Mathematics Advanced • Year 12 • Module 7 • Lesson 15

Extra Repayments, Offset Accounts and Redraw

Build fluency computing time and interest saved by extra repayments and offset balances.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the formula for the time-to-clear with a higher repayment M_new:

n_new = ____________________

Q1.2 An offset account holds $50,000 against a $400,000 loan. State the effective loan balance.

Effective balance = ____________________

Q1.3 In one sentence, state the mathematical difference (if any) between an offset and a redraw — and the practical difference.

Stuck? Revisit lesson § Formula Reference and § Offset vs Redraw.

2. Worked example — Time saved by paying $2,500 instead of $2,148

Follow every line. Each step has a short reason.

Problem. P = $400,000 at 5% p.a. compounded monthly. The minimum monthly repayment is M_old = $2,148 over 30 years (n_old = 360). The borrower pays M_new = $2,500 instead. Find n_new and the dollars of interest saved.

Step 1 — Monthly rate.

r = 0.05 ÷ 12 = 0.004167

Step 2 — Compute Pr ÷ M_new.

Pr ÷ M_new = 400,000 × 0.004167 ÷ 2,500 = 1,666.67 ÷ 2,500 = 0.6667

Step 3 — Solve for n_new.

n_new = −ln(1 − 0.6667) ÷ ln(1.004167)

= −ln(0.3333) ÷ ln(1.004167)

= 1.0986 ÷ 0.004158

≈ 264.2 months (≈ 22.0 years)

Step 4 — Interest saved.

Old total repaid = 2,148 × 360 = $773,280 → Old interest ≈ $373,280

New total repaid = 2,500 × 264 = $660,000 → New interest ≈ $260,000

Interest saved ≈ $373,280 − $260,000 = $113,280

Conclusion. An extra $352/month pays the loan off ~8 years sooner and saves about $113,000 in interest.

3. Faded example — fill in the missing steps

A $500,000 loan at 5.2% p.a. compounded monthly. Minimum M = $2,748/month. The borrower pays $3,200/month instead. Find n_new. 4 marks

Step 1 — Monthly rate:

r = 0.052 ÷ 12 = ______________

Step 2 — Compute Pr ÷ M_new.

Pr ÷ M_new = 500,000 × ______________ ÷ 3,200 = ______________

Step 3 — Plug into n_new = −ln(1 − Pr ÷ M_new) ÷ ln(1 + r).

n_new = −ln(1 − ______________) ÷ ln(______________)

= −ln(______________) ÷ ______________

= ______________ ÷ ______________ ≈ ______________ months

Conclusion. ≈ ______________ years (vs 30 at the minimum). Saves ______________ years.

Stuck? Revisit lesson § Worked Example — Try It Now.

4. Graduated practice

Show the substitution and final value (to nearest cent or nearest month unless stated).

Foundation — direct substitution (4 questions)

Q	Scenario	Working & answer
4.1 1	A $400,000 loan with $50,000 in offset. State the effective balance.
4.2 1	Old M = $2,148, n = 360. State old total repaid.
4.3 1	New M = $2,500, n = 264. State new total repaid.
4.4 1	State the interest saved using the answers to 4.2 and 4.3.

Standard — typical HSC difficulty (6 questions)

Show at least one substitution line and one evaluation line.

4.5 $300,000 loan at 6% p.a. monthly, minimum M = $1,791. Borrower pays $2,000 instead. Find n_new (months and years). 2 marks

4.6 $250,000 loan at 5% p.a. monthly, minimum M = $1,342. Borrower pays $1,700/month. Find n_new. 2 marks

4.7 A $400,000 loan has $80,000 in an offset account at r = 0.004167 per month. Compute the monthly interest charged (a) with offset, (b) without offset, and (c) the monthly saving. 2 marks

4.8 A $200,000 loan at 4.8% p.a. monthly with M = $1,400. Find the interest paid in Month 1 and the principal reduction. 2 marks

4.9 A borrower puts $30,000 in offset against a $350,000 loan at 4.8% p.a. monthly. State the annual interest saving (12 × monthly saving). 2 marks

4.10 An old loan satisfies M_old = $2,500, n_old = 360 (30 yr). The borrower's new schedule satisfies M_new = $3,000, n_new = 210 (17.5 yr). Compute the total interest saved. 2 marks

Extension — combine concepts (2 questions)

4.11 A $400,000 loan at 5% p.a. monthly. The borrower has $60,000 in offset. Compare (a) "offset for 5 years then withdraw" against (b) "no offset". Estimate how much interest is saved over those 5 years by keeping the $60,000 in offset. (Assume the savings each month ≈ 60,000 × r.) 3 marks

4.12 A borrower with a $400,000 / 5% / 30-yr loan considers either (i) putting $20,000 spare cash into an offset, or (ii) depositing it into a savings account paying 4% p.a. taxable at 32.5%. Compare the after-tax annual benefit of each option. State which wins and by how much. 3 marks

Stuck on 4.12? Offset benefit = 20,000 × 0.05 = $1,000/yr tax-free. Savings net = 20,000 × 0.04 × (1 − 0.325).

