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 \ $$ \begin{array}{l} ext { Convert an interest rate of } 10 \% ext { p.a. compounded quarterly to an annual effective } \ ext { interest rate. } \ ext { A lawnmower was traded in for R236 after } 10 ext { years. The depreciation was } \ ext { calculated at } 9,2 \% ext { p.a. on a straight-line basis. } \ ext { Determine the cost of the lawnmower } 10 ext { years ago. }\end{array} $$ \ \hline 10$$ ] $$\end{tabular}

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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 \ $$ \begin{array}{l}	ext { Convert an interest rate of } 10 \% 	ext { p.a. compounded quarterly to an annual effective } \ 	ext { interest rate. } \ 	ext { A lawnmower was traded in for R236 after } 10 	ext { years. The depreciation was } \ 	ext { calculated at } 9,2 \% 	ext { p.a. on a straight-line basis. } \ 	ext { Determine the cost of the lawnmower } 10 	ext { years ago. }\end{array} $$ \ \hline 10$$ ] $$\end{tabular}

UpStudy AI · Accepted Answer

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)

$$
iₑ = \left(1 + \frac{0.10}{4}\right)^4 - 1 = (1.025)^4 - 1 \approx 1.10381289 - 1 = 0.10381289 \text{ or } 10.38\%
$$

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#### **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**

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### **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.
After 7 years, the investment grows to approximately R57,923.45. An interest rate of 10% per annum compounded quarterly is equivalent to an effective annual rate of 10.38%. Ten years ago, the lawnmower originally cost R2,950.

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