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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}
  1. Find the total amount Maya pays for her purchase if repaying in full on 30 March 2026.   (1 mark)
  2. 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)

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.    \(r = 16\% = 0.16, \quad n = \dfrac{8 \times 7}{365} = \dfrac{56}{365}\)

\(I= Prn = 120 \times 0.16 \times \dfrac{56}{365}= 2.945\ldots = \$2.95\)

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

Algebra, STD2 EQ-Bank 25

A local bowling club sold memberships for the new season.

Senior memberships \((s)\) cost $70 each and junior memberships \((j)\) cost $45 each.

On a particular day, a total of 19 memberships were sold, with sales totalling $1030.

  1. Write two equations, in terms of \(s\) and \(j\), to represent this information.   (1 mark)

  2. Determine the exact number of senior and junior memberships sold.   (2 marks)
Show Answers Only

a.    \(70s+45j=1030\ …\ (1)\)

\(s+j=19\ …\ (2)\)

b.    \(s=7,\ \ j=12\)

Show Worked Solution

a.    \(70s+45j=1030\ …\ (1)\)

\(s+j=19\ …\ (2)\)
 

b.    \(\text{Rearranging (2) above:}\)

\(s=19-j\)

\(\text{Substitute}\ \ s=19-j\ \ \text{into (1):}\)

\(70(19-j)+45j\) \(=1030\)
\(1330-70j+45j\) \(=1030\)
\(25j\) \(=1330-1030=300\)
\(j\) \(=12\)

 
\(\text{Substitute}\ \ j=12\ \ \text{into (2):}\)

\(s=19-12=7\)

\(\therefore\ \text{7 senior and 12 junior memberships were sold.}\)

Algebra, STD2 EQ-Bank 24

A bookshop sold a number of biographies at a weekend sale.

Hardcover biographies \((h)\) sold for $25 each and paperback biographies \((p)\) sold for $12 each.

A total of 11 biographies were sold, with total sales revenue of $210.

  1. Write two equations, in terms of \(h\) and \(p\), to represent this information.   (1 mark)

  2. Determine the number of hardcover biographies and the number of paperback biographies sold.   (2 marks)

Show Answers Only

\(h=6, \ p=5\)

Show Worked Solution

a.    \(25h+12p=210\ …\ (1)\)

\(h+p=11\ …\ (2)\)
 

b.    \(\text{Rearranging (2) above:}\)

\(h=11-p\)

\(\text{Substitute}\ \ h=9-p\ \ \text{into (1):}\)

\(25(11-p)+12p\) \(=210\)
\(275-25p+12p\) \(=210\)
\(13p\) \(=275-210=65\)
\(p\) \(=5\)

 
\(\text{Substitute}\ \ p=5\ \ \text{into (2):}\)

\(h=11-5=6\)

\(\therefore\ \text{6 hardcover and 5 paperback biographies were sold.}\)

Algebra, STD2 EO-Bank 30

The number of monthly subscribers \((S)\) to a print magazine is decreasing. The number of subscribers is modelled using

\(S=k(1.06)^{-t}\)  for  \( t \geq 0,\)

where \(t\) is time in months.

The number of subscribers today is 12 000.

  1. Use the model to estimate the number of subscribers 10 months from today.   (2 marks)

  2. The number of subscribers needs to always be above 3000.
  3. Explain why this model is NOT appropriate to use in this case.   (1 mark)

Show Answers Only

a.    \(S=12\,000(1.06)^{-10}=6700.73…=6701\ \text{(nearest subscriber)}\)

b.    \(\text{When}\ \ t=30\ \ \Rightarrow\ \ S \approx 2089\)

\(\text{Since the number of subscribers needs to always be over 3000, it is not appropriate.}\)

Show Worked Solution

a.    \(\text{When}\ \ t=0,\ \ S=12\,000:\)

\(12\,000=k(1.06)^{0}\ \ \Rightarrow\ \ k=12\,000\)

\(\text{Find}\ S\ \text{when}\ \ t=10:\)

\(S=12\,000(1.06)^{-10}=6700.73…=6701\ \text{(nearest subscriber)}\)
 

b.    \(\text{When}\ \ t=30:\)

\(S=12\,000(1.06)^{-30} \approx 2089\)

\(\text{Since the number of subscribers needs to always be over 3000, it is not appropriate.}\)

Algebra, STD2 EQ-Bank 30

The number \((F)\) of fish in Lake Mulloway is decreasing. The number of fish is modelled using

\(F=k(1.05)^{-t}\)  for  \( t \geq 0,\)

where \(t\) is time in years.

The number of fish in Lake Mulloway after 1 year is 4620.

