Question
QUESTION 1
1.1 Calculate the effective annual interest rate of a nominal rate of 20%
per annum compounded quarterly. (4)
1.2 A washing machine is sold after 5 years for R9 500. What was the
purchase price if the depreciation is 18% p.a. on a reducing balance? (3)
1.3 Vadi invests R5 500 in a savings account. The interest rate for the
first 5 years is 7% p.a. compounded monthly. After the first five-year
period, the interest rate is changed to 10% p.a. compounded
semi-annually for the next 5 years.
Determine the amount of money Vadi has in her savings account at
the end of this period. (6)
1.4 Exactly 8 years ago, Ethan invested R30 000 in an account earning an
interest of 6,5% p.a. compounded monthly.
He withdrew R10 000 three years after making the initial deposit
and re-invested R10 000 five years later after making the initial
deposit.
Calculate the final amount that he received after 8 years. (6)
[19]
1.1 Calculate the effective annual interest rate of a nominal rate of 20%
per annum compounded quarterly. (4)
1.2 A washing machine is sold after 5 years for R9 500. What was the
purchase price if the depreciation is 18% p.a. on a reducing balance? (3)
1.3 Vadi invests R5 500 in a savings account. The interest rate for the
first 5 years is 7% p.a. compounded monthly. After the first five-year
period, the interest rate is changed to 10% p.a. compounded
semi-annually for the next 5 years.
Determine the amount of money Vadi has in her savings account at
the end of this period. (6)
1.4 Exactly 8 years ago, Ethan invested R30 000 in an account earning an
interest of 6,5% p.a. compounded monthly.
He withdrew R10 000 three years after making the initial deposit
and re-invested R10 000 five years later after making the initial
deposit.
Calculate the final amount that he received after 8 years. (6)
[19]
Answers
GPT-4o mini
Let's solve each question step-by-step.
### 1.1 Calculate the effective annual interest rate of a nominal rate of 20% per annum compounded quarterly.
To calculate the effective annual interest rate (EAR) from a nominal interest rate (i) compounded quarterly (m), we use the formula:
\[
EAR = \left(1 + \frac{i}{m}\right)^{m} - 1
\]
Where:
- \(i = 0.20\) (20% expressed as a decimal)
- \(m = 4\) (since the interest is compounded quarterly)
Substituting the values:
\[
EAR = \left(1 + \frac{0.20}{4}\right)^{4} - 1
\]
\[
EAR = \left(1 + 0.05\right)^{4} - 1
\]
\[
EAR = (1.05)^{4} - 1
\]
Calculating \( (1.05)^{4} \):
\[
(1.05)^{4} \approx 1.21550625
\]
Now subtracting 1:
\[
EAR \approx 1.21550625 - 1 \approx 0.21550625
\]
To express it as a percentage:
\[
EAR \approx 21.55\%
\]
### Answer for 1.1: 21.55%
---
### 1.2 A washing machine is sold after 5 years for R9 500. What was the purchase price if the depreciation is 18% p.a. on a reducing balance?
Let \( P \) be the purchase price. The formula for the value of an asset after depreciation is:
\[
V = P(1 - r)^t
\]
Where:
- \( V = 9500 \) (value after 5 years)
- \( r = 0.18 \) (18% depreciation)
- \( t = 5 \) (years)
Setting up the equation:
\[
9500 = P(1 - 0.18)^5
\]
Calculating \( (1 - 0.18)^5 \):
\[
(0.82)^5 \approx 0.413161
\]
Now substituting back:
\[
9500 = P \times 0.413161
\]
Solving for \( P \):
\[
P = \frac{9500}{0.413161} \approx 22941.47
\]
### Answer for 1.2: R22,941.47 (approximately)
---
### 1.3 Vadi invests R5 500 in a savings account. The interest rate for the first 5 years is 7% p.a. compounded monthly. After the first five-year period, the interest rate is changed to 10% p.a. compounded semi-annually for the next 5 years. Determine the amount of money Vadi has in her savings account at the end of this period.
