Savings/loan scenarios with supplied models | Year 10

01 / LEARN

Understand the idea.

Write the order of events

Write the sequence of events before calculating. If interest is applied before an end deposit $d$ , use $B_{n + 1} = B_{n} (1 + i) + d$ . For interest before an end repayment $R$ , use $L_{n + 1} = L_{n} (1 + i) - R$ .

A beginning deposit would instead earn interest immediately: $(B_{n} + d) (1 + i)$ . The two formulas differ because transaction timing changes the interest-bearing balance. Use the order specified in the supplied model.

Each close becomes the next opening

Each closing balance becomes the next opening balance. Carrying forward the updated amount is essential: the next period earns or owes interest on what is actually present after the preceding transaction.

Follow the stated rounding convention. Near the end of a loan, compare the amount owed after interest with the scheduled repayment; a negative closing debt indicates that the final payment should be reduced to the amount needed to clear it.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

End deposits

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

Worked solution

1
Calculate interest on the first opening balance.
$1000 (1.05) = 1050 dollars$
The year-end deposit has not yet been added, so it earns no first-year interest.
2
Add the first end deposit.
$B_{1} = 1050 + 100 = 1150 dollars$
The stated order puts the deposit after interest.
3
Carry the close forward and repeat.
$B_{2} = 1150 (1.05) + 100 = 1307.50 dollars$
The first close is the second opening, so both the earlier deposit and interest can now earn interest.

Answer

1150 dollars; 1307.50 dollars.

EXAMPLE 02

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.

Worked solution

1
Apply first-month interest, then the payment.
$L_{1} = 1000 (1.01) - 250 = 760 dollars$
The repayment occurs after $10$ dollars of monthly interest has been added.
2
Use the updated balance next month.
$760 (1.01) = 767.60 dollars$
The second interest charge is based on the remaining debt of $760$ dollars.
3
Subtract the second payment.
$L_{2} = 767.60 - 250 = 517.60 dollars$
After the second interest charge and repayment, this is the amount still owed.

Answer

517.60 dollars.

03 / YOUR TURN

Try it yourself.

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 & reasoning

540 dollars

Apply the $2 %$ interest before the end deposit: $500 (1.02) + 30 = 510 + 30 = 540$ dollars. The deposited $30$ does not earn interest in this period.

Keep practising

6 worksheet questions · 16 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Write the order of events

Each close becomes the next opening

See the method in action.

End deposits

Calculate interest on the first opening balance.

Add the first end deposit.

Carry the close forward and repeat.

A loan rule

Apply first-month interest, then the payment.

Use the updated balance next month.

Subtract the second payment.

Try it yourself.

Check your understanding

Keep practising