Question
A = 500(1+0.10)^3 represents a $500 investment that Kim put in a savings account for 3 years earning an interest rate of 10%, compounded annually To the nearest cent, how much more interest will Kim earn if she keeps her investment for 6 years instead of 3 years
Answers
Answered by
GPT-4o mini
To find out how much more interest Kim will earn by keeping her investment for 6 years instead of 3, we'll calculate the total value of her investment at the end of both periods and then find the difference in the interest earned.
1. **Calculate the amount after 3 years**:
\[
A = 500(1 + 0.10)^3
\]
\[
A = 500(1.10)^3
\]
\[
A = 500 \times 1.331 = 665.50
\]
Interest earned after 3 years:
\[
\text{Interest}_3 = A - \text{Principal} = 665.50 - 500 = 165.50
\]
2. **Calculate the amount after 6 years**:
\[
A = 500(1 + 0.10)^6
\]
\[
A = 500(1.10)^6
\]
\[
A = 500 \times 1.771561 = 885.78
\]
Interest earned after 6 years:
\[
\text{Interest}_6 = A - \text{Principal} = 885.78 - 500 = 385.78
\]
3. **Calculate the difference in interest earned**:
\[
\text{Difference in Interest} = \text{Interest}_6 - \text{Interest}_3
\]
\[
\text{Difference in Interest} = 385.78 - 165.50 = 220.28
\]
Thus, if Kim keeps her investment for 6 years instead of 3 years, she will earn **$220.28** more in interest.
1. **Calculate the amount after 3 years**:
\[
A = 500(1 + 0.10)^3
\]
\[
A = 500(1.10)^3
\]
\[
A = 500 \times 1.331 = 665.50
\]
Interest earned after 3 years:
\[
\text{Interest}_3 = A - \text{Principal} = 665.50 - 500 = 165.50
\]
2. **Calculate the amount after 6 years**:
\[
A = 500(1 + 0.10)^6
\]
\[
A = 500(1.10)^6
\]
\[
A = 500 \times 1.771561 = 885.78
\]
Interest earned after 6 years:
\[
\text{Interest}_6 = A - \text{Principal} = 885.78 - 500 = 385.78
\]
3. **Calculate the difference in interest earned**:
\[
\text{Difference in Interest} = \text{Interest}_6 - \text{Interest}_3
\]
\[
\text{Difference in Interest} = 385.78 - 165.50 = 220.28
\]
Thus, if Kim keeps her investment for 6 years instead of 3 years, she will earn **$220.28** more in interest.
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