Answer :
Certainly! Let's tackle these questions step-by-step.
### 1. Brenda's Future Savings Calculation
Brenda deposits RS 5000 into a savings account. The interest rate structure changes as follows:
- For the first 3 years, the interest rate is 14% per annum, compounded semi-annually.
- For the next 4 years, the interest rate is 12% per annum, compounded monthly.
#### Step 1: Calculate the future value of the savings after the first 3 years
Given the interest rate is compounded semi-annually, we have:
- Principal: [tex]\( P = RS 5000 \)[/tex]
- Annual interest rate [tex]\( r = 14\% = 0.14 \)[/tex]
- Number of times interest is compounded per year [tex]\( n = 2 \)[/tex]
- Number of years [tex]\( t = 3 \)[/tex]
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the values:
[tex]\[ A_1 = 5000 \left(1 + \frac{0.14}{2}\right)^{2 \times 3} \][/tex]
[tex]\[ A_1 = 5000 \left(1 + 0.07\right)^6 \][/tex]
[tex]\[ A_1 = 5000 \left(1.07\right)^6 \][/tex]
After calculating, we get:
[tex]\[ A_1 = 7503.65 \][/tex]
#### Step 2: Calculate the future value of the savings after the next 4 years
Now, the interest rate changes to 12% per annum, compounded monthly. Using the amount calculated from Step 1 as the new principal, we have:
- New Principal: [tex]\( P = 7503.65 \)[/tex]
- Annual interest rate [tex]\( r = 12\% = 0.12 \)[/tex]
- Number of times interest is compounded per year [tex]\( n = 12 \)[/tex]
- Number of years [tex]\( t = 4 \)[/tex]
Using the compound interest formula again:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the values:
[tex]\[ A_2 = 7503.65 \left(1 + \frac{0.12}{12}\right)^{12 \times 4} \][/tex]
[tex]\[ A_2 = 7503.65 \left(1 + 0.01\right)^{48} \][/tex]
[tex]\[ A_2 = 7503.65 \left(1.01\right)^{48} \][/tex]
After calculating, we get:
[tex]\[ A_2 = 12097.58 \][/tex]
Thus, the future value of Brenda's savings at the end of the seventh year is:
[tex]\[ RS 12097.58 \][/tex]
### 2. R40 000 Investment Calculation
R40 000 is invested for 5 years at 16% per annum compounded monthly.
#### Part (a): Calculate the future value using the nominal rate
Given:
- Principal: [tex]\( P = R 40,000 \)[/tex]
- Annual interest rate [tex]\( r = 16\% = 0.16 \)[/tex]
- Number of times interest is compounded per year [tex]\( n = 12 \)[/tex]
- Number of years [tex]\( t = 5 \)[/tex]
Using the compound interest formula:
[tex]\[ FV = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the values:
[tex]\[ FV = 40000 \left(1 + \frac{0.16}{12}\right)^{12 \times 5} \][/tex]
[tex]\[ FV = 40000 \left(1 + 0.0133333\right)^{60} \][/tex]
[tex]\[ FV = 40000 \left(1.0133333\right)^{60} \][/tex]
After calculating, we get:
[tex]\[ FV = 88552.28 \][/tex]
Thus, the future value of the investment using the nominal rate is:
[tex]\[ R 88552.28 \][/tex]
#### Part (b): Convert the nominal rate to the equivalent effective annual rate
The formula to convert a nominal rate compounded [tex]\( n \)[/tex] times per year to an effective annual rate [tex]\( (EAR) \)[/tex] is:
[tex]\[ EAR = \left(1 + \frac{r}{n}\right)^n - 1 \][/tex]
Substituting the values:
[tex]\[ EAR = \left(1 + \frac{0.16}{12}\right)^{12} - 1 \][/tex]
[tex]\[ EAR = \left(1 + 0.0133333\right)^{12} - 1 \][/tex]
[tex]\[ EAR = 1.1722708 - 1 \][/tex]
After calculating, we get:
[tex]\[ EAR = 0.1722708 \][/tex]
Thus, the effective annual rate is:
[tex]\[ 17.23\% \][/tex]
#### Part (c): Use the annual effective rate to calculate the future value
Using the effective annual rate, we recalculate the future value over 5 years:
- Principal: [tex]\( P = R 40,000 \)[/tex]
- Effective annual rate [tex]\( r_{eff} = 17.23\% = 0.1722708 \)[/tex]
- Number of years [tex]\( t = 5 \)[/tex]
Using the compound interest formula:
[tex]\[ FV = P \left(1 + r_{eff}\right)^t \][/tex]
Substituting the values:
[tex]\[ FV = 40000 \left(1 + 0.1722708\right)^5 \][/tex]
[tex]\[ FV = 40000 \left(1.1722708\right)^5 \][/tex]
After calculating, we get:
[tex]\[ FV = 88552.28 \][/tex]
Thus, the future value of the investment using the annual effective rate is:
[tex]\[ R 88552.28 \][/tex]
### 1. Brenda's Future Savings Calculation
Brenda deposits RS 5000 into a savings account. The interest rate structure changes as follows:
- For the first 3 years, the interest rate is 14% per annum, compounded semi-annually.
