Reducing-balance loans, amortisation tables and total interest for HSC Maths Standard 2

Syllabus dot point

“Use recurrence relations and amortisation tables to calculate loan repayments, outstanding balances and the total interest paid on a reducing-balance loan”

HSCMaths Standard 2Year 12: Financial Mathematics14 min readUpdated 20 Jun 2026

Quick answer

A reducing-balance loan satisfies $B_n = B_{n-1}(1 + r) - M$ with closed form $B_n = P(1 + r)^n - M \cdot \frac{(1 + r)^n - 1}{r}$ ; build an amortisation table by computing interest on each opening balance, subtracting that from the payment to get principal repaid, and rolling forward; total interest is $nM - P$ .

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA wants you to do five things with a reducing-balance loan. Model it with a recurrence relation, which is a rule that gives each new balance from the one before. Build an amortisation table by hand for a few periods. Find the outstanding balance after $n$ payments using the closed-form formula. Split a payment into its interest and principal parts. You should also be able to find the total interest paid over the life of the loan.

The answer

How a reducing-balance loan works

A loan of $P$ dollars is repaid by equal payments $M$ at the end of each compounding period. Each period two things happen in order:

Interest is added to the opening balance at the per-period rate $r$ .
The payment $M$ is subtracted.

The closing balance is what is still owed, and it becomes next period's opening balance. The balance shrinks a little each period, so the interest charged on it shrinks too. The payment $M$ stays the same, so a steadily larger slice of it goes to paying down the principal (the amount you still owe). That is why it is called a reducing-balance loan: the balance the interest is charged on reduces over time. The chart below shows a real loan being paid off.

The closing balance is what is still owed. Let $B_n$ be the balance just after the $n$ th payment.

B_n = B_{n-1}(1 + r) - M.

This is the recurrence model, with $B_0 = P$ .

Key fact

Each period, interest is charged on the opening balance at the per-period rate, then the payment is subtracted: $B_n = B_{n-1}(1 + r) - M$ , with $B_0 = P$ . The interest portion of a payment is $I_k = r \, B_{k-1}$ and the principal portion is $M - I_k$ . Because the balance falls each period, the interest portion shrinks and the principal portion grows, so early payments are mostly interest and late payments are mostly principal.

Building the amortisation table, row by row

An amortisation table tracks each period in five columns: opening balance, interest, payment, principal repaid, and closing balance. The safest way to build one is one row at a time, carrying each closing balance down to be the next opening balance. Take a $20000 car loan at $0.5\%$ per month (a $6\%$ p.a. rate) with $400 monthly repayments, and build the first three months stage by stage.

Stage 1, the first month. The opening balance is the full loan, $20000. Interest is $0.005 \times 20000 = 100.00$ , i.e. $100.00. The payment is $400, so the principal repaid is $400 - 100 = 300.00$ , i.e. $300.00, and the closing balance is $20000 + 100 - 400 = 19700.00$ , i.e. $19700.00.

Month	Opening	Interest ( $\times 0.005$ )	Payment	Principal repaid	Closing
$1$	$20000.00$	$100.00$	$400.00$	$300.00$	$19700.00$

Stage 2, carry the closing balance down. Month $2$ opens at last month's close, $19700.00. Interest is $0.005 \times 19700 = 98.50$ , i.e. $98.50 (already lower, because the balance is lower). Principal repaid is $400 - 98.50 = 301.50$ , i.e. $301.50, slightly more than last month, and the closing balance is $19700 + 98.50 - 400 = 19398.50$ , i.e. $19398.50.

Month	Opening	Interest ( $\times 0.005$ )	Payment	Principal repaid	Closing
$1$	$20000.00$	$100.00$	$400.00$	$300.00$	$19700.00$
$2$	$19700.00$	$98.50$	$400.00$	$301.50$	$19398.50$

Stage 3, repeat the pattern. Month $3$ opens at $19398.50. Interest is $0.005 \times 19398.50 = 96.99$ , i.e. $96.99, principal repaid is $400 - 96.99 = 303.01$ , i.e. $303.01, and the closing balance is $19095.49.

