}{1}\right)^{1 \cdot t} = 2700 \left(1.04\right)^t$

For Person B:

• $P_B = 2500$ (the principal deposit)
• $r_B = 0.05$ (5% annual interest)
• $n_B = 12$ (compounded monthly)

So the future value formula for Person B becomes: $A_B = 2500 \left(1 + \frac{0.05}{12}\right)^{12 \cdot t} = 2500 \left(1 + \frac{0.05}{12}\right)^{12t}$

Both of these equations give the future value of the respective investments after $t$ years.

Would you like me to calculate these for a specific number of years (e.g., 1 year, 5 years, etc.), or provide further details on the comparison?

5 Related Questions:

1. How does the frequency of compounding impact the total interest earned?
2. How do different interest rates affect the growth of investments over time?
3. What happens if Person A and Person B both deposit additional amounts each year?
4. Can we calculate the exact crossover point where Person B’s account exceeds Person A’s due to compounding?
5. How would inflation impact the real value of these investments?

Tip: The more frequently interest is compounded, the faster your investment grows, even if the interest rate is the same!