Loans

Financial Maths, STD2 EQ-Bank 21

Maya uses a buy now, pay later payment option to make a purchase of $120. Her repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Maya misses her final payment on 16 March 2026 and is charged a late fee of $19. Maya's payment schedule is shown, with her balance totalling $49.

\begin{array} {|l|c|c|c|}
\hline \textbf{Payment} & \textbf{Due Date} & \textbf{Amount} & \textbf{Status} \\
\hline \text{1st} & \text{2 February 2026} & \$30 & \text{Paid} \\
\hline \text{2nd} & \text{16 February 2026} & \$30 & \text{Paid} \\
\hline \text{3rd} & \text{2 March 2026} & \$30 & \text{Paid} \\
\hline \text{4th} & \text{16 March 2026} & \$30 & \text{Not Paid} \\
\hline \text{Outstanding Due} & \text{30 March 2026} & \$49\ \text{including late fee} & \\
\hline \end{array}

Find the total amount Maya pays for her purchase if repaying in full on 30 March 2026. (1 mark)
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Maya's bank offers short-term loans where simple interest is charged at 16% per annum.
Suppose Maya had borrowed $120 from the bank to make this purchase on 2 February 2026 and repaid it in full 8 weeks later.
How much would Maya have saved using this approach instead of the buy now, pay later option? (2 marks)
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Show Answers Only

a. $$139$

b. $$16.05$

Show Worked Solution

a. $\text{If total owing paid on 30 March:}$

$\text{Total paid} = 30+30+30+49=$139$

b. $r=16\%=0.16,\ \ n=\dfrac{8 \times 7}{365} = \dfrac{56}{365}$

$I=Prn=120 \times 0.16 \times \dfrac{56}{365} = 2.945… = $2.95 $

$\text{Amount saved} = 19-2.95=$16.05$

Financial Maths, STD2 EQ-Bank 20

Kimberley uses a buy now, pay later payment option to make a purchase of $100. Her repayments are split across 4 equal payments over 6 weeks. No interest is charged.

Kimberley misses her final payment and is charged a late fee of $17. Kimberley’s payment schedule is shown, with her balance totalling $42.

Find the total amount Kimberley pays for her purchase if repaying in full on 27 July 2024. (1 mark)
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Kimberley’s bank offers short-term loans where simple interest is charged at 18% per annum.
Suppose Kimberley had borrowed $100 from the bank to make this purchase on 1 June 2024 and repaid it in full 8 weeks later.
How much would Kimberley have saved using this approach instead of the buy now, pay later option? (2 marks)
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a. $$117$

b. $$14.24$

Show Worked Solution

a. $\text{If total owing paid on 27 July:}$

$\text{Total paid} = 25+25+25+42=$117$

b. $r=18\%=0.18,\ \ n=\dfrac{8 \times 7}{365} = \dfrac{56}{365}$

$I=Prn=100 \times 0.18 \times \dfrac{56}{365} = 2.761… = $2.76 $

$\text{Amount saved} = 17-2.76=$14.24$

Financial Maths, STD2 F4 2025 GEN1 7

Declan is a filmmaker and content creator.

He has taken out a reducing balance loan to fund a new production.

Interest is calculated monthly and Declan makes monthly repayments.

Three rows of the amortisation table for Declan’s loan are shown below.

\begin{array}{|c|c|c|c|c|}
\hline
\hline \rule{0pt}{2.5ex}\quad \textbf{Payment} \quad & \quad\textbf{Payment} \quad & \quad\textbf{Interest} \quad& \textbf{Principal} & \quad\textbf{Balance}\quad\\
\textbf{number} & \textbf{(\$)} \rule[-1ex]{0pt}{0pt}& \textbf{(\$)} & \quad\textbf{reduction (\$)} \quad& \textbf{(\$)}\\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 0.00 & 0.00 & 0.00 & 850\,000.00 \\
\hline \hline \rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & 15\,730.88 & 2975.00 & 12\,755.88 & 837\,244.12 \\
\hline \hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & 2930.35 & 12\,800.53 & 824\,443.59 \\
\hline
\end{array}

