24. Max invests \( R x \) in an account that pays interest at \( 8.5 \% \) per annum compounded yearly. After three years, he deposits a further \( R 8200 \) in the account. After 8 years, the amount in the account is R 24813.91 . The value of \( x \) is closest to:

To solve for \( x \), we need to break down the problem into two parts: the initial investment and the additional deposit. 1. **Initial Investment**: Max invests \( R x \) for 3 years at an interest rate of \( 8.5\% \) compounded yearly. The formula for compound interest is: \[ A = P(1 + r)^n \] 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 years the money is invested or borrowed. For the initial investment: \[ A_1 = x(1 + 0.085)^3 \] Calculating \( (1 + 0.085)^3 \): \[ (1.085)^3 \approx 1.272 \] So, \[ A_1 \approx x \cdot 1.272 \] 2. **Additional Deposit**: After 3 years, Max deposits \( R 8200 \). This amount will be invested for the remaining 5 years (8 years total - 3 years). The amount after 5 years for this deposit is: \[ A_2 = 8200(1 + 0.085)^5 \] Calculating \( (1 + 0.085)^5 \): \[ (1.085)^5 \approx 1.503 \] So, \[ A_2 \approx 8200 \cdot 1.503 \approx 12324.6 \] 3. **Total Amount After 8 Years**: The total amount in the account after 8 years is the sum of \( A_1 \) and \( A_2 \): \[ A_1 + A_2 = 24813.91 \] Substituting the values we found: \[ x \cdot 1.272 + 12324.6 = 24813.91 \] Rearranging to solve for \( x \): \[ x \cdot 1.272 = 24813.91 - 12324.6 \] \[ x \cdot 1.272 = 12489.31 \] \[ x = \frac{12489.31}{1.272} \approx 9825.00 \] Thus, the value of \( x \) is closest to \( R 9825.00 \).