**Step 1: Calculate the amount after the first 5 years.**
Using the formula for compound interest:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( P = R5,500 \)
- \( r = 0.07 \) (7% expressed as a decimal)
- \( n = 12 \) (monthly compounding)
- \( t = 5 \)
Substituting the values:
\[
A = 5500 \left(1 + \frac{0.07}{12}\right)^{12 \times 5}
\]
Calculating \( \frac{0.07}{12} \):
\[
A = 5500 \left(1 + 0.00583333\right)^{60}
\]
\[
A = 5500 \left(1.00583333\right)^{60} \approx 5500 \times 1.48985 \approx 8194.17
\]
**Step 2: Now calculate the amount after the next 5 years at 10% p.a. compounded semi-annually.**
Where:
- New principal \( P = 8194.17 \)
- \( r = 0.10 \) (10% expressed as a decimal)
- \( n = 2 \) (semi-annual compounding)
- \( t = 5 \)
Using the formula again:
\[
A = 8194.17 \left(1 + \frac{0.10}{2}\right)^{2 \times 5}
\]
Calculating \( \frac{0.10}{2} \):
\[
A = 8194.17 (1 + 0.05)^{10} = 8194.17 (1.05)^{10}
\]
Calculating \( (1.05)^{10} \approx 1.62889 \):
\[
A \approx 8194.17 \times 1.62889 \approx 13352.14
\]
### Answer for 1.3: R13,352.14 (approximately)
---
### 1.4 Exactly 8 years ago, Ethan invested R30 000 in an account earning an interest of 6.5% p.a. compounded monthly. He withdrew R10 000 three years after making the initial deposit and re-invested R10 000 five years later after making the initial deposit. Calculate the final amount that he received after 8 years.
**Step 1: Calculate the amount after 8 years.**
Using the compound interest formula:
Where:
- \( P = 30000 \)
- \( r = 0.065 \)
- \( n = 12 \)
- \( t = 8 \)
\[
A = 30000 \left(1 + \frac{0.065}{12}\right)^{12 \times 8}
\]
Calculating \( \frac{0.065}{12} \):
\[
A = 30000 \left(1 + 0.00541667\right)^{96} \approx 30000 \left(1.00541667\right)^{96}
\]
Calculating \( (1.00541667)^{96} \approx 1.747422 \):
\[
A \approx 30000 \times 1.747422 \approx 52422.66
\]
**Step 2: Withdraw R10,000 after 3 years.**
Calculate the amount after 3 years:
\[
A(3) = 30000 \left(1 + \frac{0.065}{12}\right)^{12 \times 3}
\]
Calculating \( (1.00541667)^{36} \approx 1.210698 \):
\[
A(3) = 30000 \times 1.210698 \approx 36320.93
\]
Now, withdrawing R10,000:
\[
A(3, after withdrawal) = 36320.93 - 10000 = 26320.93
\]
**Step 3: Invest remaining amount for the next 5 years.**
Using the compounded value for the remaining 5 years:
\[
A = 26320.93 \left(1 + \frac{0.065}{12}\right)^{12 \times 5} = 26320.93 \left(1.00541667\right)^{60} \approx 26320.93 \times 1.48985 \approx 39288.66
\]
**Step 4: Add re-invested R10,000 for the last 3 years.**
Re-invested R10,000 from the 5th year for 3 years:
\[
A = 10000 \left(1.00541667\right)^{36} \approx 10000 \times 1.210698 = 12106.98
\]
**Final Amount:**
Combine the two amounts:
\[
Total = 39288.66 + 12106.98 \approx 51395.64
\]
### Answer for 1.4: R51,395.64 (approximately)
---
In summary:
1.1. **21.55%**
1.2. **R22,941.47**
1.3. **R13,352.14**
1.4. **R51,395.64**
### 1.1 Calculate the effective annual interest rate of a nominal rate of 20% per annum compounded quarterly.
To calculate the effective annual interest rate (EAR) from a nominal interest rate (i) compounded quarterly (m), we use the formula:
\[
EAR = \left(1 + \frac{i}{m}\right)^{m} - 1
\]
Where:
- \(i = 0.20\) (20% expressed as a decimal)
- \(m = 4\) (since the interest is compounded quarterly)
Substituting the values:
\[
EAR = \left(1 + \frac{0.20}{4}\right)^{4} - 1
\]
\[
EAR = \left(1 + 0.05\right)^{4} - 1
\]
\[
EAR = (1.05)^{4} - 1
\]
Calculating \( (1.05)^{4} \):
\[
(1.05)^{4} \approx 1.21550625
\]
Now subtracting 1:
\[
EAR \approx 1.21550625 - 1 \approx 0.21550625
\]
To express it as a percentage:
\[
EAR \approx 21.55\%
\]
### Answer for 1.1: 21.55%
---
### 1.2 A washing machine is sold after 5 years for R9 500. What was the purchase price if the depreciation is 18% p.a. on a reducing balance?
