This is a classic compound interest problem. We will use 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 (initial investment).
• $r$ is the annual interest rate (decimal).
• $n$ is the number of times the interest is compounded per year.
• $t$ is the number of years the money is invested for.

Since it is not stated that the interest is compounded more frequently than annually, we assume it is compounded once per year. Thus, $n = 1$.

Part A:

For the first scenario:

• $P = 8000$
• $r = 12\% = 0.12$
• $n = 1$
• $t = 38$ years

$A = 8000 \left(1 + \frac{0.12}{1}\right)^{1 \times 38}$

Part B:

For the second scenario:

• $P = 8000$
• $r = 12\% = 0.12$
• $n = 1$
• $t = 28$ years

$A = 8000 \left(1 + \frac{0.12}{1}\right)^{1 \times 28}$

I'll calculate both now.### Part A: If you invest $8,000 at age 23 and leave it in the account for 38 years at a 12% annual interest rate, you will have approximately $593,437.31 by the time you withdraw.

Part B:

If you wait 10 years and invest the $8,000 when you're 33, leaving it in the account for 28 years at the same 12% annual interest rate, you will have approximately $191,070.93 by the time you withdraw.

Do you have any questions or would you like more details?

Here are 5 related questions:

1. How does the frequency of compounding affect the final amount?
2. What would happen if you invested a different amount?
3. How would different interest rates (e.g., 10%, 15%) change the outcome?
4. How much would you need to invest to reach a target retirement amount?
5. What is the impact of delaying investment on future wealth?

Tip: Starting to invest earlier can significantly increase the total value due to the power of compounding.