Month	Opening	Interest ( $\times 0.005$ )	Payment	Principal repaid	Closing
$1$	$20000.00$	$100.00$	$400.00$	$300.00$	$19700.00$
$2$	$19700.00$	$98.50$	$400.00$	$301.50$	$19398.50$
$3$	$19398.50$	$96.99$	$400.00$	$303.01$	$19095.49$

Stage 4, read the trend. Watch the two middle columns across the three rows: interest falls ( $100.00 \to 98.50 \to 96.99$ ) while principal repaid rises ( $300.00 \to 301.50 \to 303.01$ ). The payment never changes, but its split shifts steadily from interest towards principal. Continuing this table to the end, the loan is fully repaid at month $58$ (the final payment is a little under $400).

Month	Opening	Interest ( $\times 0.005$ )	Payment	Principal repaid	Closing
$1$	$20000.00$	$100.00$	$400.00$	$300.00$	$19700.00$
$2$	$19700.00$	$98.50$	$400.00$	$301.50$	$19398.50$
$3$	$19398.50$	$96.99$	$400.00$	$303.01$	$19095.49$
$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$
$57$	$667.58$	$3.34$	$400.00$	$396.66$	$270.92$
$58$	$270.92$	$1.35$	$272.27$	$270.92$	$0.00$

Each row follows the same three rules: interest is the opening balance times the per-period rate; principal repaid is payment minus interest; closing balance is opening plus interest minus payment. The final payment is only $272.27, because after month $57$ just $270.92 of principal (plus a final $1.35 of interest) remains. Over the whole loan the borrower pays $23072.27, so the total interest is $23072.27 - 20000 = 3072.27$ , i.e. $3072.27.

Closed-form for the balance

Iterating the recurrence using a geometric series gives

B_n = P(1 + r)^n - M \cdot \frac{(1 + r)^n - 1}{r}.

The first term, $P(1 + r)^n$ , is what the loan would grow to if no payments were ever made. The second term is the future value of the payments made so far. Future value means what those payments are worth by the time of the $n$ th payment, once their own interest is counted. This is the same sum used for superannuation. The difference between the two terms is what is still owed. Use this closed form as a shortcut for "the balance after the $n$ th payment" when $n$ is too big for a table to be practical.

The recurrence and the formula are two views of the same loan

The table (recurrence) and the closed-form formula always agree, because the formula is just the recurrence iterated $n$ times and summed as a geometric series. Use whichever the question points to:

Recurrence / table when asked to "use an amortisation table", to show the first few months, or to find a principal-versus-interest split for a particular payment.
Closed form when asked for the balance after many payments (say $B_{12}$ or $B_{60}$ ) directly, without building every row.

If both appear, run the table for the first few rows and use the closed form to jump to the period the question wants, then sanity-check that they meet.

Interest vs principal split

For the $k$ th payment:

Interest portion: $I_k = r \cdot B_{k-1}$ (interest on the previous closing balance)
Principal portion: $P_k = M - I_k$

Early in the loan, most of each payment goes to interest. Near the end, almost all goes to principal. This is why paying extra early saves far more interest than paying the same amount later. An extra dollar paid in month $1$ avoids interest for the whole rest of the loan. The same dollar paid near the end avoids almost none, because there is little time left.

Total interest paid

If the loan is repaid by $n$ payments of $M$ :

\text{total interest} = n M - P.

That is, total cash paid out minus the original loan amount. (If the final payment is a smaller partial payment, add up the actual payments instead of using $nM$ .) Total interest is a favourite final part because it turns the abstract schedule into a single dollar figure that shows the real cost of borrowing.

Interest-only versus reducing-balance

A reducing-balance loan pays off both interest and some principal each period, so the balance falls. An interest-only loan pays only the interest each period, $M = r P$ , so the principal never changes and the balance stays flat at $P$ for the interest-only term. Two consequences worth knowing:

On an interest-only loan, every payment is identical ( $rP$ ) and no principal is ever repaid, so the total interest over $n$ periods is simply $n \times rP$ , and you still owe the whole $P$ at the end.
A reducing-balance payment must be larger than the interest-only payment, because it has to cover the interest and chip away at the principal. If a payment equals exactly the interest ( $M = rP$ ), the balance never moves; if it is less than the interest ( $M < rP$ ), the balance actually grows and the loan never finishes. NESA sometimes tests this "does the loan ever finish?" condition.

