Get the most out of your questions with the extensive resources available on IDNLearn.com. Our experts provide timely and precise responses to help you understand and solve any issue you face.
Sagot :
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]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
