Convert an interest rate of $$ 10 \% $$ p.a. compounded quarterly to an annual effecti interest rate. A lawnmower was traded in for R 236 after 10 years. The depreciation was calculated at $$ 9,2 \% $$ p.a. on a straight-line basis. Determine the cost of the lawnmower 10 years ago.

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Sure, let's tackle both problems step by step.

---

### **1. Converting the Interest Rate to an Annual Effective Rate**

**Given:**
- **Nominal Annual Interest Rate (r):** 10% per annum.
- **Compounding Frequency:** Quarterly.

**Objective:**
Convert the nominal interest rate to the **Annual Effective Interest Rate (iₑₓₚ)**.

**Formula:**
The Annual Effective Rate can be calculated using the formula:

$$
i_{\text{eff}} = \left(1 + \frac{r}{n}\right)^n - 1
$$

where:
- $$ r $$ = nominal annual interest rate (in decimal).
- $$ n $$ = number of compounding periods per year.

**Calculation:**

1. **Convert the percentage to decimal:**
   $$
   r = 10\% = 0.10
   $$
   
2. **Determine the number of compounding periods:**
   $$
   n = 4 \quad (\text{since it's compounded quarterly})
   $$
   
3. **Plug the values into the formula:**
   $$
   i_{\text{eff}} = \left(1 + \frac{0.10}{4}\right)^4 - 1 = \left(1 + 0.025\right)^4 - 1
   $$
   
4. **Calculate the expression:**
   $$
   \left(1.025\right)^4 \approx 1.10381289
   $$
   
5. **Subtract 1 to find the effective rate:**
   $$
   i_{\text{eff}} \approx 1.10381289 - 1 = 0.10381289
   $$
   
6. **Convert back to percentage:**
   $$
   i_{\text{eff}} \approx 10.38\%
   $$

**Answer:**
The **Annual Effective Interest Rate** is **approximately **10.38%**.

---

### **2. Determining the Original Cost of the Lawnmower**

**Given:**
- **Depreciated Value after 10 years (V₁₀):** R236.
- **Depreciation Rate (d):** 9.2% per annum.
- **Depreciation Method:** Straight-line.

**Objective:**
Find the **original cost (C)** of the lawnmower 10 years ago.

**Understanding Straight-Line Depreciation:**
In the straight-line method, the asset loses the same amount of value each year. The depreciation is calculated as a fixed percentage of the **original cost** annually.

**Formula:**
$$
\text{Depreciation per year} = d \times C
$$
$$
\text{Total Depreciation over 10 years} = 10 \times (d \times C) = 10d \times C
$$
$$
\text{Remaining Value after 10 years} = C - 10d \times C = C(1 - 10d)
$$

**Calculation:**

1. **Convert the depreciation rate to decimal:**
   $$
   d = 9.2\% = 0.092
   $$
   
2. **Express the remaining value after 10 years:**
   $$
   V_{10} = C(1 - 10 \times 0.092) = C(1 - 0.92) = C \times 0.08
   $$
   
3. **Set up the equation with the given depreciated value:**
   $$
   C \times 0.08 = 236
   $$
   
4. **Solve for $$ C $$:**
   $$
   C = \frac{236}{0.08} = 2950
   $$
   
**Answer:**
The **original cost** of the lawnmower **10 years ago** was **R2,950**.

---

I hope this clarifies both problems! If you have any further questions, feel free to ask.
**1. Interest Rate Conversion:** - **Nominal Annual Rate:** 10% per annum, compounded quarterly. - **Annual Effective Interest Rate:** Approximately **10.38%**. **2. Lawnmower Cost:** - **Depreciated Value after 10 Years:** R236. - **Depreciation Rate:** 9.2% per annum. - **Original Cost 10 Years Ago:** R2,950.

Convert an interest rate of p.a. compounded quarterly to an annual effecti
interest rate.
A lawnmower was traded in for R 236 after 10 years. The depreciation was
calculated at p.a. on a straight-line basis.
Determine the cost of the lawnmower 10 years ago.

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Convert an interest rate of p.a. compounded quarterly to an annual effecti interest rate. A lawnmower was traded in for R 236 after 10 years. The depreciation was calculated at p.a. on a straight-line basis. Determine the cost of the lawnmower 10 years ago.