What amount, in dollars, did Declan borrow? (1 mark)
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Why is the interest associated with payment 2 lower than the interest associated with payment 1? (1 mark)
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The interest rate on Declan’s loan is 4.2% per annum, compounding monthly.
Using the values in the table, complete the table below.
Round all values to the nearest cent. (2 marks)

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a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

Show Worked Solution

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Monthly interest}=\dfrac{4.2}{12}=0.35\%$

$\text{Calculating missing values in the table:}$

$\text{Interest}=\dfrac{0.35}{100} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

♦ Mean mark (c) 40%.

Financial Maths, STD2 F4 2025 HSC 33

A used car has a sale price of $24 200. In addition to the sale price, the following costs are charged:

transfer of registration $50
stamp duty which is calculated at $3 for every $100, or part thereof, of the sale price.

Kat borrows the total amount to be paid for the car, including transfer of registration and stamp duty. Simple interest at the rate of 6.8% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 3 years.

Calculate Kat’s monthly repayment. (5 marks)

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Show Answers Only

$$835.31$

Show Worked Solution

$\text{Stamp Duty} =\dfrac{24\ 200}{100}\times 3=$726$

$\text{Total Cost}\ $	$=\ \text{Price + Transfer + Stamp Duty}$
	$=24\ 200+50+726$
	$=$24\ 976$

$\text{Interest}=Prn=24\,976\times 0.068\times 3=$5095.104$

$\text{Loan amount}\ $	$=\ \text{Total Cost + Interest}$
	$=24\ 976+5095.104$
	$=$30\ 071.104$

$\text{3 years}= 3 \times 12=36\ \text{months}$

$\text{Monthly repayment}$	$=\dfrac{30\ 071.104}{36}$
	$=$835.308444$
	$\approx $835.31$

Financial Maths, STD2 F4 2024 GEN1 21 MC

Lee took out a loan of $121 000, with interest compounding monthly. He makes monthly repayments of $2228.40 for five years until the loan is repaid in full.

The total interest paid by Lee is closest to

$4434
$5465
$10 539
$12 704

Show Answers Only

$D$

Show Worked Solution

$\text{Repayments}$	$ =2228.40\times 12\times 5$
	$=$133\,704$

$\text{Interest}$	$=133\,704-121\,000$
	$=$12\,704$

$\Rightarrow D$

Mean mark 56%.

Financial Maths, STD2 F4 2024 HSC 21

William has a reducing balance loan on which he owes $5590. He makes monthly repayments of $110.

The loan company charges interest at 24% per annum, compounded monthly.

The spreadsheet shows some of the information for the next two months of the loan.

Complete the entries in the spreadsheet to show the balance owing on the loan at the end of two months. (2 marks)

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Explain why the loan will never be repaid if William continues to make repayments of $110 per month. (1 mark)

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b. $\text{The interest charged exceeds the amount that is repaid.}$

$\text{Interest charges will gradually increase with the monthly repayment staying the same \(\Rightarrow$ loan will never be repaid.}\)

Show Worked Solution

$\text{Calculations:}$

$\text {Cell E2 }= 5590+111.80-110=\$ 5591.80$

$\text{Cell B3 }=\text{ Cell E2}$

$\text{Cell C3}=5591.80 \times 0.02 = \$111.84 \ \text{(monthly r/i = 2%)}$

$\text{Cell E3}=5591.80+111.84-110 = \$5593.64$

b. $\text{The interest charged exceeds the amount that is repaid.}$

$\text{Interest charges will gradually increase with the monthly repayment staying the same \(\Rightarrow$ loan will never be repaid.}\)

Financial Maths, STD2 F4 2024 HSC 27

A couple borrows $30 000 to be repaid in equal monthly repayments of $280 over 10 years.

