kercise 8 mple Interest 1. Lehlohonolo opens a savings account that bears \( 15 \% \) simple interest per annum. If he deposits R 5700 how much money will he have after 11 years and 7 months? 2. How long would it take for R8 400 to grow to R15 000 at a simple interest rate of \( 7.5 \% \) p.a.? Give your answer correct to the nearest month. 3. How much should be invested now at \( 8 \% \) simple interest per annum to accumulate to R12 000 in 6 years? 4. A car worth R145 000 depreciates at the rate of \( 16 \% \) annually according to the straight line method. What will the value of the car be after 4.5 years? 5. How long would it take for the car in the previous question to depreciate in value to R100 000 ? Give your answer correct to the nearest month. 6. If Sipho buys a camera for R899 at the beginning of the year 2010 and intends to sell it for R350 at the end of 2014 , what would the rate of depreciation be assuming simple interest decay? 7. Kwena opts for an investment with an interest rate of \( 8 \% \) per annum compounded quarterly. How much will he have after 6 years if he invests R2000? 8. Juju invests R18 000 into a savings account with an interest rate of \( 9 \% \) per annum compounded monthly. If he invests a further R10 000 into the account after 4 years, how much will he have after 6 years and 8 months? 9. How much should be invested now at \( 8 \% \) interest per annum compounded annually to accumulate to R12 000 in 6 years? 10. What is the annual interest rate if R8 700 is compounded semi-annually and grows to R12 500 after 4 years? 11. A car worth R145 000 depreciates at a compound rate of \( 16 \% \) per annum. Calculate the book value of the car after 4,5 years. 12. A car dealer uses the straight-line method to calculate a car's value. The car was originally bought for R150 000 . The car dealer calculates the car to be worth R85 667.89 as it is now 5 years old.

Let's solve each of the problems step by step. ### 1. Simple Interest Calculation **Formula:** \[ A = P(1 + rt) \] Where: - \( A \) = total amount after time \( t \) - \( P \) = principal amount (initial deposit) - \( r \) = annual interest rate (as a decimal) - \( t \) = time in years **Given:** - \( P = 5700 \) - \( r = 0.15 \) - \( t = 11 + \frac{7}{12} = 11.5833 \) years **Calculation:** \[ A = 5700(1 + 0.15 \times 11.5833) = 5700(1 + 1.738) = 5700 \times 2.738 \approx 15609.60 \] **Answer:** Lehlohonolo will have approximately R15,609.60 after 11 years and 7 months. --- ### 2. Time Calculation for Simple Interest **Formula:** \[ A = P(1 + rt) \implies t = \frac{A - P}{Pr} \] **Given:** - \( P = 8400 \) - \( A = 15000 \) - \( r = 0.075 \) **Calculation:** \[ t = \frac{15000 - 8400}{8400 \times 0.075} = \frac{6600}{630} \approx 10.4762 \text{ years} \] Convert to months: \( 10.4762 \times 12 \approx 125.71 \) months, which is approximately 126 months or 10 years and 6 months. **Answer:** It would take approximately 10 years and 6 months. --- ### 3. Present Value Calculation for Simple Interest **Formula:** \[ P = \frac{A}{1 + rt} \] **Given:** - \( A = 12000 \) - \( r = 0.08 \) - \( t = 6 \) **Calculation:** \[ P = \frac{12000}{1 + 0.08 \times 6} = \frac{12000}{1 + 0.48} = \frac{12000}{1.48} \approx 8108.11 \] **Answer:** Approximately R8,108.11 should be invested now. --- ### 4. Depreciation Calculation **Formula:** \[ V = P(1 - rt) \] **Given:** - \( P = 145000 \) - \( r = 0.16 \) - \( t = 4.5 \) **Calculation:** \[ V = 145000(1 - 0.16 \times 4.5) = 145000(1 - 0.72) = 145000 \times 0.28 \approx 40600 \] **Answer:** The value of the car after 4.5 years will be approximately R40,600. --- ### 5. Time Calculation for Depreciation **Formula:** \[ V = P(1 - rt) \implies t = \frac{P - V}{Pr} \] **Given:** - \( P = 145000 \) - \( V = 100000 \) - \( r = 0.16 \) **Calculation:** \[ t = \frac{145000 - 100000}{145000 \times 0.16} = \frac{45000}{23200} \approx 1.9388 \text{ years} \] Convert to months: \( 1.9388 \times 12 \approx 23.27 \) months, which is approximately 23 months. **Answer:** It would take approximately 23 months. --- ### 6. Rate of Depreciation Calculation **Formula:** \[ \text{Depreciation Rate} = \frac{C - S}{C \times t} \] **Given:** - \( C = 899 \) - \( S = 350 \) - \( t = 5 \) years **Calculation:** \[ \text{Depreciation Rate} = \frac{899 - 350}{899 \times 5} = \frac{549}{4495} \approx 0.122 \text{ or } 12.2\% \] **Answer:** The rate of depreciation is approximately \( 12.2\% \) per annum. --- ### 7. Future Value Calculation for Compound Interest **Formula:** \[