To calculate the compounded amount for a principal of $1000 deposited at an interest rate of 12.5% per annum compounded monthly for 2 years, we can use the compound interest formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
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 number of years the money is invested or borrowed.
Given values:
- \( P = 1000 \)
- \( r = 12.5% = 0.125 \)
- \( n = 12 \) (since the interest is compounded monthly)
- \( t = 2 \)
Now plugging in the values into the formula:
\[ A = 1000 \left(1 + \frac{0.125}{12}\right)^{12 \times 2} \]
First, calculate \( \frac{0.125}{12} \):
\[ \frac{0.125}{12} \approx 0.0104166667 \]
Then calculate \( 12 \times 2 = 24 \).
Now plug it back into the equation:
\[ A = 1000 \left(1 + 0.0104166667\right)^{24} \]
Calculating the expression inside the parentheses:
\[ 1 + 0.0104166667 \approx 1.0104166667 \]
Now raise this to the power of 24:
\[ A = 1000 \left(1.0104166667^{24}\right) \]
Calculating \( 1.0104166667^{24} \):
\[ 1.0104166667^{24} \approx 1.2682417958 \]
Finally, multiply by the principal:
\[ A \approx 1000 \times 1.2682417958 \approx 1268.24 \]
Therefore, the compounded amount after 2 years is approximately $1268.24.
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