Question
R2 000 is deposited into an account for n years at an interest rate of 16% per annum compounded quarterly. After two years the interest rate changes to 13% per annum, compounded monthly. At the end of the term the value of the investment is R4 200. Calculate the value of n.
Answers
GPT-4o mini
To calculate the value of \( n \) for the investment, we will break the problem into two parts: the first two years with an interest rate of 16% compounded quarterly and the subsequent years with an interest rate of 13% compounded monthly.
### Step 1: Calculate the amount after the first 2 years
The formula for compound interest is:
\[
A = P \left(1 + \frac{r}{m}\right)^{mt}
\]
where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount).
- \( r \) is the annual interest rate (decimal).
- \( m \) is the number of times that interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed.
For the first 2 years:
- \( P = R2000 \)
- \( r = 0.16 \)
- \( m = 4 \) (quarterly)
- \( t = 2 \)
Plugging in these values:
\[
A_2 = 2000 \left(1 + \frac{0.16}{4}\right)^{4 \times 2}
\]
Calculating the components:
\[
A_2 = 2000 \left(1 + 0.04\right)^{8}
\]
\[
A_2 = 2000 \left(1.04\right)^{8}
\]
Calculating \( (1.04)^{8} \):
\[
(1.04)^{8} \approx 1.368569
\]
Now calculating \( A_2 \):
\[
A_2 \approx 2000 \times 1.368569 \approx 2737.14
\]
### Step 2: Calculate the amount after changing interest rates
Now the new principal is \( R2737.14 \), and it will be compounded at 13% per annum compounded monthly for \( n - 2 \) years.
- New \( P = R2737.14 \)
- \( r = 0.13 \)
- \( m = 12 \) (monthly)
- \( t = n - 2 \)
So the formula becomes:
\[
A = 2737.14 \left(1 + \frac{0.13}{12}\right)^{12(n - 2)}
\]
We know the final amount \( A \) after both periods is \( R4200 \):
\[
4200 = 2737.14 \left(1 + \frac{0.13}{12}\right)^{12(n - 2)}
\]
\[
4200 = 2737.14 \left(1 + 0.0108333\right)^{12(n - 2)}
\]
\[
4200 = 2737.14 \left(1.0108333\right)^{12(n - 2)}
\]
Now, divide both sides by \( 2737.14 \):
\[
\frac{4200}{2737.14} = \left(1.0108333\right)^{12(n - 2)}
\]
\[
1.535 = \left(1.0108333\right)^{12(n - 2)}
\]
### Step 3: Solve for \( n \)
Taking the natural logarithm of both sides:
\[
\ln(1.535) = 12(n - 2) \ln(1.0108333)
\]
\[
0.430 = 12(n - 2) \cdot 0.010736
\]
\[
0.430 = 0.128832(n - 2)
\]
Now isolate \( n - 2 \):
\[
n - 2 = \frac{0.430}{0.128832} \approx 3.34
\]
Now add 2 to solve for \( n \):
\[
n \approx 3.34 + 2 = 5.34
\]
Since \( n \) must be a whole number, we round to the nearest whole number:
\[
n \approx 5
\]
Thus, the value of \( n \) is approximately \( 5 \) years.
### Step 1: Calculate the amount after the first 2 years
The formula for compound interest is:
\[
A = P \left(1 + \frac{r}{m}\right)^{mt}
\]
where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount).
- \( r \) is the annual interest rate (decimal).
- \( m \) is the number of times that interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed.
For the first 2 years:
- \( P = R2000 \)
- \( r = 0.16 \)
- \( m = 4 \) (quarterly)
- \( t = 2 \)
Plugging in these values:
\[
A_2 = 2000 \left(1 + \frac{0.16}{4}\right)^{4 \times 2}
\]
Calculating the components:
\[
A_2 = 2000 \left(1 + 0.04\right)^{8}
\]
\[
A_2 = 2000 \left(1.04\right)^{8}
\]
Calculating \( (1.04)^{8} \):
\[
(1.04)^{8} \approx 1.368569
\]
Now calculating \( A_2 \):
\[
A_2 \approx 2000 \times 1.368569 \approx 2737.14
\]
### Step 2: Calculate the amount after changing interest rates
Now the new principal is \( R2737.14 \), and it will be compounded at 13% per annum compounded monthly for \( n - 2 \) years.
- New \( P = R2737.14 \)
- \( r = 0.13 \)
- \( m = 12 \) (monthly)
- \( t = n - 2 \)
So the formula becomes:
\[
A = 2737.14 \left(1 + \frac{0.13}{12}\right)^{12(n - 2)}
\]
We know the final amount \( A \) after both periods is \( R4200 \):
\[
4200 = 2737.14 \left(1 + \frac{0.13}{12}\right)^{12(n - 2)}
\]
\[
4200 = 2737.14 \left(1 + 0.0108333\right)^{12(n - 2)}
\]
\[
4200 = 2737.14 \left(1.0108333\right)^{12(n - 2)}
\]
Now, divide both sides by \( 2737.14 \):
\[
\frac{4200}{2737.14} = \left(1.0108333\right)^{12(n - 2)}
\]
\[
1.535 = \left(1.0108333\right)^{12(n - 2)}
\]
### Step 3: Solve for \( n \)
Taking the natural logarithm of both sides:
\[
\ln(1.535) = 12(n - 2) \ln(1.0108333)
\]
\[
0.430 = 12(n - 2) \cdot 0.010736
\]
\[
0.430 = 0.128832(n - 2)
\]
Now isolate \( n - 2 \):
\[
n - 2 = \frac{0.430}{0.128832} \approx 3.34
\]
Now add 2 to solve for \( n \):
\[
n \approx 3.34 + 2 = 5.34
\]
Since \( n \) must be a whole number, we round to the nearest whole number:
\[
n \approx 5
\]
Thus, the value of \( n \) is approximately \( 5 \) years.