$$ 00: 49 $$ Online MSc Cyber Security - April 2025 Entry Ad • Benefit from strong industry ties, a dynamic alumni community, and outstanding resources. Get ready for real-world care... QUESTION 2 [24] 2.1 Boitumelo invests an amount of R28 000 for a period of 7 years. For the first two (3) 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 Convert an interest rate of $$ 10 \% $$ p.a. compounded quarterly to an annual effective (3) interest rate. 2.3 A lawnmower was traded in for R236 after 10 years. The depreciation was (4) calculated at $$ 9,2 \% $$ p.a. on a straight-line basis. Determine the cost of the lawnmower 10 years ago. $$ [10] $$

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Sure, let's tackle each part of the question step by step.

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### **Question 2.1**

**Boitumelo invests R28,000 for 7 years with varying interest rates:**
- **First 2 years:** 12% p.a. compounded semi-annually.
- **Next 5 years:** 10% p.a. compounded quarterly.

**Step 1: Calculate the amount after the first 2 years.**

- **Interest Rate per Half-Year:** $$ \frac{12\%}{2} = 6\% = 0.06 $$
- **Number of Periods:** $$ 2 \text{ years} \times 2 = 4 $$ semi-annual periods.
  
$$
A_1 = P \times \left(1 + \frac{r}{n}\right)^{nt} = 28,000 \times (1 + 0.06)^4
$$

$$
A_1 = 28,000 \times (1.06)^4 \approx 28,000 \times 1.2625 = 35,350 \text{ Rands}
$$

**Step 2: Calculate the amount after the next 5 years.**

- **Interest Rate per Quarter:** $$ \frac{10\%}{4} = 2.5\% = 0.025 $$
- **Number of Periods:** $$ 5 \text{ years} \times 4 = 20 $$ quarterly periods.
  
$$
A_2 = A_1 \times \left(1 + \frac{r}{n}\right)^{nt} = 35,350 \times (1 + 0.025)^{20}
$$

$$
A_2 = 35,350 \times (1.025)^{20} \approx 35,350 \times 1.6386 = 57,924 \text{ Rands}
$$

**Final Amount after 7 Years:** **≈ R57,924**

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### **Question 2.2**

**Convert an interest rate of 10% p.a. compounded quarterly to an annual effective interest rate.**

- **Interest Rate per Quarter:** $$ \frac{10\%}{4} = 2.5\% = 0.025 $$
- **Number of Compounding Periods per Year:** 4

$$
\text{Effective Annual Rate (EAR)} = \left(1 + \frac{r}{n}\right)^n - 1 = (1 + 0.025)^4 - 1
$$

$$
\text{EAR} = (1.025)^4 - 1 \approx 1.1038 - 1 = 0.1038 = 10.38\%
$$

**Annual Effective Interest Rate:** **≈ 10.38%**

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### **Question 2.3**

**Determine the original cost of the lawnmower given:**
- **Depreciated Value after 10 years:** R236
- **Depreciation Rate:** 9.2% p.a. (Straight-Line Basis)

**Straight-Line Depreciation Formula:**

$$
\text{Depreciated Value} = \text{Original Cost} - (\text{Depreciation Rate} \times \text{Original Cost} \times \text{Time})
$$

$$
236 = C - (0.092 \times C \times 10) \
236 = C - 0.92C \
236 = 0.08C \
C = \frac{236}{0.08} \
C = 2,950 \text{ Rands}
$$

**Original Cost of the Lawnmower:** **R2,950**

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### **Summary of Answers**

2.1. **The investment grows to approximately R57 924 after 7 years.**

2.2. **An annual effective interest rate of approximately 10.38 %**

2.3. **The lawnmower originally cost R2 950 ten years ago.**
After 7 years, the investment grows to approximately R57,924. The annual effective interest rate is about 10.38%. The lawnmower originally cost R2,950 ten years ago.

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