Australian mortgage context

A typical Sydney mortgage in 2025: $700000 borrowed at around $6.2\%$ per annum (typical major-bank variable rate, RBA cash rate $\sim 4.35\%$ plus bank margin). Over $25$ years monthly:

$r = 0.062/12 \approx 0.005167$ , $n = 300$ .

The repayment formula (also on the reference sheet) gives $M \approx 4596$ , i.e. about $4596 per month. Total paid over $25$ years is about $1.38 million. Total interest is about $678800, almost as much as the $700000 first borrowed. This is the headline fact about long mortgages: over a few decades you can pay close to double what you borrowed.

How exam questions ask about reducing-balance loans

Each wording maps to one of the tools above:

"Complete the next row of the table" or "find the balance after the 3rd repayment." A recurrence / amortisation-table question: interest on opening balance, principal = payment minus interest, roll forward.
"Find the balance owing after the 12th (or $n$ th) payment." Use the closed-form $B_n = P(1+r)^n - M\frac{(1+r)^n - 1}{r}$ .
"How much of the 5th payment is interest / pays off the loan?" A split question: $I_k = rB_{k-1}$ , principal $= M - I_k$ . Find $B_{k-1}$ first (table or formula).
"Find the monthly repayment / instalment." Use the repayment formula $M = \dfrac{Pr}{1 - (1+r)^{-n}}$ .
"Find the total interest paid over the life of the loan." Total interest $= nM - P$ (or sum the actual payments if the last is partial).
"After how many months is the loan repaid?" Set $B_n = 0$ and solve for $n$ with logs; round up to the next whole period (the final payment is partial).
"Why does so little of an early payment reduce the principal?" An explanation question: interest is charged on the large early balance, so most of $M$ is consumed by interest.

Amortisation by hand, loan term, and Australian mortgage repayment

Build an amortisation schedule for a $50000 loan at $0.5\%$ per month with $1100 payments, use the closed-form balance formula to find when the loan is paid off, then calculate monthly repayments on a $650000 mortgage at $6.0\%$ p.a. over $30$ years.

Pay off a $50000 loan by hand and by formula

Compute the first three months of the amortisation schedule: Each month, interest is charged on the current (opening) balance before the repayment reduces it. Because the balance falls slightly each month, the interest charge also falls, so a larger slice of the fixed repayment goes towards principal. Tracking this for the first few months reveals the pattern.

Month	Opening balance	Interest ( $\times 0.005$ )	Payment	Principal repaid	Closing balance
1	$50000$	$250.00$	$1100$	$850.00$	$49150.00$
2	$49150$	$245.75$	$1100$	$854.25$	$48295.75$
3	$48295.75$	$241.48$	$1100$	$858.52$	$47437.23$

Notice the principal portion grows each month (from $850$ to $854.25$ to $858.52$ ) and the interest portion shrinks correspondingly.

Apply the closed-form balance formula at $n = 60$ : Repeating the table for 60 rows by hand is impractical. The closed-form formula computes the balance after $n$ periods directly. We substitute $n = 60$ to check whether the loan survives five years.

(1.005)^{60} \approx 1.34885

B_{60} = 50000 \times 1.34885 - 1100 \times \frac{0.34885}{0.005}

= 67442.50 - 1100 \times 69.770 = 67442.50 - 76747.00 = -9304.50

A negative balance means the repayments overshoot zero before $60$ months have passed, so the loan is fully repaid before the five-year mark.

Solve for the exact number of months: Setting the balance formula equal to zero and solving for $n$ gives the precise payoff month. We rearrange to isolate $(1.005)^n$ on one side, then take logarithms.

50000(1.005)^n = 1100 \times \frac{(1.005)^n - 1}{0.005}

50000(1.005)^n = 220000\!\left((1.005)^n - 1\right)

(1.005)^n (50000 - 220000) = -220000

(1.005)^n = \frac{220000}{170000} = 1.2941

n = \frac{\log 1.2941}{\log 1.005} = \frac{0.1119}{0.00217} \approx 51.6 \text{ months}

The loan is paid off in $52$ months (the final payment is slightly less than $1100).

