Kuda bought a laptop for R10 000 on 31 January 2019. He will replace if win a new one in 5 years' time on 31 January 2024 . 6.2 .1 The value of the old Iaptop depreciates annually at a rate of $$ 20 \% $$ p.a. according to the straight-line method. After how many years will the laptop have a valuc of R0? 6.2 .2 Kuda will buy a laptop that costs R20 000 . In order to cover the cost price, he made his first monthly deposit into a savings account on 28 February 2019. He will make his $$ 60^{ ext {a }} $$ monthly deposit on 31 Jamuary 2024. The savings account pays interest at $$ 8,7 \% $$ p.3., compounded monthly. Calculate Kuda's monthly deposit into this account.

Question

Kuda bought a laptop for R10 000 on 31 January 2019. He will replace if win a new one in 5 years' time on 31 January 2024 . 6.2 .1 The value of the old Iaptop depreciates annually at a rate of $$ 20 \% $$ p.a. according to the straight-line method. After how many years will the laptop have a valuc of R0? 6.2 .2 Kuda will buy a laptop that costs R20 000 . In order to cover the cost price, he made his first monthly deposit into a savings account on 28 February 2019. He will make his $$ 60^{	ext {a }} $$ monthly deposit on 31 Jamuary 2024. The savings account pays interest at $$ 8,7 \% $$ p.3., compounded monthly. Calculate Kuda's monthly deposit into this account.

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Let's solve the problem step by step.

### 6.2.1: Depreciation of the Laptop

Kuda's laptop depreciates at a rate of $$ 20\% $$ per annum using the straight-line method. The initial value of the laptop is R10,000.

1. **Calculate the annual depreciation:**
   $$
   \text{Annual Depreciation} = \text{Initial Value} \times \text{Depreciation Rate} = 10,000 \times 0.20 = 2,000
   $$

2. **Determine how many years it will take for the laptop to reach a value of R0:**
   Since the laptop depreciates by R2,000 each year, we can set up the equation:
   $$
   \text{Value after } n \text{ years} = \text{Initial Value} - n \times \text{Annual Depreciation}
   $$
   Setting this equal to R0:
   $$
   10,000 - n \times 2,000 = 0
   $$
   Solving for $$ n $$:
   $$
   n \times 2,000 = 10,000 \implies n = \frac{10,000}{2,000} = 5
   $$

Thus, it will take **5 years** for the laptop to have a value of R0.

### 6.2.2: Monthly Deposit Calculation

Kuda wants to buy a new laptop costing R20,000 and will make monthly deposits into a savings account starting from 28 February 2019 until 31 January 2024. This is a total of 60 deposits.

1. **Identify the interest rate and compounding frequency:**
   - Interest rate: $$ 8.7\% $$ per annum
   - Monthly interest rate: 
   $$
   r = \frac{8.7\%}{12} = \frac{0.087}{12} = 0.00725
   $$

2. **Use the future value of an annuity formula to find the monthly deposit $$ P $$:**
   The future value $$ FV $$ of an annuity can be calculated using the formula:
   $$
   FV = P \times \frac{(1 + r)^n - 1}{r}
   $$
   Where:
   - $$ FV = 20,000 $$
   - $$ r = 0.00725 $$
   - $$ n = 60 $$

Rearranging the formula to solve for $$ P $$:
   $$
   P = \frac{FV \times r}{(1 + r)^n - 1}
   $$

Now, let's calculate $$ P $$.

### Calculation of Monthly Deposit $$ P $$

Using the values:
- $$ FV = 20,000 $$
- $$ r = 0.00725 $$
- $$ n = 60 $$

We will perform the calculation.

Let's calculate $$ P $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{20000\times 0.00725}{\left(\left(1+0.00725\right)^{60}-1\right)}$$
- step1: Remove the parentheses:
  $$\frac{20000\times 0.00725}{\left(1+0.00725\right)^{60}-1}$$
- step2: Add the numbers:
  $$\frac{20000\times 0.00725}{1.00725^{60}-1}$$
- step3: Convert the expressions:
  $$\frac{20000\times 0.00725}{\left(\frac{4029}{4000}\right)^{60}-1}$$
- step4: Multiply the numbers:
  $$\frac{145}{\left(\frac{4029}{4000}\right)^{60}-1}$$
- step5: Subtract the numbers:
  $$\frac{145}{\frac{4029^{60}-4000^{60}}{4000^{60}}}$$
- step6: Multiply by the reciprocal:
  $$145\times \frac{4000^{60}}{4029^{60}-4000^{60}}$$
- step7: Multiply:
  $$\frac{145\times 4000^{60}}{4029^{60}-4000^{60}}$$
The calculation for Kuda's monthly deposit $$ P $$ yields approximately R267.26.

### Summary of Results

1. **Depreciation of the Laptop:**
   - The laptop will have a value of R0 after **5 years**.

2. **Monthly Deposit Calculation:**
   - Kuda's monthly deposit into the savings account should be approximately **R267.26** to cover the cost of the new laptop.
After 5 years, the laptop will be worth R0. Kuda needs to deposit approximately R267.26 each month into his savings account to afford the new R20,000 laptop by January 2024.

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