Mr Isaacs decided to start saving money for his daughter's future education. He immediately deposited R4 000 into a savings account. Three years later, he deposited a further \( R 5000 \) into the account. Five years after thr first deposit, he withdrew R2000 in order to pay Trevineo. The interest rate for the first three years was \( 15 \% \mathrm{p} \).a. compounded monthly. The interest rate for the remaining 8 years was \( 16 \% \) per annum compounded quarterly. Caluclate the value of his investment at the end of the \( 8^{\text {th }} \) year.

To calculate the value of Mr. Isaacs' investment at the end of the 8th year, we will break the problem into parts based on the deposits, withdrawals, and interest rates. ### Step 1: Calculate the value of the initial deposit after 3 years **Initial Deposit:** \( R4,000 \) **Interest Rate:** \( 15\% \) per annum compounded monthly **Time:** 3 years The formula for compound interest is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) = the amount of money accumulated after n years, including interest. - \( P \) = the principal amount (the initial deposit). - \( r \) = annual interest rate (decimal). - \( n \) = number of times that interest is compounded per year. - \( t \) = the number of years the money is invested or borrowed. For the first deposit: - \( P = 4000 \) - \( r = 0.15 \) - \( n = 12 \) (monthly) - \( t = 3 \) Calculating: \[ A = 4000 \left(1 + \frac{0.15}{12}\right)^{12 \times 3} \] \[ A = 4000 \left(1 + 0.0125\right)^{36} \] \[ A = 4000 \left(1.0125\right)^{36} \] Calculating \( (1.0125)^{36} \): \[ (1.0125)^{36} \approx 1.490665 \] Now, calculate \( A \): \[ A \approx 4000 \times 1.490665 \approx 5962.66 \] ### Step 2: Calculate the value after the second deposit **Second Deposit:** \( R5,000 \) **Time until withdrawal:** 5 years (2 years after the second deposit) The total amount after the first 3 years is \( R5962.66 \). After 3 years, he deposits \( R5000 \). Total amount before withdrawal: \[ Total = 5962.66 + 5000 = 10962.66 \] ### Step 3: Calculate the value after the withdrawal **Withdrawal:** \( R2,000 \) After the withdrawal: \[ Total = 10962.66 - 2000 = 8962.66 \] ### Step 4: Calculate the value of the investment for the remaining 5 years **Interest Rate:** \( 16\% \) per annum compounded quarterly **Time:** 5 years For the remaining amount: - \( P = 8962.66 \) - \( r = 0.16 \) - \( n = 4 \) (quarterly) - \( t = 5 \) Calculating: \[ A = 8962.66 \left(1 + \frac{0.16}{4}\right)^{4 \times 5} \] \[ A = 8962.66 \left(1 + 0.04\right)^{20} \] \[ A = 8962.66 \left(1.04\right)^{20} \] Calculating \( (1.04)^{20} \): \[ (1.04)^{20} \approx 2.208040 \] Now, calculate \( A \): \[ A \approx 8962.66 \times 2.208040 \approx 19893.45 \] ### Final Result The value of Mr. Isaacs' investment at the end of the 8th year is approximately **R19,893.45**.