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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)

Show Answers Only

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

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 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.
 

  1. What are the values of `A` and `B`?  (2 marks)
  2. 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.
  3. How much less will Frankie pay overall by making the lump sum payment?  (3 marks)
Show Answers Only
  1. `A=$1198.29,\ \ B=$199\ 139.86`
  2. `$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)

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 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
  1. `A = $19\ 000,quadB = $17\ 551.33`
  2. `$74.56`
  3. `$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 SM-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

A.   $191

B.   $446

C.   $612

D.   $752

Show Answers Only

`D`

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

`=>D`

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\ 230`

`= $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
  1. `text{$130 000 (over 30 years)}`
  2. `$45\ 964.80`
Show Worked Solution

i.  `text(From table)`

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

 

ii.  `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
  1. `$1162.50`
  2. `$129\ 000`
  3. `$36\ 270`
Show Worked Solution

i.   `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`

 

ii.  `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`

 

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

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 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
  1. `$528\ 000`
  2. `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.

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

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

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

 

ii.    `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`