After following this repayment plan for 5 years, they decide to decrease their monthly repayment to $250. As a result, it will take them an additional two years to pay off their loan.

How much more will they repay in total by making this change? (3 marks)

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$\text{Extra repaid}\ =$4200$

Show Worked Solution

$\text{Original plan:}$

$\text{Total repayments}\ =10 \times 12=120$

$\text{Amount repaid}\ =120 \times 280=$33\ 600$

$\text{New plan:}$

$\text{Repayments at \$280}\ =5 \times 12=60$

$\text{Repayments at \$250}\ =7 \times 12=84$

$\text{Amount repaid}\ = 60 \times 280 + 84 \times 250 =$37\ 800$

$\text{Extra repaid}\ =37\ 800-33\ 600=$4200$

Financial Maths, STD2 F4 2023 HSC 29

The table shows monthly repayments for each $1000 borrowed.

A couple borrows $520 000 to buy a house at 8% per annum over 25 years.
How much does the couple repay in total for this loan? (3 marks)

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Chris borrows some money at 7% per annum. Chris will repay the loan over 15 years, paying $3596 per month.
How much money does Chris borrow? (1 mark)

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`$1\ 204\ 320`
`400\ 000`

Show Worked Solution

a. `text{8.0% interest over a 25 year loan}`

`text{Monthly repayments to borrow $1000 = $7.72}`

`text{Total months}\ = 25 xx 12 = 300`

`text{Monthly repayments}`	`=520 xx 7.72`
	`=$4014.40`

`:.\ text{Total repayments}`	`= 4014.40 xx 300`
	`=$1\ 204\ 320`

♦ Mean mark (a) 50%.

b. `text{7.0% interest over a 15 year loan}`

`text{Monthly repayments to borrow $1000 = $8.99}`

`:.\ text{Amount borrowed}`	`=3596/8.99 xx $1000`
	`=400\ 000`

♦♦♦ Mean mark (b) 11%.

Financial Maths, STD2 F4 2022 HSC 36

Frankie borrows $200 000 from a bank. The loan is to be repaid over 23 years at a rate of 7.2% per annum, compounded monthly. The repayments have been set at $1485 per month.

The interest charged and the balance owing for the first three months of the loan are shown in the spreadsheet below.

What are the values of `A` and `B`? (2 marks)
After 50 months of repaying the loan, Frankie decides to make a lump sum payment of $ 40 000 and to continue making the monthly repayments of $1485. The loan will then be fully repaid after a further 146 monthly repayments.
How much less will Frankie pay overall by making the lump sum payment? (3 marks)

Show Answers Only

`A=$1198.29,\ \ B=$199\ 139.86`
`$78\ 800`

Show Worked Solution

a. `text{Monthly interest rate}\ =7.2/12=0.6text{%}`

`A`	`=199\ 715 xx 0.6/100`
	`=$1198.29`

`B`	`=P+I-R`
	`=199\ 428.29 + 1196.57-1485`
	`=$199\ 139.86`

b. `text{Total payments if lump sum not paid}`

`= (23xx12) xx 1485`

`=$409\ 860`

`text{Total payments if lump sum paid}`

`=40\ 000 + (50 + 146) xx 1485`

`=$331\ 060`

`text{Savings by paying the lump sum}`

`=409\ 860-331\ 060`

`=$78\ 800`

♦♦♦ Mean mark (b) 17%.

Financial Maths, STD2 F4 2018 HSC 29e

Andrew borrowed $20 000 to be repaid in equal monthly repayments of $243 over 10 years. Having made this monthly repayment for 4 years, he increased his monthly repayment to $281. As a result, Andrew paid off the loan one year earlier.

How much less did he repay altogether by making this change? (2 marks)

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Show Answers Only

`$636`

Show Worked Solution

`text(Total original repayments)= 10 xx 12 xx 243= $29\ 160`

`text(Actual repayments)= 4 xx 12 xx 243\ +\ 5 xx 12 xx 281= $28\ 524`

`:.\ text(Savings)= 29\ 160-28\ 524= $636`

Financial Maths, STD2 F4 2017 HSC 28c

Michelle borrows $100 000. The interest rate charged is 12% per annum compounded monthly. The monthly payment is $1029 and the first repayment is made after one month.