Repayment and total interest on an Australian mortgage

For the mortgage we use the repayment formula directly. The monthly rate is $6.0\% \div 12 = 0.5\%$ , so $r = 0.005$ and $n = 360$ for a 30-year term. Computing $(1.005)^{-360}$ first keeps the arithmetic manageable.

$r = 0.005$ , $n = 360$ .

(1.005)^{-360} \approx 0.16604, \quad \text{so } 1 - (1.005)^{-360} \approx 0.83396

M = \frac{Pr}{1-(1+r)^{-n}} = \frac{650000 \times 0.005}{0.83396} = \frac{3250}{0.83396} \approx 3897.08 \text{ per month}

i.e. about $3897.08 per month.

Find the total interest paid: Multiplying the monthly repayment by the number of payments gives the total amount paid to the bank. Subtracting the original loan reveals how much of that total was interest, not principal.

\text{Total paid} = 360 \times 3897.08 \approx 1402948.80

\text{Total interest} = 1402948.80 - 650000 = 752948.80

In dollars, the total paid is about $1402948.80 and the total interest about $752948.80.

Final answer: The $50000 loan is paid off in $52$ months. The $650000 mortgage at $6.0\%$ p.a. over $30$ years requires a monthly repayment of approximately $3897.08, with total interest of $752948.80.

Edge case: interest-only versus reducing-balance, same loan

Take the $650000 at $6.0\%$ p.a. ( $r = 0.005$ per month) from the worked example. An interest-only payment would be $M = rP = 0.005 \times 650000 = 3250$ , i.e. $3250 per month. Compare:

Interest-only: pay $3250 a month, the balance stays at $650000 forever, and after, say, $5$ years ( $60$ months) you have paid $60 \times 3250 = 195000$ , i.e. $195000 in interest and still owe the full $650000.
Reducing-balance: pay $3897.08 a month (about $647 more), and the loan is gone in $30$ years with the principal fully cleared.

The extra $647 a month is exactly what buys down the principal. This is why interest-only periods lower the payment in the short term but cost more overall: nothing is being repaid.

Exam technique

Lay the amortisation table out with the five standard columns (opening, interest, payment, principal repaid, closing) and label them; markers award method marks for the correct structure even if one number slips. Compute interest on the opening balance, not the closing one. Carry full cents down each row, and round only the final reported figure. For "balance after the $n$ th payment" use the closed form and show the substitution; for "complete the table" use the recurrence. State total interest as $nM - P$ with a sentence saying it is total payments minus the amount borrowed. If asked whether a loan ever finishes, compare the payment $M$ with the first interest charge $rP$ : if $M \le rP$ it never does.

Common traps

Using simple interest: Loan interest is added to the balance each period, then the next interest is calculated on the new balance. Compound, not simple.
Computing interest on the closing balance: Interest is charged on the opening balance for the period, before the payment is subtracted.
Forgetting to convert the rate: Monthly compounding requires the monthly rate $R/12$ . Using the annual rate gives a wildly wrong answer.
Wrong sign on the closed form: The balance formula subtracts the payment term. Sign errors mean you compute the wrong direction.
Treating principal repaid as constant: Early payments are mostly interest; the principal repaid grows each period, so the balance does not fall in equal steps.
Ignoring the no-completion case: If $M \le P r$ , the loan never finishes because the payment does not even cover the interest. A worded problem may test this.

Note

A reducing-balance loan is the normal kind of loan you use to buy a car or a house. You pay the same amount every month, but that money does two jobs at once. Part of it pays the rent on the money you still owe (the interest), and the rest pays back a bit of what you borrowed. Each month you owe a little less, so the interest part gets smaller and the part that pays off the loan gets bigger. That is why paying extra early helps so much: it cuts down the amount you are charged interest on for years to come.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style5 marks

A car loan of $25000 at

7.2\%

per annum compounded monthly is repaid by monthly instalments of $498.21. Find the balance owing immediately after the 12th payment. Round to the nearest cent.

Show worked answer →

Per-period rate: $r = \frac{0.072}{12} = 0.006$ . Monthly payment $M = 498.21$ .

