To determine which of the two investment rates yields a larger amount after 2 years, we can calculate the final amount for both options.
Option 1: 11% Compounded Monthly
For monthly compounding, the formula used is:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
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:
- \( P = 11000 \)
- \( r = 0.11 \) (11%)
- \( n = 12 \) (monthly compounding)
- \( t = 2 \)
Plugging the values into the formula:
\[ A = 11000 \left(1 + \frac{0.11}{12}\right)^{12 \times 2} \]
Calculating \( \frac{0.11}{12} \):
\[ \frac{0.11}{12} \approx 0.00916667 \]
Now substituting into the equation:
\[ A = 11000 \left(1 + 0.00916667\right)^{24} \] \[ A = 11000 \left(1.00916667\right)^{24} \]
Calculating \( (1.00916667)^{24} \):
\[ (1.00916667)^{24} \approx 1.233 \]
Now calculating \( A \):
\[ A \approx 11000 \times 1.233 \approx 13563 \]
Option 2: 10.92% Compounded Continuously
For continuous compounding, the formula is:
\[ A = Pe^{rt} \]
Where:
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
- All other variables are as defined previously.
Given:
- \( r = 0.1092 \) (10.92%)
Substituting in the values:
\[ A = 11000 \cdot e^{0.1092 \times 2} \]
Calculating the exponent:
\[ 0.1092 \times 2 = 0.2184 \]
Now calculating \( e^{0.2184} \):
\[ e^{0.2184} \approx 1.243 \]
Now calculating \( A \):
\[ A \approx 11000 \cdot 1.243 \approx 13673 \]
Summary of Results
- For 11% compounded monthly: \( A \approx 13563 \)
- For 10.92% compounded continuously: \( A \approx 13673 \)
Conclusion
The option of 10.92% compounded continuously yields a larger amount after 2 years compared to 11% compounded monthly.