Savings/loan scenarios with supplied models | Year 10

01 / LEARN

The idea, step by step

Write the order of events

Interest followed by an end deposit $d$ gives $B_{n + 1} = B_{n} (1 + i) + d$ . Interest followed by an end repayment $R$ gives $L_{n + 1} = L_{n} (1 + i) - R$ .

Beginning-of-period transactions need a different formula because they change the amount that earns interest. State the timing before substituting.

Each close becomes the next opening

Carry the closing amount into the following period. A repayment first covers interest, so the amount owed does not fall by the full payment.

Use the rounding convention supplied. A negative model balance near repayment means the scheduled final payment needs adjustment, not that a negative debt should be left unexplained.

A common trap

A 250-dollar repayment reduces the original principal by 250 dollars.

Instead: Part of the payment covers the period’s interest. Apply the events in the given order.

02 / WORKED EXAMPLES

See the thinking

Example 1

End deposits

A model starts at 1000 dollars, earns 5% annually, then receives 100 dollars at each year end. Find two closing balances.

$B_{1} = 1000 (1.05) + 100 = 1150$ .
$B_{2} = 1150 (1.05) + 100 = 1307.5$ .

1150 dollars; 1307.50 dollars.

Example 2

A loan rule

A loan starts at 1000 dollars. Monthly interest is 1%, followed by a 250-dollar repayment. Find the amount owed after 2 months.

$L_{1} = 1000 (1.01) - 250 = 760$ .
$L_{2} = 760 (1.01) - 250 = 517.6$ .

517.60 dollars.

Check your understanding

An opening saving balance is 500 dollars, the period rate is 2%, and the end deposit is 30 dollars. Find the close.

Show answer

540 dollars

$500 (1.02) + 30 = 540$ .

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Use $B_{n + 1} = 1.04 B_{n} + 50$ from $B_{0} = 200$ . Find $B_{1}$ .
2 marks
Continue that model to its second close.
2 marks
Use $L_{n + 1} = 1.02 L_{n} - 100$ from $L_{0} = 600$ . Find two closes.
3 marks
A 100-dollar deposit is made before 5% interest on 1000 dollars. Compare with an end deposit.
3 marks
A 2000-dollar opening loan balance gains 1% interest, then a 300-dollar payment is made. How much does the amount owed fall below its original value?
3 marks
Why does repayment before interest need a different formula?
3 marks

Read the worked solutions

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

Use $B_{n + 1} = 1.04 B_{n} + 50$ from $B_{0} = 200$ . Find $B_{1}$ .
2 marks
$1.04 (200) + 50 = 258$ .
258 dollars
Continue that model to its second close.
2 marks
$1.04 (258) + 50 = 318.32$ .
318.32 dollars
Use $L_{n + 1} = 1.02 L_{n} - 100$ from $L_{0} = 600$ . Find two closes.
3 marks
$600 (1.02) - 100 = 512$ ; $512 (1.02) - 100 = 422.24$ .
512 dollars; 422.24 dollars.
A 100-dollar deposit is made before 5% interest on 1000 dollars. Compare with an end deposit.
3 marks
Before: $(1000 + 100) (1.05) = 1155$ . End: $1000 (1.05) + 100 = 1150$ .
1155 dollars before; 1150 dollars at the end.
A 2000-dollar opening loan balance gains 1% interest, then a 300-dollar payment is made. How much does the amount owed fall below its original value?
3 marks
Interest is 20 dollars, so the reduction below 2000 is $300 - 20 = 280$ .
280 dollars
Why does repayment before interest need a different formula?
3 marks
That ordering gives $(L_{n} - R) (1 + i)$ , not $L_{n} (1 + i) - R$ .
It reduces the amount on which interest is charged.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Simple versus compound growth; compound interest; depreciation; repeated percentage change; savings/loan scenarios with supplied models; changing assumptions and checking results; spreadsheets/digital models.

Official curriculum source ↗

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