The balance after $n$ payments is

$B_n = P(1 + r)^n - M \cdot \frac{(1 + r)^n - 1}{r}$ .

$(1.006)^{12} \approx 1.07442$ .

$B_{12} = 25000 \times 1.07442 - 498.21 \times \frac{0.07442}{0.006}$

$= 26860.50 - 498.21 \times 12.4033$

$= 26860.50 - 6179.45 \approx 20680.79$ , i.e. $20680.79.

Markers reward the outstanding-balance formula, accurate intermediate values, and the answer rounded to cents.

2021 HSC-style4 marks

Use an amortisation table to find the balance after

3

monthly repayments of $1200 on a $60000 loan at

6\%

per annum compounded monthly.

Show worked answer →

Per-period rate: $r = 0.005$ .

Month $1$ : opening $60000$ , interest $60000 \times 0.005 = 300$ , payment $1200$ , principal repaid $900$ , closing $59100$ .

Month $2$ : opening $59100$ , interest $59100 \times 0.005 = 295.50$ , payment $1200$ , principal repaid $904.50$ , closing $58195.50$ .

Month $3$ : opening $58195.50$ , interest $58195.50 \times 0.005 \approx 290.98$ , payment $1200$ , principal repaid $909.02$ , closing $\approx 57286.48$ , i.e. $57286.48.

Markers reward a clean amortisation table layout, interest computed on the opening balance each period, principal repaid as payment minus interest, and the closing balance.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation1 mark

The first three rows of the amortisation table for a $15000 personal loan with monthly repayments of $450 are shown. | Month | Opening | Interest | Payment | Principal repaid | Closing | |---|---|---|---|---|---| |

1

15000.00

75.00

450.00

375.00

14625.00

| |

2

14625.00

73.13

450.00

376.87

14248.13

| |

3

14248.13

71.24

450.00

| ? |

13869.37

| Read the interest charged in month

3

from the table, then find the principal repaid in month

3

(the value marked ?).

Show worked solution →

Read the interest from the table. The month $3$ row shows the interest charged is $71.24.

Find the principal repaid. Principal repaid is the payment minus the interest:

450 - 71.24 = 378.76,

i.e. $378.76.

State the answer. The principal repaid in month $3$ is $378.76.

Check. The closing balance equals the opening balance minus the principal repaid: $14248.13 - 378.76 = 13869.37$ , which matches the table.

foundation2 marks

A personal loan of $18000 is charged interest at

0.5\%

per month and repaid by monthly instalments of $500. Complete the first row of the amortisation table: find the interest charged in month

1

, the principal repaid, and the closing balance.

Show worked solution →

Interest on the opening balance. Month $1$ opens at the full loan, $18000. Interest is the opening balance times the per-period rate:

I_1 = 0.005 \times 18000 = 90.00,

i.e. $90.00.

Principal repaid. The payment is $500, and principal repaid is payment minus interest:

500 - 90 = 410.00,

i.e. $410.00.

Closing balance. The closing balance is opening plus interest minus payment:

18000 + 90 - 500 = 17590.00,

i.e. $17590.00.

Check. Closing balance equals opening minus principal repaid: $18000 - 410 = 17590$ , which matches.

foundation2 marks

A $9000 personal loan is fully repaid by

24

equal monthly instalments of $415.50. Find the total interest paid over the life of the loan.

Show worked solution →

Find the total amount paid. With every instalment the same, the total paid is the number of payments times the payment:

24 \times 415.50 = 9972.00,

i.e. $9972.00.

Subtract the amount borrowed. Total interest is total paid minus the principal:

9972.00 - 9000 = 972.00,

i.e. $972.00.

State the answer. The total interest paid over the life of the loan is $972.00.

Check. The interest is a small fraction of the $9000 borrowed, which is reasonable for a short $2$ year loan.

foundation3 marks

Mia has a reducing-balance loan charged at

5.4\%

per annum compounded monthly, repaid by monthly instalments of $620. Just before her next repayment the balance owing is $12350. Find the interest portion and the principal portion of that repayment, and the new balance owing afterwards.

Show worked solution →

Convert the rate. Monthly compounding uses the monthly rate:

r = \frac{0.054}{12} = 0.0045.

