QUESTION 2 2.1 \begin{tabular}{l} Boitumelo invests an amount of R28 000 for a period of 7 years. For the first two \\ years the interest was \( 12 \% \) p.a. compounded semi-annually. The interest was then \\ changed to \( 10 \% \) p.a. compounded quarterly. \\ Determine the value of the investment after the 7 -year period. \\ 2.2 \\ \( \begin{array}{l}\text { Convert an interest rate of } 10 \% \text { p.a. compounded quarterly to an annual effective } \\ \text { interest rate. } \\ \text { A lawnmower was traded in for R236 after } 10 \text { years. The depreciation was } \\ \text { calculated at } 9,2 \% \text { p.a. on a straight-line basis. } \\ \text { Determine the cost of the lawnmower } 10 \text { years ago. }\end{array} \) \\ \hline 10\( ] \)\end{tabular}

Certainly! Let's tackle both parts of **Question 2** step by step. --- ### **2.1 Investment Growth Calculation** **Given:** - **Initial Investment (P₁):** R28,000 - **First 2 Years:** - **Interest Rate:** 12% per annum - **Compounding Frequency:** Semi-annually (2 times a year) - **Next 5 Years:** - **Interest Rate:** 10% per annum - **Compounding Frequency:** Quarterly (4 times a year) - **Total Investment Period:** 7 years **Objective:** Determine the value of the investment after 7 years. --- **Step 1: Calculate Future Value for the First 2 Years** - **Number of Compounding Periods (n₁):** 2 years × 2 = 4 - **Interest Rate per Period (r₁):** 12% / 2 = 6% or 0.06 Using the compound interest formula: \[ A₁ = P₁ \times (1 + r₁)^{n₁} \] \[ A₁ = 28,000 \times (1 + 0.06)^4 \] \[ A₁ = 28,000 \times 1.26247696 \approx R35,349.36 \] --- **Step 2: Calculate Future Value for the Next 5 Years** - **Number of Compounding Periods (n₂):** 5 years × 4 = 20 - **Interest Rate per Period (r₂):** 10% / 4 = 2.5% or 0.025 Using the compound interest formula: \[ A₂ = A₁ \times (1 + r₂)^{n₂} \] \[ A₂ = 35,349.36 \times (1 + 0.025)^{20} \] \[ A₂ = 35,349.36 \times 1.638616 \approx R57,923.45 \] --- ****Final Value After 7 Years:** **≈ R57,923.45** --- ### **2.2 Financial Conversions and Depreciation** #### **a) Converting Nominal to Effective Annual Interest Rate** **Given:** - **Nominal Interest Rate (iₙ):** 10% per annum - **Compounding Frequency:** Quarterly (4 times a year) **Objective:** Find the **Annual Effective Interest Rate (iₑ)**. **Formula:** \[ iₑ = \left(1 + \frac{iₙ}{m}\right)^m - 1 \] Where \( m = 4 \) (quarterly) \[ iₑ = \left(1 + \frac{0.10}{4}\right)^4 - 1 = (1.025)^4 - 1 \approx 1.10381289 - 1 = 0.10381289 \text{ or } 10.38\% \] --- #### **b) Determining the Original Cost of the Lawnmower** **Given:** - **Final Value (After Depreciation):** R236 - **Depreciation Rate:** 9.2% per annum - **Depreciation Method:** Straight-line - **Time Period:** 10 years **Objective:** Find the **Initial Cost (C)** of the lawnmower. **Straight-Line Depreciation Formula:** \[ \text{Final Value} = C - (C \times d \times t) \] Where: - \( d = 0.092 \) (9.2% per annum) - \( t = 10 \) years \[ 236 = C - (C \times 0.092 \times 10) \] \[ 236 = C - 0.92C \] \[ 236 = 0.08C \] \[ C = \frac{236}{0.08} = R2,950 \] --- ****Initial Cost of the Lawnmower 10 Years Ago:** **R2,950** --- ### **Summary of Answers** - **2.1:** The investment grows to **≈ R57,923.45** after 7 years. - **2.2a:** An interest rate of **10% p.a. compounded quarterly** is equivalent to an **effective annual rate of 10.38%**. - **2.2b:** The lawnmower originally cost **R2,950** ten years ago.