1. **Problem 19.1:** Calculate the cash needed to replace the tractor in 5 years. - Present tractor value: R800000 - Value after 5 years (trade-in): R200000 - Replacement cost increases at 8% per year Calculate the replacement cost after 5 years: $$\text{Replacement price} = 800000 \times (1 + 0.08)^5$$ Calculate: $$800000 \times 1.08^5 = 800000 \times 1.469328 = 1175462.4$$ Trade-in value is R200000, so cash needed: $$\text{Cash needed} = 1175462.4 - 200000 = 975462.4$$ **Answer:** R975462.40 2. **Problem 19.2:** Find the monthly deposit \(X\) so that after 60 months (5 years) at 12% p.a. compounded monthly, the amount equals the cash needed R975462.40. - Interest per month: $$i = \frac{0.12}{12} = 0.01$$ - Number of months: $$n = 60$$ - Future value of ordinary annuity formula: $$FV = X \times \frac{(1+i)^n - 1}{i}$$ Set $$FV = 975462.4$$ and solve for $$X$$: $$X = \frac{975462.4 \times 0.01}{(1.01)^{60} - 1}$$ Calculate \( (1.01)^{60} \): $$1.01^{60} = 1.8194$$ So: $$X = \frac{9754.624}{1.8194 - 1} = \frac{9754.624}{0.8194} = 11900.9$$ **Answer:** The monthly deposit $$X \approx 11900.90$$ rands 3. **Problem 19.3:** New monthly deposit with 5 annual withdrawals of R5000 after the first 12 months. - Withdrawals happen once every 12 months, total 5 withdrawals - Each withdrawal reduces amount saved by $$5000 \times (1 + i)^{k}$$, adjusting for time value (assuming withdrawals reduce future value) We use the annuity formula with withdrawals: The withdrawals total to an annuity of 5 terms, each withdrawal at times $$t = 12, 24, 36, 48, 60$$ months. Calculate the present value of withdrawals at month 60: Effective monthly interest rate: $$i = 0.01$$ The future value of withdrawals at month 60 is: $$FV_{withdrawals} = 5000 \times \left[(1.01)^{48} + (1.01)^{36} + (1.01)^{24} + (1.01)^{12} + (1.01)^0\right]$$ Calculate each: - $$1.01^{48} = 1.6084$$ - $$1.01^{36} = 1.4324$$ - $$1.01^{24} = 1.2682$$ - $$1.01^{12} = 1.1268$$ - $$1.01^{0} = 1$$ Sum: $$1.6084 + 1.4324 + 1.2682 + 1.1268 + 1 = 6.4358$$ So: $$FV_{withdrawals} = 5000 \times 6.4358 = 32179$$ Total amount to accumulate to buy the tractor: $$975462.4 + 32179 = 1009641.4$$ Now solve for new monthly deposit $$X'$$: $$X' = \frac{1009641.4 \times 0.01}{1.8194 - 1} = \frac{10096.4}{0.8194} = 12325.4$$ **Answer:** New monthly deposit $$\approx 12325.40$$ rands