Interest portion. Interest is charged on the balance owing before the payment:

I = 0.0045 \times 12350 = 55.575 \approx 55.58,

i.e. $55.58.

Principal portion. Principal repaid is the payment minus the interest:

620 - 55.58 = 564.42,

i.e. $564.42.

New balance. Subtract the principal repaid from the old balance:

12350 - 564.42 = 11785.58,

i.e. $11785.58.

Check. The two portions add to the payment: $55.58 + 564.42 = 620.00$ , which matches.

core4 marks

A $25000 car loan is charged interest at

0.5\%

per month with normal monthly repayments of $600. In the very first month the borrower pays an extra $1000, so the month

1

payment is $1600. (a) Find the balance owing after the month

1

payment of $1600. (b) Find the interest charged in month

2

, and state how much less it is than the $122.63 that would have been charged without the extra repayment.

Show worked solution →

Part (a), interest in month 1. Interest is charged on the opening balance $25000 at $r = 0.005$ :

I_1 = 0.005 \times 25000 = 125.00,

i.e. $125.00.

Closing balance after the $1600 payment. Closing is opening plus interest minus the payment:

25000 + 125 - 1600 = 23525.00,

i.e. $23525.00.

Part (b), interest in month 2. Month $2$ opens at $23525.00, so

I_2 = 0.005 \times 23525 = 117.625 \approx 117.63,

i.e. $117.63.

Compare with no extra repayment: The interest is $122.63 - 117.63 = 5.00$ less, i.e. $5.00 less.
State the answer: The balance after month $1$ is $23525.00, and the month $2$ interest is $117.63, which is $5.00 less than it would have been.
Check: The extra $1000 of principal saves one month's interest of $0.005 \times 1000 = 5.00$ , matching the $5.00 reduction.

core4 marks

A $30000 car loan is charged interest at

0.5\%

per month and repaid by monthly instalments of $650. Use an amortisation table to find the balance owing immediately after the third repayment, showing each row.

Show worked solution →

Set up the rule for each row: The per-period rate is $r = 0.005$ . For each month: interest is the opening balance times $0.005$ , principal repaid is $650$ minus that interest, and the closing balance is opening plus interest minus $650$ . Carry each closing balance down as the next opening balance.
Month 1: Opening $30000. Interest $0.005 \times 30000 = 150.00$ . Principal repaid $650 - 150 = 500.00$ . Closing $30000 + 150 - 650 = 29500.00$ .
Month 2: Opening $29500. Interest $0.005 \times 29500 = 147.50$ . Principal repaid $650 - 147.50 = 502.50$ . Closing $29500 + 147.50 - 650 = 28997.50$ .
Month 3: Opening $28997.50. Interest $0.005 \times 28997.50 = 144.9875 \approx 144.99$ . Principal repaid $650 - 144.99 = 505.01$ . Closing $28997.50 + 144.99 - 650 = 28492.49$ .

Month	Opening	Interest	Payment	Principal repaid	Closing
$1$	$30000.00$	$150.00$	$650.00$	$500.00$	$29500.00$
$2$	$29500.00$	$147.50$	$650.00$	$502.50$	$28997.50$
$3$	$28997.50$	$144.99$	$650.00$	$505.01$	$28492.49$

State the answer. The balance owing immediately after the third repayment is $28492.49.

Check. Interest falls each month ( $150.00 \to 147.50 \to 144.99$ ) while principal repaid rises ( $500.00 \to 502.50 \to 505.01$ ), as expected for a reducing-balance loan.

core4 marks

A $40000 loan is charged interest at

6.6\%

per annum compounded monthly and repaid by monthly instalments of $780. Use the formula

B_n = P(1+r)^n - M \cdot \dfrac{(1+r)^n - 1}{r}

to find the balance owing immediately after the

12

th repayment. Round to the nearest cent.

Show worked solution →

Convert the rate and list the values. Monthly compounding gives

r = \frac{0.066}{12} = 0.0055,

with $P = 40000$ , $M = 780$ and $n = 12$ .

Find the growth factor.

(1.0055)^{12} \approx 1.068034.

Substitute into the formula.

B_{12} = 40000 \times 1.068034 - 780 \times \frac{1.068034 - 1}{0.0055}.