What is the amount outstanding immediately after the SECOND monthly repayment is made? (3 marks)

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Show Answers Only

`$99\ 941.71`

Show Worked Solution

`text(Interest per month)`

`= text(12%)/12= 1text(%)`

`text(Let)\ \ A=\ text(amount owing)`

♦ Mean mark 41%.

`text(After 1st repayment:)`

`A_1`	`= (100\ 000 + text(1%) xx 100\ 000)-1029`
	`= $99\ 971`

`text(After 2nd repayment:)`

`A_2`	`= (99\ 971 + text(1%) xx 99\ 971)-1029`
	`= $99\ 941.71`

Financial Maths, STD2 F4 2016 HSC 27d

Marge borrowed $19 000 to buy a used car. Interest on the loan was charged at 4.8% pa at the end of each month. She made a repayment of $436 at the end of every month. The table below sets out her monthly repayment schedule for the first four months of the loan.

Some values in the table are missing. Write down the values for `A` and `B`. (2 marks)

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Calculate the value of `X`. (2 marks)

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Marge repaid this loan over four years.

What is the total amount that Marge repaid? (1 mark)

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`A = $19\ 000,quadB = $17\ 551.33`
`$74.56`
`$20\ 928`

Show Worked Solution

i.	`A + 76-436`	`= 18\ 640`
	`:. A`	`= $19\ 000`

`17\ 915.67 + 71.66-436 = B`

`:. B = $17\ 551.33`

ii. `18\ 640 + X-436 = 18\ 278.56`

`:. X`	`= 18\ 278.56 + 436-18\ 640`
	`= $74.56`

♦♦ Mean mark part (iii) 28%.
COMMENT: Read carefully whether total paid or total interest paid is required.

iii. `text(Total amount repaid)`

`= 48 xx 436= $20\ 928`

Financial Maths, STD2 F4 EQ-Bank 2 MC

Sally purchased an electronic game machine on hire purchase. She paid $140 deposit and then $25.50 per month for two years.

The total amount that Sally paid is

$191
$446
$612
$752

Show Answers Only

`D`

Show Worked Solution

`text(Total paid)`	`= 140 + 25.50 xx 2 xx12`
	`= $752`

`=>D`

Financial Maths, STD2 F4 EQ-Bank 1 MC

Ernie took out a reducing balance loan to buy a new family home.

He correctly graphed the amount paid off the principal of his loan each year for the first five years.

The shape of this graph (for the first five years of the loan) is best represented by

Show Answers Only

`B`

Show Worked Solution

`text(A reducing balance loan means that the amount of)`

`text(interest paid out decreases each year, and therefore)`

`text(the amount paid off the principal will not only increase)`

`text(each year, but will do so at an increasing rate.)`

`B\ text(correctly shows this trend.)`

`=> B`

Financial Maths, STD2 F4 2015 HSC 29b

Jamal borrowed $350 000 to be repaid over 30 years, with monthly repayments of $1880. However, after 10 years he made a lump sum payment of $80 000. The monthly repayment remained unchanged. The graph shows the balances owing over the period of the loan.

Over the period of the loan, how much less did Jamal pay by making the lump sum payment? (2 marks)

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Show Answers Only

`$100\ 480`

Show Worked Solution

`text(Without the lump sum payment)`

♦ Mean mark 34%.

`text(Total repayments)`	`= 30 xx 12 xx $1880`
	`= $676\ 800`

`text(With the lump sum payment)`

`text(Total repayments)`	`= (22 xx 12 xx $1880) + $80\ 000`
	`= $496\ 320 + $80\ 000`
	`= $576\ 320`

`:.\ text(Amount Jamal saved)`

`= 676\ 800-576\320`

`= $100\ 480`

Financial Maths, STD2 F4 2006 HSC 27a

Liliana wants to borrow money to buy a house. The bank sent her an email with the following table attached.

