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Financial Maths, STD1 F3 2025 HSC 24

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)

Show Answers Only

\($835.31\)

Show Worked Solution

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

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

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

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

 
\(\text{3 years}= 3 \times 12=36\ \text{months}\)

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

♦♦ Mean mark 46%.

Financial Maths, STD1 F3 2021 HSC 27

Tracy takes out a 30-year reducing balance loan of $680 000 to buy a house. Interest is charged at 0.25% per month. The loan is to be repaid in equal monthly instalments of $2866.91 over a term of 30 years.

Part of a spreadsheet used to model the reducing balance loan is shown.
 

   

  1. Find the amount owing at the end of the second month.   (2 marks)

  2. Suppose that the interest rate reduces to 0.15% per month and the monthly instalments remain as $2866.91.
  3. What will happen to the term of the loan? Explain your answer without using calculations.   (2 marks)

Show Answers Only

a.    `$677\ 663.26`

b.    `text(Loan term will decrease.)`

`text(The repayment will pay down more of the principal each)`

`text(month, reducing the term of the loan.)`

Show Worked Solution

a.   `text(In month 2:)`

♦♦ Mean mark part (a) 31%.

`text(Interest) = 678\ 833.09 xx 0.0025 = $1697.08`

`text(Repayment) = $2866.91`

`text(Balance owing)` `= 678\ 833.09 + 1697.08-2866.91`
  `= $677\ 663.26`

 

b.   `text(The term of the loan will decrease.)`

♦♦♦ Mean mark part (b) 18%.

`text(If interest rate reduces, the monthly interest amount)`

`text(payable decreases.)`

`text(The repayment will pay down more of the principal each)`

`text(month, reducing the term of the loan.)`

Financial Maths, STD1 F3 2020 HSC 30

Colin takes out a 5-year reducing balance loan of $19 000 with interest charged at 6% per annum.

He uses this money to buy a car valued at $19 000.

The table shows some of the output from a spreadsheet used to model the reducing balance loan.
 


 

Colin's car is depreciated using the declining-balance method, with a depreciation rate of 20% per annum.

At the end of 3 years, after making the third repayment on the loan, Colin sells the car at its salvage value. He uses the money from the sale of the car to repay the amount owing on the loan at the end of the third year.

How much money will he have left over?   (4 marks)

Show Answers Only

`$1458.43`

Show Worked Solution

`V_0 = 19\ 000 \ , \ r = 20text(%) \ , \ n=3`

♦ Mean mark 23%.
`S` `= V_0 (1-r)^n`
  `= 19\ 000 (1-0.2)^3`
  `= $9728`

 

`text{Find the amount owing on the loan after 3 years:}`

`text(Using the table,)`

`text{Interest (year 3)} = 0.06 xx 12\ 056.70 = $ 723.40`

`text{Amount owing (end of year 3)`

`= 12\ 056.70 + 723.40-4510.53`

`= $ 8269.57`

 

`therefore \ text{Money left over}`

`= 9728-8269.57`

`= $1458.43`

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)

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)

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. 
 

2ug-2016-hsc-q27_1

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

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

  3. Marge repaid this loan over four years.

     

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

Show Answers Only

i.    `A = $19\ 000,quadB = $17\ 551.33`

ii.   `$74.56`

iii.  `$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 13 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.
 

2015 29b

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

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.

2006 27a

  1. 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)

  2. 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)

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.

2004 27a

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

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

  3. 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)

Show Answers Only

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?

  1. $418 496
  2. $418 520
  3. $445 952
  4. $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.
 

2UG-2005-10MC
 

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

What is her monthly loan repayment?

  1. $143.40
  2. $669.20
  3. $8030.40
  4. $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.
 

VCAA 2008 15 mc
 

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

  1. $3200 
  2. $5600
  3. $19 200
  4. $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.

2010 25b

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

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.

 2010 28a

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

  2. 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)

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

     

    Graphs of their loan balances are shown.
     
          2010 28a2

    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)

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.

2UG 2011 22

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

  1. $6364.32
  2. $6366.11
  3. $6369.67
  4. $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.

2009 26c 

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

  2. Find the values of `A` and `B`.   (2 marks)

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.
 

2012 24 mc

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?

  1. $51 596
  2. $166 823
  3. $210 000
  4. $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`