To solve this problem, we'll use the formula for compound interest:

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

where:

• $A$ is the amount of money accumulated after $n$ years, including interest.
• $P$ is the principal amount (the initial amount of money).
• $r$ is the annual interest rate (decimal).
• $n$ is the number of times that interest is compounded per year.
• $t$ is the time the money is invested or borrowed for, in years.

For Chang's investment:

• $P = 1200$
• $r = 0.04$ (4.0%)
• $n = 1$ (compounded annually)
• $t = 10$ and $t = 30$ years

After 10 years:

$A = 1200 \left(1 + \frac{0.04}{1}\right)^{1 \times 10} = 1200 \times \left(1.04\right)^{10}$

Let's calculate this.

After 30 years:

$A = 1200 \left(1 + \frac{0.04}{1}\right)^{1 \times 30} = 1200 \times \left(1.04\right)^{30}$

Now let's compute both values.

Calculation

1. For 10 years: $A = 1200 \times (1.04)^{10} \approx 1200 \times 1.48024 \approx 1776.29$ So, after 10 years, Chang will have approximately $1776.29.

2. For 30 years: $A = 1200 \times (1.04)^{30} \approx 1200 \times 3.2434 \approx 3892.08$ So, after 30 years, Chang will have approximately $3892.08.

Summary:

• After 10 years, Chang will have a balance of approximately $1776.29.
• After 30 years, Chang will have a balance of approximately $3892.08.

Would you like to see the detailed calculation for Kio's investment or discuss the differences?

Related Questions:

1. How much more will Kio's account be worth after 10 years compared to Chang's?
2. What is the difference in the interest earned by Chang and Kio after 30 years?
3. How would the results change if both investments were compounded semi-annually instead of annually?
4. What is the effective annual rate (EAR) for Chang’s investment compared to Kio's?
5. If Chang had invested $2000 instead of $1200, what would be his balance after 30 years?

Tip:

Always remember that even small differences in interest rates can significantly impact the final balance, especially over long periods of time due to the power of compound interest.