Liliana decides that she can afford $1000 per month on repayments.

What is the maximum amount she can borrow, and how many years will she have to repay the loan? (1 mark)

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Zali intends to borrow $160 000 over 15 years from the same bank.

If she chooses to borrow $160 000 over 20 years instead, how much more interest will she pay? (2 marks)

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Show Answers Only

a. `text{$130 000 (over 30 years)}`

b. `$45\ 964.80`

Show Worked Solution

a. `text(From table)`

`text{$130 000 (over 30 years)}`

b. `text(Total repayments over 15 years)`

`=1529.04 xx 180`

`= $275\ 227.20`

`text(Total repayments over 20 years)`

`=1338.30 xx 240`

`= $321\ 192.00`

`:.\ text(Extra interest over 20 years)`

`= 321\ 192.00-275\ 227.20`

`= $45\ 964.80`

Financial Maths, STD2 F4 2004 HSC 27a

Aaron decides to borrow $150 000 over a period of 20 years at a rate of 7.0% per annum.

Using the Monthly Repayment Table, calculate Aaron’s monthly repayment. (2 marks)

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How much interest does he pay over the 20 years? (2 marks)

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Aaron calculates that if he repays the loan over 15 years, his total repayments would be `$242\ 730`.

How much interest would he save by repaying the loan over 15 years instead of 20 years? (2 marks)

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a. `$1162.50`

b. `$129\ 000`

c. `$36\ 270`

Show Worked Solution

a. `text(Using the table:)`

`text(Monthly repayment on $1000 at 7.0% over 20 years = $7.75)`

`:.\ text(Monthly repayment on $150 000 loan)`

`= 150 xx 7.75`

`= $1162.50`

b. `text(Total repayments over 20 years)`

`= 20 xx 12 xx 1162.50= $279\ 000`

`:.\ text(Interest paid over 20 years)`

`= 279\ 000-150\ 000= $129\ 000`

c. `text(Savings)`	`=\ text{Total paid (20 years) – Total paid (15 years)`
	`= 279\ 000-242\ 730`
	`= $36\ 270`

Financial Maths, STD2 F4 2006 HSC 21 MC

Bill borrows $420 000 to buy a house. Interest is charged at 7.2% per annum, compounded monthly.

How much does he owe at the end of the first month, after he has made a $4000 repayment?

$418 496
$418 520
$445 952
$446 240

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ L\ =\ text(Amount of the loan after 1 month)`

`r= (7.2 text(%))/12=0.6 text(%)\ =0.006`

`text(Using)\ \ FV=PV(1+r)^n`

`L`	`= 420\ 000 (1 + 0.006)^1-text(repayment)`
	`= 420\ 000 (1.006)-4000`
	`= $418\ 520`

`=> B`

Financial Maths, STD2 F4 2005 HSC 10 MC

The table is used to calculate monthly loan repayments.

Samantha has borrowed $70 000 at 8% per annum for 15 years.

What is her monthly loan repayment?

$143.40
$669.20
$8030.40
$10 038.00

Show Answers Only

`B`

Show Worked Solution

`text(Monthly repayment of $1000 at 8% for 15 years)`

`= $9.56`

`:.\ text(Monthly repayment of $70 000)`

`= 70 × $9.56`

`= $669.20`

`=> B`

Financial Maths, STD2 F4 2008 HSC 15 MC

Ali is buying a speedboat at Betty’s Boats.

What is the amount of interest Ali will have to pay if he chooses to buy the boat on terms?

