8.1 The interest rate on an investment is 𝑥% per annum compounded

Question

monthly. Calculate the value of 𝑥 given that the corresponding
effective interest rate is 9,92%. (3)
8.2 Liyabona bought a computer for R4 700. The value of the computer
depreciated at a rate of 18% p.a. Using the reducing-balance
method, calculate the book value of the computer 4 years after it
was bought. (3)
8.3 Ethan invested R30 000 for 6 years. The investment earned interest
at 12% p.a., compounded monthly for the first two years.
Thereafter, the interest rate changed to 10,8% p.a., compounded
semi-annually for the rest of the period.
Calculate the value of the investment at the end of 6 years. (4)
8.4 Fatima deposited R25 000 into a savings account with an interest
rate of 18% p.a., compounded monthly. Fatima withdrew R8 000
from the account 2 years after depositing the initial amount. She
deposited another R4 000 into this account 3½ years after the
initial deposit.
What amount will Fatima have 5 years after making the initial
deposit in this account? (6)

Answer 1

Let's solve each part of the question step by step.

8.1 Effective Interest Rate Calculation

The formula to calculate the effective interest rate from the nominal interest rate \( x % \) compounded monthly is:

\[ \text{Effective Interest Rate} = \left(1 + \frac{x}{100 \times 12}\right)^{12} - 1 = 0.0992 \]

We need to solve for \( x \):

1. Rearranging gives:

\[ \left(1 + \frac{x}{1200}\right)^{12} = 1.0992 \]

2. Taking the 12th root:

\[ 1 + \frac{x}{1200} = (1.0992)^{\frac{1}{12}} \]

Calculating \( (1.0992)^{\frac{1}{12}} \):

\[ (1.0992)^{\frac{1}{12}} \approx 1.00797 \]

3. Now, solve for \( x \):

\[ \frac{x}{1200} = 1.00797 - 1 \quad \Rightarrow \quad \frac{x}{1200} \approx 0.00797 \]

\[ x \approx 0.00797 \times 1200 \quad \Rightarrow \quad x \approx 9.564 \]

So, rounded to two decimal places, the interest rate \( x \approx 9.56% \).

8.2 Depreciation Calculation Using the Reducing-Balance Method

Using the formula for the book value after \( n \) years with reducing balance depreciation:

\[ \text{Book Value} = P \times (1 - r)^n \]

Where \( P = 4700 \), \( r = 0.18 \), and \( n = 4 \):

\[ \text{Book Value} = 4700 \times (1 - 0.18)^4 \]

Calculating \( (1 - 0.18)^4 \):

\[ (0.82)^4 \approx 0.4524 \]

Now calculate the book value:

\[ \text{Book Value} = 4700 \times 0.4524 \approx 2126.68 \]

So, the book value of the computer after 4 years is approximately R2126.68.

8.3 Investment Value Calculation

First we calculate the value of Ethan’s investment for the first 2 years at 12% compounded monthly:

1. The formula for compound interest is:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

Where:

• \( P = 30000 \)
• \( r = 0.12 \)
• \( n = 12 \) (monthly)
• \( t = 2 \)

Calculating:

\[ A = 30000 \left(1 + \frac{0.12}{12}\right)^{12 \times 2} = 30000 \left(1 + 0.01\right)^{24} \approx 30000 \times (1.01^{24}) \approx 30000 \times 1.2682418 \approx R38047.25 \]

Next, for the next 4 years at 10.8% compounded semi-annually:

• \( r = 0.108 \)
• \( n = 2 \)

Using the amount from the first 2 years as the principal:

\[ A = 38047.25 \left(1 + \frac{0.108}{2}\right)^{2 \times 4} \]

Calculating:

\[ A = 38047.25 \left(1 + 0.054\right)^{8} \approx 38047.25 \times (1.054^{8}) \approx 38047.25 \times 1.48024 \approx R56498.29 \]

So, at the end of 6 years, the value of the investment is approximately R56498.29.

8.4 Final Account Balance Calculation

First, calculate the amount in the savings account after 5 years at 18% compounded monthly:

1. After 2 years:

\[ A_1 = 25000 \left(1 + \frac{0.18}{12}\right)^{12 \times 2} = 25000 \left(1 + 0.015\right)^{24} = 25000 \times (1.015^{24}) \approx 25000 \times 1.404112 \approx R35103 \]

2. After withdrawing R8000:

\[ A_2 = 35103 - 8000 = R27103 \]

3. Compound this for another 3 years (3 to 5 years):

\[ A_2' = 27103 \left(1 + \frac{0.18}{12}\right)^{12 \times 3} = 27103 \left(1.015^{36}\right) \approx 27103 \times 1.72033 \approx R46683.53 \]

4. Now, deposit R4000 after 3.5 years, 0.5 years before the total time (from year 3.5 to year 5):

\[ A_3 = 4000 \left(1 + \frac{0.18}{12}\right)^{12 \times 1.5} = 4000 \left(1.015^{18}\right) \approx 4000 \times 1.30464 \approx R5218.56 \]

5. Add this to the previous total after 5 years:

\[ A_{final} = 46683.53 + 5218.56 \approx R51892.09 \]

So, the total amount Fatima will have after 5 years is approximately R51892.09.