Evaluate each term. The first term is $40000 \times 1.068034 = 42721.36$ . The annuity factor is $\dfrac{0.068034}{0.0055} = 12.3698$ , so the second term is $780 \times 12.3698 = 9648.41$ .

B_{12} = 42721.36 - 9648.41 = 33072.95.

State the answer. The balance owing immediately after the $12$ th repayment is $33072.95.

Check. Twelve payments total $12 \times 780 = 9360$ , but the balance has fallen by only $40000 - 33072.95 = 6927.05$ ; the difference $9360 - 6927.05 = 2432.95$ is the interest charged over the first year, which is positive and less than a full year's interest on the opening balance ( $0.066 \times 40000 = 2640$ ), as expected since the balance is shrinking.

core3 marks

A $20000 car loan at

0.5\%

per month is repaid by monthly instalments of $400. The amortisation table shows the balance after the fourth repayment is $18790.97. Find the interest portion and the principal portion of the fifth repayment.

Show worked solution →

Identify the balance the interest is charged on. The fifth payment's interest is charged on the balance after the fourth payment, which is the opening balance for month $5$ : $18790.97.

Interest portion. Interest is the opening balance times the per-period rate $r = 0.005$ :

I_5 = 0.005 \times 18790.97 = 93.95485 \approx 93.95,

i.e. $93.95.

Principal portion. Principal repaid is the payment minus the interest:

400 - 93.95 = 306.05,

i.e. $306.05.

State the answer. Of the fifth repayment, $93.95 is interest and $306.05 reduces the principal.

Check. The two portions add to the payment: $93.95 + 306.05 = 400.00$ , which matches.

exam5 marks

The Tran family have a $300000 home loan charged at

6.0\%

per annum compounded monthly, with normal monthly repayments of $1934. (a) Construct the first two rows of the amortisation table, showing the opening balance, interest, payment, principal repaid and closing balance for each month. (b) In month

3

the family use a tax refund to pay an extra $5000 on top of their normal $1934 repayment. Find the balance owing immediately after the month

3

payment. (c) Hence find the interest charged in month

4

, and state how much less it is than the $1493.46 that would have been charged in month

4

without the extra repayment.

Show worked solution →

Part (a), set up the monthly rate: Monthly compounding gives $r = \dfrac{0.06}{12} = 0.005$ . For each row: interest is the opening balance times $0.005$ , principal repaid is $1934$ minus that interest, and closing is opening plus interest minus $1934$ .
Month 1: Interest $0.005 \times 300000 = 1500.00$ . Principal repaid $1934 - 1500 = 434.00$ . Closing $300000 + 1500 - 1934 = 299566.00$ .
Month 2: Opening $299566.00. Interest $0.005 \times 299566 = 1497.83$ . Principal repaid $1934 - 1497.83 = 436.17$ . Closing $299566 + 1497.83 - 1934 = 299129.83$ .

Month	Opening	Interest	Payment	Principal repaid	Closing
$1$	$300000.00$	$1500.00$	$1934.00$	$434.00$	$299566.00$
$2$	$299566.00$	$1497.83$	$1934.00$	$436.17$	$299129.83$

Part (b), month 3 with the extra repayment. Month $3$ opens at $299129.83. Interest $0.005 \times 299129.83 = 1495.65$ . The payment is $1934 + 5000 = 6934$ , so

299129.83 + 1495.65 - 6934 = 293691.48,

i.e. $293691.48.

Part (c), month 4 interest. Month $4$ opens at $293691.48, so

I_4 = 0.005 \times 293691.48 = 1468.46,

i.e. $1468.46. This is $1493.46 - 1468.46 = 25.00$ less, i.e. $25.00 less.

State the answer. After month $3$ the balance is $293691.48, and the month $4$ interest is $1468.46, which is $25.00 less than without the extra repayment.

Check. The extra $5000 of principal saves one month's interest of $0.005 \times 5000 = 25.00$ , matching the $25.00 reduction.

exam5 marks

The Nguyen family borrow $360000 for a home at

6.0\%

per annum compounded monthly, repaid over

25

years. (a) Use the repayment formula

M = \dfrac{Pr}{1 - (1+r)^{-n}}

to find the monthly repayment, to the nearest cent. (b) Find the total amount of interest paid over the life of the loan.

