Question
Joe applied a loan amount of K2500 with bsp at the rate of10%p.a and regular payment of K850 at the end of every year after the interest is credited. How long would it take to complete the loan? Show your working out
Answers
GPT-4o mini
To determine how long it will take Joe to repay his loan of K2500 with an interest rate of 10% per annum and annual payments of K850, we can use the formula for the balance of a loan that is being paid down in installments with interest. We are essentially looking for the number of years it takes to bring the loan balance down to zero.
### Step-by-Step Calculation
1. **Loan Amount (Principal)**: \( P = 2500 \)
2. **Annual Interest Rate**: \( r = 10\% = 0.10 \)
3. **Annual Payment**: \( PMT = 850 \)
The balance of a loan after \( n \) years can be calculated using the formula:
\[
B_n = P(1 + r)^n - PMT \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \( B_n \) is the balance of the loan after \( n \) years
- \( P \) is the principal amount
- \( r \) is the interest rate
- \( PMT \) is the payment amount
- \( n \) is the number of years
We want to find \( n \) such that \( B_n = 0 \):
\[
0 = P(1 + r)^n - PMT \times \frac{(1 + r)^n - 1}{r}
\]
Substituting the known values into the formula:
\[
0 = 2500(1 + 0.10)^n - 850 \times \frac{(1 + 0.10)^n - 1}{0.10}
\]
Simplifying this:
\[
0 = 2500(1.10)^n - 850 \times 10 \left((1.10)^n - 1\right)
\]
\[
0 = 2500(1.10)^n - 8500((1.10)^n - 1)
\]
\[
0 = 2500(1.10)^n - 8500(1.10)^n + 8500
\]
\[
0 = (-6000)(1.10)^n + 8500
\]
Reordering terms gives us:
\[
6000(1.10)^n = 8500
\]
\[
(1.10)^n = \frac{8500}{6000}
\]
\[
(1.10)^n = \frac{17}{12}
\]
### Taking the Logarithm of Both Sides:
\[
n \cdot \log(1.10) = \log\left(\frac{17}{12}\right)
\]
\[
n = \frac{\log\left(\frac{17}{12}\right)}{\log(1.10)}
\]
### Calculating \( n \):
Using a calculator:
1. Calculate \( \frac{17}{12} \):
\[
\frac{17}{12} \approx 1.4167
\]
2. Calculate logarithms:
\[
\log(1.10) \approx 0.0414
\]
\[
\log(1.4167) \approx 0.1520
\]
3. Finally, calculate \( n \):
\[
n \approx \frac{0.1520}{0.0414} \approx 3.67
\]
### Conclusion:
Since \( n \) represents the number of years, Joe should expect to take approximately **4 years** to fully pay off the loan when rounded up to the nearest whole year.
### Step-by-Step Calculation
1. **Loan Amount (Principal)**: \( P = 2500 \)
2. **Annual Interest Rate**: \( r = 10\% = 0.10 \)
3. **Annual Payment**: \( PMT = 850 \)
The balance of a loan after \( n \) years can be calculated using the formula:
\[
B_n = P(1 + r)^n - PMT \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \( B_n \) is the balance of the loan after \( n \) years
- \( P \) is the principal amount
- \( r \) is the interest rate
- \( PMT \) is the payment amount
- \( n \) is the number of years
We want to find \( n \) such that \( B_n = 0 \):
\[
0 = P(1 + r)^n - PMT \times \frac{(1 + r)^n - 1}{r}
\]
Substituting the known values into the formula:
\[
0 = 2500(1 + 0.10)^n - 850 \times \frac{(1 + 0.10)^n - 1}{0.10}
\]
Simplifying this:
\[
0 = 2500(1.10)^n - 850 \times 10 \left((1.10)^n - 1\right)
\]
\[
0 = 2500(1.10)^n - 8500((1.10)^n - 1)
\]
\[
0 = 2500(1.10)^n - 8500(1.10)^n + 8500
\]
\[
0 = (-6000)(1.10)^n + 8500
\]
Reordering terms gives us:
\[
6000(1.10)^n = 8500
\]
\[
(1.10)^n = \frac{8500}{6000}
\]
\[
(1.10)^n = \frac{17}{12}
\]
### Taking the Logarithm of Both Sides:
\[
n \cdot \log(1.10) = \log\left(\frac{17}{12}\right)
\]
\[
n = \frac{\log\left(\frac{17}{12}\right)}{\log(1.10)}
\]
### Calculating \( n \):
Using a calculator:
1. Calculate \( \frac{17}{12} \):
\[
\frac{17}{12} \approx 1.4167
\]
2. Calculate logarithms:
\[
\log(1.10) \approx 0.0414
\]
\[
\log(1.4167) \approx 0.1520
\]
3. Finally, calculate \( n \):
\[
n \approx \frac{0.1520}{0.0414} \approx 3.67
\]
### Conclusion:
Since \( n \) represents the number of years, Joe should expect to take approximately **4 years** to fully pay off the loan when rounded up to the nearest whole year.