Questions: Scherita wants 6,000 saved in 3 years to make a down payment on a house. How much money should she invest now at 3.2% compounded annually in order to meet her goal?

Solution Steps

To determine how much Scherita should invest now, we need to use the formula for compound interest. The formula is \( A = P(1 + r/n)^{nt} \), 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, and \( t \) is the time the money is invested for in years. In this case, we need to solve for \( P \), given \( A = 6000 \), \( r = 0.032 \), \( n = 1 \), and \( t = 3 \).

We are given the following values:

• Future value \( A = 6000 \)
• Annual interest rate \( r = 0.032 \)
• Compounding frequency \( n = 1 \) (annually)
• Time period \( t = 3 \)

We will use the formula for compound interest to find the present value \( P \): \[ A = P(1 + \frac{r}{n})^{nt} \] Rearranging the formula to solve for \( P \): \[ P = \frac{A}{(1 + \frac{r}{n})^{nt}} \]

Substituting the known values into the equation: \[ P = \frac{6000}{(1 + \frac{0.032}{1})^{1 \cdot 3}} = \frac{6000}{(1 + 0.032)^{3}} = \frac{6000}{(1.032)^{3}} \]

Calculating \( (1.032)^{3} \): \[ (1.032)^{3} \approx 1.0995 \] Now substituting back to find \( P \): \[ P \approx \frac{6000}{1.0995} \approx 5458.9882 \]

Final Answer