Show worked solution →

Part (a), convert the rate and term. Monthly compounding over $25$ years gives

r = \frac{0.06}{12} = 0.005, \qquad n = 25 \times 12 = 300.

Find the discount factor.

(1.005)^{-300} \approx 0.223966, \quad \text{so } 1 - (1.005)^{-300} \approx 0.776034.

Substitute into the repayment formula.

M = \frac{360000 \times 0.005}{0.776034} = \frac{1800}{0.776034} \approx 2319.49,

i.e. about $2319.49 per month.

Part (b), total interest. Total interest is total payments minus the amount borrowed, with $\text{total paid} = nM$ :

\text{total paid} = 300 \times 2319.49 = 695847.00,

\text{total interest} = 695847.00 - 360000 = 335847.00.

State the answer. The monthly repayment is about $2319.49, and the total interest paid is about $335847.00.

Check. The total interest is just under the amount borrowed, which is typical for a $25$ -year loan at this rate, so the figure is the right order of magnitude.

exam4 marks

A borrower takes a $50000 loan at

9.0\%

per annum compounded monthly and plans to repay it with monthly instalments of $370. By comparing the repayment with the first month's interest, explain whether this loan will ever be repaid, and describe what happens to the balance.

Show worked solution →

Convert the rate. Monthly compounding gives

r = \frac{0.09}{12} = 0.0075.

Find the first month's interest. Interest is charged on the opening balance $50000:

I_1 = 0.0075 \times 50000 = 375.00,

i.e. $375.00.

Compare the repayment with the interest. The repayment is $370, which is less than the interest of $375. So the payment does not even cover one month's interest. Formally, the loan finishes only if $M > rP$ ; here $M = 370$ and $rP = 375$ , so $M < rP$ .

Describe what happens to the balance. Since the payment is $5 short of the interest in month $1$ , the unpaid $5 of interest is added to the balance:

50000 + 375 - 370 = 50005.00,

i.e. the balance rises to $50005.00. The balance grows every month, so the interest charged grows too, and the gap widens.

State the answer. The loan is never repaid: because the $370 repayment is smaller than the $375 monthly interest, the balance increases each month instead of falling.

Check. The condition $M \le rP$ for a loan that never finishes is satisfied ( $370 \le 375$ ), confirming the conclusion.

exam5 marks

A $45000 loan is charged interest at

6.0\%

per annum compounded monthly and repaid by monthly instalments of $900. (a) By setting the balance to zero and using logarithms, find the number of months needed to repay the loan. (b) Find the value of the smaller final payment.

Show worked solution →

Part (a), set up the equation. With $r = 0.005$ , $P = 45000$ and $M = 900$ , set the closed-form balance to zero:

45000(1.005)^n - 900 \cdot \frac{(1.005)^n - 1}{0.005} = 0.

Rearrange to isolate the power. Note $\dfrac{900}{0.005} = 180000$ , so

45000(1.005)^n = 180000\!\left((1.005)^n - 1\right).

Collecting the $(1.005)^n$ terms gives

(1.005)^n(45000 - 180000) = -180000,

(1.005)^n = \frac{180000}{135000} = 1.3333.

Take logarithms.

n = \frac{\log 1.3333}{\log 1.005} = \frac{0.124939}{0.002166} \approx 57.68.

Since the $57$ th payment leaves a small balance still owing, a $58$ th (smaller) payment is needed, so the loan is repaid in $58$ months.

Part (b), find the final payment. After $57$ full payments the balance owing is $609.56 (from the closed form with $n = 57$ ). The final payment must clear this balance plus its one month of interest:

\text{final interest} = 0.005 \times 609.56 = 3.05,

\text{final payment} = 609.56 + 3.05 = 612.61,

i.e. $612.61.

State the answer. The loan is repaid in $58$ months, with a smaller final payment of $612.61.

Check. The final payment $612.61 is less than the regular $900, as expected for a partial last instalment, and applying it to the month- $58$ opening balance of $609.56 leaves $609.56 + 3.05 - 612.61 = 0.00$ , so the loan clears exactly.

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA