Year 10 Maths Advanced Earning Money

Term Payments

Q: How do you calculate flat-rate interest?

Use I = Prn, where P is the amount borrowed, r is the rate per year as a decimal and n is the time in years. The interest is the same for every year of the loan.

20 practice questions 1 video lesson Theory + worked examples

Buying on terms means paying a deposit and then regular repayments instead of the cash price. The extra cost is usually charged as flat-rate interest on the balance owed, so the total price is higher than the cash price. In NSW Year 10 Advanced maths you find repayments, total costs and the flat interest rate, and compare finance offers (NESA outcome MA5‑FIN‑C‑01).

You will calculate deposits, interest on the balance and weekly or monthly instalments, then work backwards from repayments to the interest rate or the cash price. You will check interest-free and payment-free deals, include establishment fees, and compare store terms with a personal loan. Harder questions find the smallest deposit for a budget, read a loan from its graph, and rearrange the repayment formula to find the rate. It is part of Year 10 Advanced Earning Money.

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Theory

Buying on terms means paying a deposit, then paying off the balance plus flat-rate interest in equal instalments. This Year 10 Advanced maths page shows how to find the instalment, the total cost and the flat interest rate, and how to form equations to work back to the cash price or the length of a loan.

The cash price is what an item costs if you pay in full now. On terms you pay a deposit now and borrow the rest, called the balance.

Flat-rate interest is simple interest, $I = Prn$, worked out on the amount borrowed for the whole time of the loan, even though the debt shrinks as you repay it.

The balance plus the interest is repaid in equal instalments (repayments), usually weekly, fortnightly or monthly. The total cost on terms is the deposit plus all the instalments, plus any fees.

An interest-free offer charges no interest if the full price is paid within a set time. A deferred payment plan has an opening period with no repayments, but the remaining repayments still have to cover the whole cost.

Buying on terms costs more than paying cash. The extra is the interest (plus any fees).

With flat-rate interest the total owed is a straight line. It starts at the amount borrowed and rises by $Pr$ each year.

With $P$ the amount borrowed, $r$ the flat rate per year as a decimal and $n$ the time in years:

\[I = Prn \qquad \text{amount owed} = P + Prn = P(1 + rn)\]

I = P r n, A = P (1 + r n)

Instalments and total cost:

\[\text{instalment} = \dfrac{\text{balance} + \text{interest}}{\text{number of instalments}} \qquad \text{total cost} = \text{deposit} + \text{instalments} + \text{fees}\]

instalment = \frac{balance + interest}{number}

For monthly repayments $M$ over $n$ years there are $12n$ of them, so:

\[M = \dfrac{P(1 + rn)}{12n} \qquad r = \dfrac{I}{Pn}\]

M = \frac{P (1 + r n)}{12 n}

Repaid	Instalments in $n$ years
weekly	$52n$
fortnightly	$26n$
monthly	$12n$

Rate per month. If the flat rate is given per month, either use months for the time, or multiply by \(12\) to get the rate per annum: \(1.2\%\) a month is \(14.4\%\) p.a.

The graph is a straight line. Total owed \(= P + (Pr)n\). The vertical intercept is the amount borrowed and the gradient is the interest per year, so \(r = \dfrac{\text{gradient}}{P}\).

How to solve a term payment problem

Find the deposit and the balance. A percentage deposit is a percentage of the cash price.
Calculate the interest with $I = Prn$ on the amount the question says, usually the balance. Time is in years unless the rate is per month.
Add the interest to the balance, then divide by the number of instalments.
Total the cost: deposit plus all instalments plus any fees. The extra over the cash price is the true cost of borrowing.

Working backwards and comparing

Flat rate from repayments: interest $=$ total repaid $-$ amount borrowed, then $r = \dfrac{I}{Pn}$ (Example 2). Treat fees as part of the interest if the question says so.
Cash price unknown: call it $P$, write the amount owed as a number times $P$, and set it equal to the total of the repayments (Example 3).
Time unknown: write the interest in terms of the number of months, $m$, and solve balance $+$ interest $=$ repayment $\times\, m$ (Example 4).
Largest repayment you can afford: the most you can repay in total limits the balance. Solve $\text{balance} \times (1 + rn) \le \text{total}$, then the smallest deposit is the price minus the largest balance.
Comparing offers: find the total cost of each. The lower instalment is not always the cheaper deal. Two offers cost the same when their totals are equal; with a price of $P$ both totals are a number times $P$, so $P$ cancels.
Interest-free and deferred plans: divide the price by the number of months in the free period to find the smallest repayment. With a payment-free period, count only the months you actually pay.
Rearranging the formula: undo the operations in reverse order. To make $r$ the subject of $A = P(1 + rn)$, divide by $P$, subtract $1$, then divide by $n$: $r = \dfrac{A - P}{Pn}$. A formula for the monthly repayment is rearranged the same way.