$3200
$5600
$19 200
$21 600

Show Answers Only

`B`

Show Worked Solution

`text(Deposit)= text(15%) xx 16\ 000= $2400`

`text(Payments)= 320 xx 5 xx 12= $19\ 200`

`text(Total paid)= 2400 + 19\ 200= $21\ 600`

`:.\ text(Interest)= 21\ 600-16\ 000=$5600`

`=> B`

Financial Maths, STD2 F4 2010 HSC 25b

William wants to buy a car. He takes out a loan for $28 000 at 7% per annum interest for four years.

Monthly repayments for loans at different interest rates are shown in the spreadsheet.

How much interest does William pay over the term of this loan? (2 marks)

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Show Answers Only

`$4183.52`

Show Worked Solution

♦ Mean mark 42%
MARKER’S COMMENT: An incorrect table value used correctly in the following calculations received half-marks here. Show your working!

`text(Loan) = $28\ 000,\ \ \ \ r =\ text(7% p.a.)`

`text(Monthly repayment = $670.49`

`text(# Repayments) = 4 xx 12 = 48`

`text(Total repaid)`	`= 48 xx 670.49`
	`= $32\ 183.52`

`:.\ text(Interest paid)`	`=32\ 183.52-28\ 000`
	`=$4183.52`

Financial Maths, STD2 F4 2010 HSC 28a

The table shows monthly home loan repayments with interest rate changes from February to October 2009.

What is the change in monthly repayments on a $250 000 loan from February 2009 to April 2009? (1 mark)

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Xiang wants to borrow $307 000 to buy a house.

Xiang’s bank approves loans for customers if their loan repayments are no more than 30% of their monthly gross salary.

Xiang’s monthly gross salary is $6500.

If she had applied for the loan in October 2009, would her bank have approved her loan?

Justify your answer with suitable calculations. (3 marks)

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Jack took out a loan at the same time and for the same amount as Xiang.

Graphs of their loan balances are shown.

Identify TWO differences between the graphs and provide a possible explanation for each difference, making reference to interest rates and/or loan repayments. (2 marks)

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Show Answers Only

a. `text(Monthly repayments decrease by $15)`

b. `text(S)text(ince repayments of 1987.29 > 1950, the loan)`

`text(would not have been approved.)`

c. `text(Differences)`

`text(1. Jack’s loan balance falls more sharply for first 12 years)`

`text(2. Jack’s loan balance falls less sharply between years 12-30.)`

`text(Explanation)`

`text(1. Jack made larger repayments for first 12 years, OR)`

`text(Jack made the same repayments but had a lower interest rate)`

`text(for the first 12 years.)`

`text(2. Jack made smaller repayments in years 12 – 30.)`

Show Worked Solution

a.	`text(Repayment)\ text{(Feb 09)}`	`= 1588`
	`text(Repayment)\ text{(Apr 09)}`	`= 1573`

`text(Difference) = 1588-1573 = 15`

`:.\ text(Monthly repayments decrease by $15)`

♦ Mean mark 39%
MARKER’S COMMENT: Borrowing $307,000 can be achieved by borrowing $300,000, and then `7` times the table repayment value for borrowing $1000.

b. `text(Loan) = $307\ 000`

`text{Repayments (Oct 09)}`	`= 1942 + (7 xx 6.47)`
	`= 1942 + 45.29`
	`= $1987.29\ text(per month)`

`text(30% Gross salary)`	`= 6500 xx\ text(30%)`
	`= $1950\ text(per month)`

`:.\ text(S)text(ince repayments of $1987.29 > $1950, the loan)`

`text(would not have been approved.)`

c. `text(Differences)`

`text(1. Jack’s loan balance falls more sharply for first 12 years.)`

`text(2. Jack’s loan balance falls less sharply between years 12-30.)`

`text{Explanation(s)}`

♦ Mean mark 36%
MARKER’S COMMENT: Explanations were generally poor and many failed to refer directly to the graphs shown, or reference Xiang or Jack directly.

`text(1. Jack made larger repayments for 1st 12 years, OR)`

`text(Jack made the same repayments but had a)`

`text(lower interest rate for the first 12 years.)`

`text(2. Jack made smaller repayments in years 12-30.)`

Financial Maths, STD2 F4 2011 HSC 22 MC

Ying borrowed $250 000 to buy a house. The interest rate and monthly repayment for her loan are shown in the spreadsheet.

What is the total interest charged for the first four months of this loan?

$6364.32
$6366.11
$6369.67
$6376.25

Show Answers Only

`A`

Show Worked Solution

`text(Month 3)`

`P+I-R`	`=251\ 032.04-1871.94`
	`=249\ 160.10`

`text(Month 4)`

`P`	`=249\ 160.10`
`:.I`	`=249\ 160.10xx0.0765/12`
	`=1588.40`

`:.\ text(Total interest)`	`=1593.75+1591.98+1590.19+1588.40`
	`=6364.32`

`=> A`

Financial Maths, STD2 F4 2009 HSC 26c

Margaret borrowed $300 000 to buy an apartment. The interest rate is 6% per annum, compounded monthly. The repayments were set by the bank at $2200 per month for 20 years.

The loan balance sheet shows the interest charged and the balance owing for the first month.

What is the total amount that is to be paid for this loan over the 20 years? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Find the values of `A` and `B`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `$528\ 000`

b. `A = $1496.50, B = $298\ 596.50`

Show Worked Solution

♦ Mean mark 39%
MARKER’S COMMENT: 1 mark allocation flags the answer should not be too involved.

a. `text(Monthly repayment) = $2200`

`text(# Repayments)\ = 20 xx 12 = 240`

`:.\ text(Total paid)`	`= 2200 xx 240`
	`= $528\ 000`

b. `text(Interest rate monthly)\ = text(6%)/12=\ text(0.5%)`

`A`	`= text(Principal at start of month) xx 0.5/100`
	`= 299\ 300 xx 0.5/100`
	`= $1496.50`

`B`	`=\ text(Principal + interest – repayment)`
	`= 299\ 300 + 1496.50-2200`
	`= $298\ 596.50`

Financial Maths, STD2 F4 2012 HSC 24 MC

A $400 000 loan can be repaid by making either monthly or fortnightly repayments.

The graph shows the loan balances over time using these two different methods of repayment.

The monthly repayment is $2796.86 and the fortnightly repayment is $1404.76.

What is the difference in the total interest paid using the two different methods of
repayment, to the nearest dollar?

$51 596
$166 823
$210 000
$234 936

Show Answers Only

`B`

Show Worked Solution

`text(Monthly repayment)`	`= $2796.86`
`text(# Repayments)`	`= 30 xx 12 = 360`
`text(Total repaid)`	`= 360 xx 2796.86`
	`= $1\ 006\ 869.60`
`text(Total interest)`	`= 1\ 006\ 869.60-400\ 000`
	`=$606\ 869.60`

`text(Fortnightly payment)`	`= $1404.76`
`text(# Repayments)`	`= 23 xx 26 = 598`
`text(Total repaid)`	`= 598 xx 1404.76`
	`=$840\ 046.48`
`text(Total interest)`	`= 840\ 046.48-400\ 000`
	`= $440\ 046.48`

`:.\ text(Difference in interest)`	`= 606\ 869.60-440\ 046.48`
	`= $166\ 823\ text((nearest dollar))`

`=> B`

\(\text{Total Cost}\ \)	\(=\ \text{Price + Transfer + Stamp Duty}\)
	\(=24\ 200+50+726\)
	\(=$24\ 976\)

\(\text{Loan amount}\ \)	\(=\ \text{Total Cost + Interest}\)
	\(=24\ 976+5095.104\)
	\(=$30\ 071.104\)

\(\text{Monthly repayment}\)	\(=\dfrac{30\ 071.104}{36}\)
	\(=$835.308444\)
	\(\approx $835.31\)

\(\text{Repayments}\)	\( =2228.40\times 12\times 5\)
	\(=$133\,704\)

\(\text{Interest}\)	\(=133\,704-121\,000\)
	\(=$12\,704\)