Let \( P \) be the purchase price. The formula for the value of an asset after depreciation is:
\[
V = P(1 - r)^t
\]
Where:
- \( V = 9500 \) (value after 5 years)
- \( r = 0.18 \) (18% depreciation)
- \( t = 5 \) (years)
Setting up the equation:
\[
9500 = P(1 - 0.18)^5
\]
Calculating \( (1 - 0.18)^5 \):
\[
(0.82)^5 \approx 0.413161
\]
Now substituting back:
\[
9500 = P \times 0.413161
\]
Solving for \( P \):
\[
P = \frac{9500}{0.413161} \approx 22941.47
\]
### Answer for 1.2: R22,941.47 (approximately)
---
### 1.3 Vadi invests R5 500 in a savings account. The interest rate for the first 5 years is 7% p.a. compounded monthly. After the first five-year period, the interest rate is changed to 10% p.a. compounded semi-annually for the next 5 years. Determine the amount of money Vadi has in her savings account at the end of this period.
**Step 1: Calculate the amount after the first 5 years.**
Using the formula for compound interest:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( P = R5,500 \)
- \( r = 0.07 \) (7% expressed as a decimal)
- \( n = 12 \) (monthly compounding)
- \( t = 5 \)
Substituting the values:
\[
A = 5500 \left(1 + \frac{0.07}{12}\right)^{12 \times 5}
\]
Calculating \( \frac{0.07}{12} \):
\[
A = 5500 \left(1 + 0.00583333\right)^{60}
\]
\[
A = 5500 \left(1.00583333\right)^{60} \approx 5500 \times 1.48985 \approx 8194.17
\]
**Step 2: Now calculate the amount after the next 5 years at 10% p.a. compounded semi-annually.**
Where:
- New principal \( P = 8194.17 \)
- \( r = 0.10 \) (10% expressed as a decimal)
- \( n = 2 \) (semi-annual compounding)
- \( t = 5 \)
Using the formula again:
\[
A = 8194.17 \left(1 + \frac{0.10}{2}\right)^{2 \times 5}
\]
Calculating \( \frac{0.10}{2} \):
\[
A = 8194.17 (1 + 0.05)^{10} = 8194.17 (1.05)^{10}
\]
Calculating \( (1.05)^{10} \approx 1.62889 \):
\[
A \approx 8194.17 \times 1.62889 \approx 13352.14
\]
### Answer for 1.3: R13,352.14 (approximately)
---
### 1.4 Exactly 8 years ago, Ethan invested R30 000 in an account earning an interest of 6.5% p.a. compounded monthly. He withdrew R10 000 three years after making the initial deposit and re-invested R10 000 five years later after making the initial deposit. Calculate the final amount that he received after 8 years.
**Step 1: Calculate the amount after 8 years.**
Using the compound interest formula:
Where:
- \( P = 30000 \)
- \( r = 0.065 \)
- \( n = 12 \)
- \( t = 8 \)
\[
A = 30000 \left(1 + \frac{0.065}{12}\right)^{12 \times 8}
\]
Calculating \( \frac{0.065}{12} \):
\[
A = 30000 \left(1 + 0.00541667\right)^{96} \approx 30000 \left(1.00541667\right)^{96}
\]
Calculating \( (1.00541667)^{96} \approx 1.747422 \):
\[
A \approx 30000 \times 1.747422 \approx 52422.66
\]
**Step 2: Withdraw R10,000 after 3 years.**
Calculate the amount after 3 years:
\[
A(3) = 30000 \left(1 + \frac{0.065}{12}\right)^{12 \times 3}
\]
Calculating \( (1.00541667)^{36} \approx 1.210698 \):
\[
A(3) = 30000 \times 1.210698 \approx 36320.93
\]
Now, withdrawing R10,000:
\[
A(3, after withdrawal) = 36320.93 - 10000 = 26320.93
\]
**Step 3: Invest remaining amount for the next 5 years.**
Using the compounded value for the remaining 5 years:
\[
A = 26320.93 \left(1 + \frac{0.065}{12}\right)^{12 \times 5} = 26320.93 \left(1.00541667\right)^{60} \approx 26320.93 \times 1.48985 \approx 39288.66
\]
**Step 4: Add re-invested R10,000 for the last 3 years.**
Re-invested R10,000 from the 5th year for 3 years:
\[
A = 10000 \left(1.00541667\right)^{36} \approx 10000 \times 1.210698 = 12106.98
\]
**Final Amount:**
Combine the two amounts:
\[
Total = 39288.66 + 12106.98 \approx 51395.64
\]
### Answer for 1.4: R51,395.64 (approximately)
---
In summary:
1.1. **21.55%**
1.2. **R22,941.47**
1.3. **R13,352.14**
1.4. **R51,395.64**