Example 1 — Finding the instalment

A guitar and amplifier cost $\$2400$ cash. On terms, Ella pays a $15\%$ deposit. Flat-rate interest of $11\%$ p.a. is charged on the balance for $2$ years, and she repays the balance plus interest in equal monthly instalments. Find:
(i) the monthly instalment
(ii) the total cost on terms.

Solution

Work out the deposit and the balance first. Interest is charged only on the balance.

$\text{Deposit}$	$=$	$0.15 \times 2400 = \$360$
$\text{Balance}$	$=$	$2400 - 360 = \$2040$
$I$	$=$	$Prn$
	$=$	$2040 \times 0.11 \times 2$
	$=$	$\$448.80$

(i) Add the interest to the balance and share it over $2 \times 12 = 24$ months.

$\text{Owed}$	$=$	$2040 + 448.80 = \$2488.80$
$\text{Instalment}$	$=$	$\dfrac{2488.80}{24}$
	$=$	$\mathbf{\$103.70}\text{ a month}$

M = \frac{2040 + 2040 \times 0.11 \times 2}{24} = 103.70

(ii)

$\text{Total cost}$	$=$	$\text{deposit} + \text{instalments}$
	$=$	$360 + 24 \times 103.70$
	$=$	$\mathbf{\$2848.80}$

360 + 24 \times 103.70 = 2848.80

That is $\$448.80$ more than the cash price, which is exactly the interest.

Example 2 — Finding the flat rate

A used car costs $\$19\,500$. Sam pays a $\$2000$ deposit, then $\$135$ a week for $3$ years ($1$ year $= 52$ weeks). Find:
(i) the interest paid
(ii) the flat interest rate p.a., correct to $1$ decimal place
(iii) the total cost as a percentage more than the cash price, correct to $1$ decimal place.

Solution

(i) The interest is everything repaid minus the amount borrowed.

$\text{Borrowed}$	$=$	$19\,500 - 2000 = \$17\,500$
$\text{Repaid}$	$=$	$135 \times 52 \times 3 = \$21\,060$
$I$	$=$	$21\,060 - 17\,500 = \mathbf{\$3560}$

(ii) Substitute into $I = Prn$ and solve for $r$.

$3560$	$=$	$17\,500 \times r \times 3$
$r$	$=$	$\dfrac{3560}{52\,500}$
	$=$	$0.06780\ldots$
	$\approx$	$\mathbf{6.8\%}\text{ p.a.}$

r = \frac{3560}{17500 \times 3} \approx 6.8 %

(iii) Compare the total cost with the cash price.

$\text{Total cost}$	$=$	$2000 + 21\,060 = \$23\,060$
$\dfrac{23\,060 - 19\,500}{19\,500} \times 100\%$	$=$	$18.25\ldots\%$
	$\approx$	$\mathbf{18.3\%}$

\frac{23060 - 19500}{19500} \times 100 \approx 18.3 %

The flat rate uses the amount borrowed. The percentage in (iii) uses the cash price, so the two answers differ.

Example 3 — Working back to the cash price

A box trailer is bought on terms with a $20\%$ deposit. Flat-rate interest of $9\%$ p.a. is charged on the balance for $3$ years, and the balance plus interest is repaid at $\$127$ a month. Find the cash price.

Solution

Let the cash price be $\$P$. Write the amount owed in terms of $P$, then match it to the repayments.

$\text{Balance}$	$=$	$P - 0.2P = 0.8P$
$\text{Owed}$	$=$	$0.8P + 0.8P \times 0.09 \times 3$
	$=$	$0.8P \times 1.27$
	$=$	$1.016P$

The repayments must cover the amount owed:

$1.016P$	$=$	$127 \times 36$
$1.016P$	$=$	$4572$
$P$	$=$	$\dfrac{4572}{1.016}$
	$=$	$\mathbf{\$4500}$

1.016 P = 4572, P = 4500

Check: deposit $\$900$, balance $\$3600$, interest $\$972$, and $3600 + 972 = 4572 = 127 \times 36$.

Example 4 — How long is the loan?

A balance of $\$3600$ is charged flat-rate interest of $10\%$ p.a. The balance plus interest is repaid at $\$130$ a month. Form and solve an equation to find how many months the loan runs for.

Solution

Let the loan run for $m$ months, which is $\dfrac{m}{12}$ years. Write the interest in terms of $m$.

$I$	$=$	$3600 \times 0.10 \times \dfrac{m}{12}$
	$=$	$30m$

Balance plus interest equals the total of the repayments:

$3600 + 30m$	$=$	$130m$
$3600$	$=$	$100m$
$m$	$=$	$\mathbf{36}\text{ months}$

3600 + 30 m = 130 m, m = 36

Check: $I = 3600 \times 0.10 \times 3 = \$1080$, and $(3600 + 1080) \div 36 = \$130$.

Common pitfalls

Interest on the right amount. Interest is usually charged on the balance, not the cash price. Read whether it is on the balance or the full price.

Count the instalments carefully. Three years of weekly payments is $3 \times 52 = 156$ payments, and two years of fortnightly payments is $52$. With a payment-free period, leave those months out.

Do not forget the deposit. The total cost on terms is the deposit plus the instalments. The instalments alone only cover the balance and the interest.

Match the rate and the time. A flat rate per year needs time in years: $18$ months is $1.5$ years. A rate per month needs time in months.

A flat rate is not the true cost of the loan. Flat-rate interest is charged on the whole amount borrowed for the whole time, even as the debt shrinks, so the real cost of borrowing is higher than the flat rate suggests.

Frequently asked questions

What does buying on terms mean?

It means paying a deposit now and repaying the rest, plus interest, in regular instalments. It lets you have the item straight away, but it usually costs more than paying cash.

How do you calculate flat-rate interest?

Use $I = Prn$, where $P$ is the amount borrowed, $r$ is the rate per year as a decimal and $n$ is the time in years. The interest is the same for every year of the loan.

How do you work out monthly repayments on a flat-rate loan?

Add the flat-rate interest to the amount borrowed, then divide by the number of months. For a $2$ year loan that is $24$ repayments.

How do you find the flat interest rate from the repayments?

Multiply the repayment by the number of repayments to get the total repaid, subtract the amount borrowed to get the interest, then use $r = \dfrac{I}{Pn}$.

Is a lower repayment always the better deal?

No. A lower repayment can come from a bigger deposit or a longer loan. Compare the total cost of each offer, not just the size of the repayment.

\(\text{Deposit}\)	\(=\)	\(0.15 \times 2400 = \$360\)
\(\text{Balance}\)	\(=\)	\(2400 - 360 = \$2040\)
\(I\)	\(=\)	\(Prn\)
\(\)	\(=\)	\(2040 \times 0.11 \times 2\)
\(\)	\(=\)	\(\$448.80\)

\(\text{Owed}\)	\(=\)	\(2040 + 448.80 = \$2488.80\)
\(\text{Instalment}\)	\(=\)	\(\dfrac{2488.80}{24}\)
\(\)	\(=\)	\(\mathbf{\$103.70}\text{ a month}\)

\(\text{Total cost}\)	\(=\)	\(\text{deposit} + \text{instalments}\)
\(\)	\(=\)	\(360 + 24 \times 103.70\)
\(\)	\(=\)	\(\mathbf{\$2848.80}\)

\(\text{Borrowed}\)	\(=\)	\(19\,500 - 2000 = \$17\,500\)
\(\text{Repaid}\)	\(=\)	\(135 \times 52 \times 3 = \$21\,060\)
\(I\)	\(=\)	\(21\,060 - 17\,500 = \mathbf{\$3560}\)

\(3560\)	\(=\)	\(17\,500 \times r \times 3\)
\(r\)	\(=\)	\(\dfrac{3560}{52\,500}\)
\(\)	\(=\)	\(0.06780\ldots\)
\(\)	\(\approx\)	\(\mathbf{6.8\%}\text{ p.a.}\)

\(\text{Total cost}\)	\(=\)	\(2000 + 21\,060 = \$23\,060\)
\(\dfrac{23\,060 - 19\,500}{19\,500} \times 100\%\)	\(=\)	\(18.25\ldots\%\)
\(\)	\(\approx\)	\(\mathbf{18.3\%}\)

\(\text{Balance}\)	\(=\)	\(P - 0.2P = 0.8P\)
\(\text{Owed}\)	\(=\)	\(0.8P + 0.8P \times 0.09 \times 3\)
\(\)	\(=\)	\(0.8P \times 1.27\)
\(\)	\(=\)	\(1.016P\)

\(1.016P\)	\(=\)	\(127 \times 36\)
\(1.016P\)	\(=\)	\(4572\)
\(P\)	\(=\)	\(\dfrac{4572}{1.016}\)
\(\)	\(=\)	\(\mathbf{\$4500}\)

\(I\)	\(=\)	\(3600 \times 0.10 \times \dfrac{m}{12}\)
\(\)	\(=\)	\(30m\)

\(3600 + 30m\)	\(=\)	\(130m\)
\(3600\)	\(=\)	\(100m\)
\(m\)	\(=\)	\(\mathbf{36}\text{ months}\)

Term Payments

📖 Theory

How to solve a term payment problem

Working backwards and comparing

Common pitfalls

Frequently asked questions

Theory