6. Ang is buying a new photocopy machine from Cannon Copiers for RM45,000. To finance his purchase, he trades his old photocopy machine for RM20,000 and takes a loan of RM25,000 from the bank at \( 6 \% \) compounded monthly. (a) If Ang intends to settle his debt in five years, how much must he repay the bank monthly? (b) Immediately after making 12 repayments, Ang wants to sell off his photocopy machine. If the bank requires him to settle his outstanding debt in one payment, how much must he repay to bank?

To solve the problem, we will use the formula for the monthly payment on an amortizing loan and the formula for the remaining balance after a certain number of payments. ### (a) Monthly Payment Calculation The formula for the monthly payment \( M \) on a loan is given by: \[ M = P \frac{r(1 + r)^n}{(1 + r)^n - 1} \] Where: - \( P \) = principal amount (loan amount) - \( r \) = monthly interest rate (annual rate / 12) - \( n \) = total number of payments (loan term in months) Given: - \( P = 25,000 \) RM - Annual interest rate = \( 6\% \) or \( 0.06 \) - Monthly interest rate \( r = \frac{0.06}{12} = 0.005 \) - Loan term = 5 years = \( 5 \times 12 = 60 \) months Substituting the values into the formula: \[ M = 25000 \frac{0.005(1 + 0.005)^{60}}{(1 + 0.005)^{60} - 1} \] Calculating \( (1 + 0.005)^{60} \): \[ (1 + 0.005)^{60} \approx 1.34885 \] Now substituting this back into the formula: \[ M = 25000 \frac{0.005 \times 1.34885}{1.34885 - 1} \] Calculating the denominator: \[ 1.34885 - 1 = 0.34885 \] Now substituting back: \[ M = 25000 \frac{0.00674425}{0.34885} \approx 25000 \times 0.01933 \approx 483.25 \] Thus, the monthly payment \( M \) is approximately **RM 483.25**. ### (b) Outstanding Debt After 12 Payments To find the outstanding balance after 12 payments, we can use the formula for the remaining balance \( B \) after \( k \) payments: \[ B = P \frac{(1 + r)^n - (1 + r)^k}{(1 + r)^n - 1} \] Where: - \( k \) = number of payments made (12 in this case) Substituting the known values: \[ B = 25000 \frac{(1 + 0.005)^{60} - (1 + 0.005)^{12}}{(1 + 0.005)^{60} - 1} \] We already calculated \( (1 + 0.005)^{60} \approx 1.34885 \). Now we need to calculate \( (1 + 0.005)^{12} \): \[ (1 + 0.005)^{12} \approx 1.061677812 \] Now substituting these values into the formula: \[ B = 25000 \frac{1.34885 - 1.061677812}{1.34885 - 1} \] Calculating the numerator: \[ 1.34885 - 1.061677812 \approx 0.287172188 \] And the denominator: \[ 1.34885 - 1 = 0.34885 \] Now substituting back: \[ B = 25000 \frac{0.287172188}{0.34885} \approx 25000 \times 0.8231 \approx 20577.75 \] Thus, the outstanding debt after 12 payments is approximately **RM 20,577.75**. ### Summary of Answers: (a) Monthly repayment: **RM 483.25** (b) Outstanding debt after 12 payments: **RM 20,577.75**