Question
b) Calculate the compound interest on an investment of K3,400.00 compounded every 4 months for 3 years
Prove your answer
Answers
GPT-4o mini
To calculate the compound interest on an investment, we can 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 (in decimal).
- \(n\) is the number of times that interest is compounded per year.
- \(t\) is the number of years the money is invested or borrowed.
Since you did not provide the annual interest rate, let's assume it as \(r\). You can replace it with the actual rate later.
Given:
- \(P = 3,400.00\)
- Interest is compounded every 4 months, which means \(n = 3\) (since there are 3 compounding periods in a year).
- \(t = 3\) years.
Now, we can rewrite our formula with the provided values:
\[
A = 3400 \left(1 + \frac{r}{3}\right)^{3 \cdot 3}
\]
This simplifies to:
\[
A = 3400 \left(1 + \frac{r}{3}\right)^9
\]
The compound interest is found by subtracting the principal from the total amount \(A\):
\[
\text{Compound Interest} = A - P
\]
Thus:
\[
\text{Compound Interest} = 3400 \left(1 + \frac{r}{3}\right)^9 - 3400
\]
To complete the calculation, you'll need to insert the specific value of the interest rate \(r\).
### Example
If we assume an annual interest rate of 6% (0.06 in decimal):
1. Plug in \(r = 0.06\):
\[
A = 3400 \left(1 + \frac{0.06}{3}\right)^{9}
\]
2. Calculate:
\[
A = 3400 \left(1 + 0.02\right)^{9} = 3400 \left(1.02\right)^{9}
\]
\[
A = 3400 \cdot 1.1975 \approx 4,058.50
\]
3. Now, calculate the compound interest:
\[
\text{Compound Interest} = 4,058.50 - 3,400 \approx 658.50
\]
Thus, if the interest rate was 6%, the compound interest would be approximately K658.50 after 3 years. Please replace \(0.06\) with your actual interest rate for a precise answer.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
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 (in decimal).
- \(n\) is the number of times that interest is compounded per year.
- \(t\) is the number of years the money is invested or borrowed.
Since you did not provide the annual interest rate, let's assume it as \(r\). You can replace it with the actual rate later.
Given:
- \(P = 3,400.00\)
- Interest is compounded every 4 months, which means \(n = 3\) (since there are 3 compounding periods in a year).
- \(t = 3\) years.
Now, we can rewrite our formula with the provided values:
\[
A = 3400 \left(1 + \frac{r}{3}\right)^{3 \cdot 3}
\]
This simplifies to:
\[
A = 3400 \left(1 + \frac{r}{3}\right)^9
\]
The compound interest is found by subtracting the principal from the total amount \(A\):
\[
\text{Compound Interest} = A - P
\]
Thus:
\[
\text{Compound Interest} = 3400 \left(1 + \frac{r}{3}\right)^9 - 3400
\]
To complete the calculation, you'll need to insert the specific value of the interest rate \(r\).
### Example
If we assume an annual interest rate of 6% (0.06 in decimal):
1. Plug in \(r = 0.06\):
\[
A = 3400 \left(1 + \frac{0.06}{3}\right)^{9}
\]
2. Calculate:
\[
A = 3400 \left(1 + 0.02\right)^{9} = 3400 \left(1.02\right)^{9}
\]
\[
A = 3400 \cdot 1.1975 \approx 4,058.50
\]
3. Now, calculate the compound interest:
\[
\text{Compound Interest} = 4,058.50 - 3,400 \approx 658.50
\]
Thus, if the interest rate was 6%, the compound interest would be approximately K658.50 after 3 years. Please replace \(0.06\) with your actual interest rate for a precise answer.