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Year 11 · Investing and borrowing · 20 min

Personal loans and comparing borrowing with saving

Compare borrowing now with saving first, using a clearly timed balance model.

01 / LEARN

The idea, step by step

Track a reducing balance

For a simplified loan with periodic rate i and repayment R at each period end, the update is Bn⁢e⁢x⁢t=B⁢(1+i)−R. Calculate interest first and then subtract the repayment because that is the supplied timing rule.

A repayment must exceed the period interest to reduce the balance that period. Keep full internal precision unless the contract model explicitly rounds each credit. A balance table makes the reduction visible.

Monthly interest is added before the repayment
MonthOpening (AUD)Interest (AUD)Repayment (AUD)Closing (AUD)
11000.0010.00200.00810.00
2810.008.10200.00618.10

Compare two decisions

Borrowing now gives earlier access but usually adds finance costs. Saving first delays the purchase and may earn interest. Compare the complete payment totals, dates and assumptions; a higher cash price at the later date can also matter.

The lowest modelled cost is not a complete personal decision. State what the model leaves out, such as uncertain income, changing prices or the importance of access. Interpret the answer in the hypothetical situation instead of prescribing a real loan.

A common trap

Subtract the repayment before applying interest even when payment is at period end.

Instead: Follow the stated cash-flow order. End-of-period repayment follows that period’s interest calculation.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

A AUD 1000 loan charges 1% monthly. A AUD 200 repayment is made at each month end. Find the balances after months 1 and 2.

  1. B1=1000×1.01−200=810.
  2. B2=810×1.01−200=618.10.
AUD 810.00; AUD 618.10
Example 2

Worked example

An item costs AUD 1500 now. Borrowing costs 18 monthly payments of AUD 90 and a AUD 30 fee. Saving first leads to a cash price of AUD 1560, paid entirely from savings. Compare purchase outlays.

  1. Borrowing=18×90+30=1650.
  2. Delayed cash outlay=1560.
  3. Difference=90;this does not value earlier use or other omitted factors.
Borrowing costs AUD 1650; saving-first purchase costs AUD 1560, AUD 90 less.

Check your understanding

A AUD 2000 balance incurs 1% interest and a AUD 150 end-of-month payment. Find the new balance.

Show answer

AUD 1,870.00

2000×1.01−150=1870.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Read the questions online
  1. Loan balance AUD 5000, monthly rate 0.6%, month-end payment AUD 250. Find the new balance.
    2 marks
  2. A AUD 10000 loan incurs 0.5% monthly interest. What is the interest in its first month?
    2 marks
  3. A AUD 10000 balance has 0.5% monthly interest and only AUD 40 repaid at month end. Does it fall? Find the new balance.
    2 marks
  4. Find the balance after 3 monthly periods on AUD 2000 at 1% per month with AUD 300 month-end payments.
    2 marks
  5. Cash price AUD 2200. A loan requires 24 payments of AUD 102 plus AUD 50 fee. Find the extra cost.
    2 marks
  6. A comparison assumes a constant loan rate and unchanged future cash price. Give two reasons its actual outcome could differ.
    2 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Loan balance AUD 5000, monthly rate 0.6%, month-end payment AUD 250. Find the new balance.
    2 marks

    5000×1.006−250=4780.

    AUD 4,780.00
  2. A AUD 10000 loan incurs 0.5% monthly interest. What is the interest in its first month?
    2 marks

    10000×0.005=50.

    AUD 50.00
  3. A AUD 10000 balance has 0.5% monthly interest and only AUD 40 repaid at month end. Does it fall? Find the new balance.
    2 marks

    Interest=50.

    10000+50−40=10010.

    No; it rises to AUD 10010.
  4. Find the balance after 3 monthly periods on AUD 2000 at 1% per month with AUD 300 month-end payments.
    2 marks

    B1=1720.

    B2=1437.20.

    B3=1437.20×1.01−300=1151.572.

    AUD 1,151.57
  5. Cash price AUD 2200. A loan requires 24 payments of AUD 102 plus AUD 50 fee. Find the extra cost.
    2 marks

    24×102+50=2498.

    2498−2200=298.

    AUD 298.00
  6. A comparison assumes a constant loan rate and unchanged future cash price. Give two reasons its actual outcome could differ.
    2 marks

    These variables are assumptions rather than guaranteed facts.

    The loan rate could change; the future cash price could change.
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: investing and borrowing. Supplied educational models cover interest, investment comparisons, shares and credit; they are not recommendations of real products or statements of personal tax liability.

Official curriculum source ↗
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