Mathematics Advanced • Year 12 • Module 7 • Lesson 14
Calculating Loan Repayments
Build fluency with the loan-repayment formula M = Pr ÷ [1 − (1+r)−n] and its transpositions for P and n.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the loan-repayment formula (M is the regular repayment, P is the principal, r is the rate per period, n is the number of periods):
M = ____________________
Q1.2 A loan of $400,000 is taken at 5% p.a. compounded monthly over 30 years. State r per month and n.
r = ____________ n = ____________
Q1.3 Write the transposed formula that gives P (the maximum borrowable) from a known repayment M, rate r and term n.
2. Worked example, Minimum monthly repayment
Follow every line. Each step has a short reason.
Problem. A home loan of P = $400,000 at 5% p.a. compounded monthly over 30 years. Find the minimum monthly repayment M.
Step 1, Convert to periodic rate and total periods.
r = 0.05 ÷ 12 = 0.004167 (per month)
n = 30 × 12 = 360 months
Step 2, Compute the numerator P × r.
P × r = 400,000 × 0.004167 = $1,666.67
Step 3, Compute (1 + r)−n and the denominator.
(1.004167)−360 = 0.22384
1 − 0.22384 = 0.77616
Step 4, Divide.
M = 1,666.67 ÷ 0.77616 = $2,147.29
Conclusion. The minimum monthly repayment is $2,147.29. Anything less and the loan never clears; anything more shortens the term and reduces total interest.
3. Faded example, fill in the missing steps
A $250,000 loan at 6% p.a. compounded monthly over 20 years. Find M. 4 marks
Step 1, Convert.
r = 0.06 ÷ 12 = ______________ n = 20 × 12 = ______________
Step 2, Numerator.
P × r = 250,000 × ______________ = $______________
Step 3, Denominator.
(1.005)−240 = ______________
1 − ______________ = ______________
Step 4, Divide.
M = ______________ ÷ ______________ = $______________
Conclusion. Minimum monthly repayment ≈ $______________.
4. Graduated practice
Show the substitution and the final value (to nearest cent unless stated). Use monthly compounding throughout unless stated.
Foundation, direct substitution (4 questions)
| Q | Scenario | Working & answer |
|---|---|---|
| 4.1 1 | 5% p.a. compounded monthly over 30 years: state r per month and n. | |
| 4.2 1 | 4.8% p.a. compounded monthly over 25 years: state r and n. | |
| 4.3 1 | Write the formula for the maximum borrowable P given M, r, n. | |
| 4.4 1 | Write the formula for n in terms of P, M and r (use natural logs). |
Standard, typical HSC difficulty (6 questions)
Show at least one substitution line and one evaluation line.
4.5 Find M for P = $300,000, r = 0.005/month, n = 240. 2 marks
4.6 Find M for P = $450,000, 5% p.a. monthly, term 30 years. 2 marks
4.7 Find M for P = $500,000, 4.8% p.a. monthly, term 30 years. 2 marks
4.8 Find the maximum P that can be borrowed if M = $1,500/month, r = 0.004/month, n = 180 (15 years). 2 marks
4.9 Find n (months) for P = $20,000, M = $400, r = 0.006/month. 2 marks
4.10 A $250,000 loan at 6% p.a. monthly over 25 years. Find M and the total amount repaid over the full term. 2 marks
Extension, combine concepts (2 questions)
4.11 A borrower can afford M = $2,800/month at 4.8% p.a. monthly. Find the maximum P for a 25-year loan and for a 30-year loan. State the dollar difference in borrowing capacity. 3 marks
4.12 A $450,000 home loan at 5% p.a. compounded monthly. Find M for the 30-year term and the 20-year term. Then compute the total interest for each, and the dollar interest saved by choosing the 20-year option. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked the method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, Loan-repayment formula
M = Pr ÷ [1 − (1 + r)−n].
