Questions: If 49600 dollars is invested at an interest rate of 10 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent. (a) Annual: (b) Semiannual: (c) Monthly: (d) Daily:

Solution Steps

For annual compounding, the formula used is:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

Substituting the values:

\[ A_{\text{annual}} = 49600 \left(1 + \frac{0.1}{1}\right)^{1 \cdot 5} = 49600 \left(1 + 0.1\right)^{5} = 49600 \cdot (1.1)^{5} \]

Calculating \( (1.1)^{5} \):

\[ (1.1)^{5} \approx 1.61051 \]

Thus,

\[ A_{\text{annual}} \approx 49600 \cdot 1.61051 \approx 79881.30 \]

For semiannual compounding, the formula remains the same, but \( n = 2 \):

\[ A_{\text{semiannual}} = 49600 \left(1 + \frac{0.1}{2}\right)^{2 \cdot 5} = 49600 \left(1 + 0.05\right)^{10} = 49600 \cdot (1.05)^{10} \]

Calculating \( (1.05)^{10} \):

\[ (1.05)^{10} \approx 1.62889 \]

Thus,

\[ A_{\text{semiannual}} \approx 49600 \cdot 1.62889 \approx 80793.17 \]

For monthly compounding, we set \( n = 12 \):

\[ A_{\text{monthly}} = 49600 \left(1 + \frac{0.1}{12}\right)^{12 \cdot 5} = 49600 \left(1 + \frac{0.1}{12}\right)^{60} = 49600 \cdot \left(1.0083333\right)^{60} \]

Calculating \( (1.0083333)^{60} \):

\[ (1.0083333)^{60} \approx 1.64701 \]

Thus,

\[ A_{\text{monthly}} \approx 49600 \cdot 1.64701 \approx 81607.32 \]

Final Answer