Question

What is the compounded amount for $1000 deposited at 12.5%p.a. monthly after 2 years

Answer 1

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**.