Q1.2, Rate and periods for 5% over 30 years monthly
r = 0.05 ÷ 12 ≈ 0.004167 per month. n = 30 × 12 = 360 months.
Q1.3, Maximum P
P = M × [1 − (1 + r)−n] ÷ r.
Q3, Faded example: $250,000 at 6% over 20 years
r = 0.005, n = 240. P × r = 250,000 × 0.005 = $1,250.00. (1.005)−240 ≈ 0.30212. Denominator = 1 − 0.30212 = 0.69788. M = 1,250 ÷ 0.69788 ≈ $1,790.79 per month.
Q4.1-5% monthly over 30 years
r = 0.05 ÷ 12 ≈ 0.004167; n = 360.
Q4.2-4.8% monthly over 25 years
r = 0.048 ÷ 12 = 0.004; n = 25 × 12 = 300.
Q4.3, P formula
P = M × [1 − (1 + r)−n] ÷ r.
Q4.4, n formula
n = −ln(1 − Pr ÷ M) ÷ ln(1 + r).
Q4.5, P = 300,000, r = 0.005, n = 240
Pr = $1,500. (1.005)−240 ≈ 0.30212. Denominator = 0.69788. M = 1,500 ÷ 0.69788 ≈ $2,148.95/month.
Q4.6, P = $450,000, 5% monthly, 30 years
r = 0.004167, n = 360. Pr = $1,875. (1.004167)−360 ≈ 0.22384. Denom = 0.77616. M = 1,875 ÷ 0.77616 ≈ $2,415.70/month.
Q4.7, P = $500,000, 4.8% monthly, 30 years
r = 0.004, n = 360. Pr = $2,000. (1.004)−360 ≈ 0.23769. Denom = 0.76231. M = 2,000 ÷ 0.76231 ≈ $2,623.55/month.
Q4.8, Max P for M = $1,500, r = 0.004, n = 180
(1.004)−180 ≈ 0.48774. 1 − 0.48774 = 0.51226. Annuity factor = 0.51226 ÷ 0.004 = 128.065. P = 1,500 × 128.065 ≈ $192,098.
Q4.9, n for P = 20,000, M = 400, r = 0.006
Pr ÷ M = 20,000 × 0.006 ÷ 400 = 0.30. n = −ln(1 − 0.30) ÷ ln(1.006) = −ln(0.70) ÷ ln(1.006) = 0.35667 ÷ 0.005982 ≈ 59.6 months (about 5 years).
Q4.10, $250,000 at 6% monthly over 25 years
r = 0.005, n = 300. Pr = $1,250. (1.005)−300 ≈ 0.22397. Denom = 0.77603. M = 1,250 ÷ 0.77603 ≈ $1,610.75/month. Total repaid = 1,610.75 × 300 = $483,225.
Q4.11, Borrowing capacity at $2,800/month, 4.8% p.a.
r = 0.004. 25-year term: n = 300; (1.004)−300 ≈ 0.30207; annuity factor = (1 − 0.30207)/0.004 = 174.483. P = 2,800 × 174.483 ≈ $488,552.
30-year term: n = 360; (1.004)−360 ≈ 0.23769; annuity factor = (1 − 0.23769)/0.004 = 190.578. P = 2,800 × 190.578 ≈ $533,618.
Extra borrowing capacity for going from 25 to 30 years ≈ $45,066 at the cost of paying for 5 extra years.
Q4.12, $450,000 at 5%, 30 vs 20 years
30-year: M ≈ $2,415.70 (from 4.6). Total repaid = 2,415.70 × 360 = $869,652. Interest = 869,652 − 450,000 = $419,652.
20-year: r = 0.004167, n = 240. (1.004167)−240 ≈ 0.36879. Denom = 0.63121. M = 1,875 ÷ 0.63121 ≈ $2,970.55. Total repaid = 2,970.55 × 240 = $712,932. Interest = $262,932.
Interest saved by 20-year option = 419,652 − 262,932 ≈ $156,720.