- For the next 4 years, the interest rate is 12% per annum, compounded monthly.
#### Step 1: Calculate the future value of the savings after the first 3 years
Given the interest rate is compounded semi-annually, we have:
- Principal: [tex]\( P = RS 5000 \)[/tex]
- Annual interest rate [tex]\( r = 14\% = 0.14 \)[/tex]
- Number of times interest is compounded per year [tex]\( n = 2 \)[/tex]
- Number of years [tex]\( t = 3 \)[/tex]
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the values:
[tex]\[ A_1 = 5000 \left(1 + \frac{0.14}{2}\right)^{2 \times 3} \][/tex]
[tex]\[ A_1 = 5000 \left(1 + 0.07\right)^6 \][/tex]
[tex]\[ A_1 = 5000 \left(1.07\right)^6 \][/tex]
After calculating, we get:
[tex]\[ A_1 = 7503.65 \][/tex]
#### Step 2: Calculate the future value of the savings after the next 4 years
Now, the interest rate changes to 12% per annum, compounded monthly. Using the amount calculated from Step 1 as the new principal, we have:
- New Principal: [tex]\( P = 7503.65 \)[/tex]
- Annual interest rate [tex]\( r = 12\% = 0.12 \)[/tex]
- Number of times interest is compounded per year [tex]\( n = 12 \)[/tex]
- Number of years [tex]\( t = 4 \)[/tex]
Using the compound interest formula again:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the values:
[tex]\[ A_2 = 7503.65 \left(1 + \frac{0.12}{12}\right)^{12 \times 4} \][/tex]
[tex]\[ A_2 = 7503.65 \left(1 + 0.01\right)^{48} \][/tex]
[tex]\[ A_2 = 7503.65 \left(1.01\right)^{48} \][/tex]
After calculating, we get:
[tex]\[ A_2 = 12097.58 \][/tex]
Thus, the future value of Brenda's savings at the end of the seventh year is:
[tex]\[ RS 12097.58 \][/tex]
### 2. R40 000 Investment Calculation
R40 000 is invested for 5 years at 16% per annum compounded monthly.
#### Part (a): Calculate the future value using the nominal rate
Given:
- Principal: [tex]\( P = R 40,000 \)[/tex]
- Annual interest rate [tex]\( r = 16\% = 0.16 \)[/tex]
- Number of times interest is compounded per year [tex]\( n = 12 \)[/tex]
- Number of years [tex]\( t = 5 \)[/tex]
Using the compound interest formula:
[tex]\[ FV = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the values:
[tex]\[ FV = 40000 \left(1 + \frac{0.16}{12}\right)^{12 \times 5} \][/tex]
[tex]\[ FV = 40000 \left(1 + 0.0133333\right)^{60} \][/tex]
[tex]\[ FV = 40000 \left(1.0133333\right)^{60} \][/tex]
After calculating, we get:
[tex]\[ FV = 88552.28 \][/tex]
Thus, the future value of the investment using the nominal rate is:
[tex]\[ R 88552.28 \][/tex]
#### Part (b): Convert the nominal rate to the equivalent effective annual rate
The formula to convert a nominal rate compounded [tex]\( n \)[/tex] times per year to an effective annual rate [tex]\( (EAR) \)[/tex] is:
[tex]\[ EAR = \left(1 + \frac{r}{n}\right)^n - 1 \][/tex]
Substituting the values:
[tex]\[ EAR = \left(1 + \frac{0.16}{12}\right)^{12} - 1 \][/tex]
[tex]\[ EAR = \left(1 + 0.0133333\right)^{12} - 1 \][/tex]
[tex]\[ EAR = 1.1722708 - 1 \][/tex]
After calculating, we get:
[tex]\[ EAR = 0.1722708 \][/tex]
Thus, the effective annual rate is:
[tex]\[ 17.23\% \][/tex]
#### Part (c): Use the annual effective rate to calculate the future value
Using the effective annual rate, we recalculate the future value over 5 years:
- Principal: [tex]\( P = R 40,000 \)[/tex]
- Effective annual rate [tex]\( r_{eff} = 17.23\% = 0.1722708 \)[/tex]
- Number of years [tex]\( t = 5 \)[/tex]
Using the compound interest formula:
[tex]\[ FV = P \left(1 + r_{eff}\right)^t \][/tex]
Substituting the values:
[tex]\[ FV = 40000 \left(1 + 0.1722708\right)^5 \][/tex]
[tex]\[ FV = 40000 \left(1.1722708\right)^5 \][/tex]
After calculating, we get:
[tex]\[ FV = 88552.28 \][/tex]
Thus, the future value of the investment using the annual effective rate is:
[tex]\[ R 88552.28 \][/tex]