5. Self-check the easy 3

Tick the first three once you have checked the method works.

For 4.1 effective balance = 400,000 − 50,000 = $350,000.

For 4.2 old total = 2,148 × 360 = $773,280.

For 4.3 new total = 2,500 × 264 = $660,000.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Time-to-clear formula

n_new = −ln(1 − Pr ÷ M_new) ÷ ln(1 + r).

Q1.2 — Effective balance with $50,000 offset against $400,000 loan

Effective balance = 400,000 − 50,000 = $350,000. Interest is charged on $350,000 only.

Q1.3 — Offset vs redraw

Mathematically: identical — both subtract from the balance on which interest is calculated. Practically: offset money is instantly accessible (ATM, card) and the offset account is a transaction account; redraw money requires a formal withdrawal request from the loan account.

Q3 — Faded example: $500,000 at 5.2%, M_new = $3,200

r = 0.052 ÷ 12 ≈ 0.004333. Pr ÷ M_new = 500,000 × 0.004333 ÷ 3,200 = 2,166.67 ÷ 3,200 ≈ 0.6771.
n_new = −ln(1 − 0.6771) ÷ ln(1.004333) = −ln(0.3229) ÷ 0.004324 = 1.1306 ÷ 0.004324 ≈ 261.5 months (≈ 21.8 years). Saves ≈ 8.2 years vs the 30-year minimum.

Q4.1 — Effective balance with $50,000 offset on $400,000

Effective = 400,000 − 50,000 = $350,000.

Q4.2 — Old total repaid at M = $2,148, n = 360

Total = 2,148 × 360 = $773,280.

Q4.3 — New total at M = $2,500, n = 264

Total = 2,500 × 264 = $660,000.

Q4.4 — Interest saved

Both schedules pay off the same $400,000, so the difference between total repayments is interest saved. Saving = 773,280 − 660,000 = $113,280.

Q4.5 — $300,000 at 6%, minimum $1,791, pays $2,000

r = 0.005. Pr ÷ M_new = 300,000 × 0.005 ÷ 2,000 = 0.75. n_new = −ln(0.25) ÷ ln(1.005) = 1.3863 ÷ 0.004988 ≈ 277.9 months (≈ 23.2 years). Saves ≈ 6.8 years vs 30-year minimum.

Q4.6 — $250,000 at 5%, M_new = $1,700

r ≈ 0.004167. Pr ÷ M_new = 250,000 × 0.004167 ÷ 1,700 = 1,041.67 ÷ 1,700 ≈ 0.6127. n_new = −ln(0.3873) ÷ ln(1.004167) = 0.9487 ÷ 0.004158 ≈ 228.2 months (≈ 19.0 years).

Q4.7 — Offset $80,000 on $400,000 at r = 0.004167

(a) With offset: interest = 320,000 × 0.004167 = $1,333.33/month.
(b) Without offset: interest = 400,000 × 0.004167 = $1,666.67/month.
(c) Saving = 1,666.67 − 1,333.33 = $333.34/month — exactly 80,000 × 0.004167.

Q4.8 — $200,000 at 4.8% monthly, M = $1,400

r = 0.004. I₁ = 200,000 × 0.004 = $800.00. P₁ = 1,400 − 800 = $600.00.

Q4.9 — Offset $30,000 on $350,000 at 4.8% p.a.

r = 0.004 per month. Monthly saving = 30,000 × 0.004 = $120. Annual saving = 120 × 12 = $1,440 per year (tax-free).

Q4.10 — Interest saved going from (M_old = 2,500, n_old = 360) to (M_new = 3,000, n_new = 210)

Old total = 2,500 × 360 = $900,000. New total = 3,000 × 210 = $630,000. Saving = 900,000 − 630,000 = $270,000.

Q4.11 — $60,000 in offset for 5 years on $400,000 / 5%

r = 0.004167 per month. Saving ≈ 60,000 × 0.004167 = $250.00/month. Over 5 years (60 months) ≈ $15,000 of interest avoided. (Strictly slightly more, because the balance also reduces faster, saving extra interest compounded — the simple estimate is a lower bound.)

Q4.12 — Offset vs savings account on $20,000

Offset on a 5% loan: tax-free saving = 20,000 × 0.05 = $1,000/yr.
Savings 4% taxable at 32.5%: after-tax interest = 20,000 × 0.04 × (1 − 0.325) = 20,000 × 0.04 × 0.675 = 20,000 × 0.027 = $540/yr.
Offset wins by $460/yr on $20,000 — equivalent to earning an effective tax-free return of 5%.