  1. Using the model, estimate the number of fish in Lake Mulloway today.   (2 marks)

  2. The local newsgroup on socials has used the model to predict the number of fish that will be in Lake Mulloway in 40 years time.
  3. If the number of fish in the lake needs to always be above 1000, explain why this model is NOT appropriate to use in this case.   (1 mark)
Show Answers Only

a.    \(F=4851\)

b.    \(\text{When}\ \ t=40\ \ \Rightarrow\ \ F \approx 689\)

\(\text{Since the number of fish needs to always be over 1000, it is not appropriate.}\)

Show Worked Solution

a.    \(\text{When}\ \ t=1,\ \ F=4620:\)

\(4620=k(1.05)^{-1}\ \ \Rightarrow\ \ k=4620 \times 1.05=4851\)

\(\text{Find}\ F\ \text{when}\ \ t=0:\)

\(F=4851(1.05)^{0}=4851\)
 

b.    \(\text{When}\ \ t=40:\)

\(F=4851(1.05)^{-40} \approx 689\)

\(\text{Since the number of fish needs to always be over 1000, it is not appropriate.}\)

Algebra, STD2 EO-Bank 29

The number \((N)\) of native birds in a wildlife reserve is decreasing. The number of birds is modelled using

\(N=k(1.08)^{-t}\)  for \( t \geq 0,\)

where \(t\) is time in years.

The number of birds in the reserve today is 8000.

  1. Use the model to estimate the number of birds in the reserve 12 years from today.   (2 marks)

  2. The number of birds in the reserve needs to always be above 1500.
  3. Explain why this model is NOT appropriate to use in this case.   (1 mark)

Show Answers Only

a.    \(N=8000(1.08)^{-12}=3176.91…=3177\ \text{(nearest bird)}\)

b.    \(\text{When}\ \ t=25:\)

\(N=8000(1.08)^{-25}=1168.14…\)

\(\text{Since the number of birds needs to always be over 1500, it is not appropriate.}\)

Show Worked Solution

a.    \(\text{When}\ \ t=0,\ \ N=8000:\)

\(8000=k(1.08)^{0}\ \ \Rightarrow\ \ k=8000\)

\(\text{Find}\ N\ \text{when}\ \ t=12:\)

\(N=8000(1.08)^{-12}=3176.91…=3177\ \text{(nearest bird)}\)
 

b.    \(\text{When}\ \ t=25:\)

\(N=8000(1.08)^{-25}=1168.14…\)

\(\text{Since the number of birds needs to always be over 1500, it is not appropriate.}\)

Statistical, STD2 EQ-Bank 30

A teacher surveyed the students in her Year 8 class to investigate the relationship between the number of hours of phone use per day and the number of hours of sleep per day.

The results for five students are shown on the scatterplot. The least-squares regression line is also shown.
 

         

  1. Calculate Pearson's correlation coefficient \((r)\), to 4 decimal places, and describe the relationship between number of hours of sleep per day and number of hours of phone use per day.   (3  marks)

  2. Find the equation of the least-squares regression line.   (2  marks)

  3. From the data, the median number of hours of phone use per day, \(a\), and the median of the number of hours of sleep per day, \(b\), are to be calculated.
  4. By finding the coordinates \((a, b)\), determine whether this point would lie on, below or above the least-squares regression line.   (3  marks)

Show Answers Only

a.    \(r=-0.9414\)

\(\text{There is a strong, negative linear relationship between number of hours of sleep}\)

\(\text{per day and number of hours of phone use per day.}\)
 

b.    \(\text{By calculator (inputting all data points):}\)

\(y=-0.591 x+9.818 \ \text{(3 d.p.)}\)
 

c.    \(x\text{-values of data points:}\ {0,2,2,3,5}\)

\(\text{Median of the number of hours of phone use = 2 hours}\)

\(y\text{-values of data points:}\ {7,8,8,9,10}\)

\(\text{Median of the number of hours of sleep = 8 hours}\)

\((a,b) = (2,8)\ \ \Rightarrow\ \ \text{this point lies below the LSRL.}\)

Show Worked Solution

a.    \(r=-0.9414\)

\(\text{There is a strong, negative linear relationship between number of hours of sleep}\)

\(\text{per day and number of hours of phone use per day.}\)
 

b.    \(\text{By calculator (inputting all data points):}\)

\(y=-0.591 x+9.818 \ \text{(3 d.p.)}\)
 

c.    \(x\text{-values of data points:}\ {0,2,2,3,5}\)

\(\text{Median of the number of hours of phone use = 2 hours}\)

\(y\text{-values of data points:}\ {7,8,8,9,10}\)

\(\text{Median of the number of hours of sleep = 8 hours}\)

\((a,b) = (2,8)\ \ \Rightarrow\ \ \text{this point lies below the LSRL.}\)

Networks, STD2 EQ-Bank 30

A weighted and directed network diagram is shown.
 

  1. What is the outflow from vertex \(C\) ?   (1 mark)

  2. Calculate the maximum flow through the network.   (2 marks)

  3. The capacity of ONE saturated edge is to be increased in order to produce a new network with the maximum possible flow.
  4. State an edge which could have an increased capacity AND find the new maximum flow through this new network.   (2 marks)

Show Answers Only

a.    \(\text{Inflow of \(C\) = Outflow of \(C\) = 14}\)

b.    \(\text{Method 1:}\)

\(\text{Minimum cut through}\ \ sC-Bt-At:\)

\(\text{Maximum flow} = 14+20+11=45\)
 

\(\text{Method 2:}\)
 

     

\(sAt\ \ 11\ \text{(leaving an excess flow capacity of } s A=11 \text { )}\)

\(sABt\ \ 11\ \text{(leaving an excess flow capacity of } AB=7 \text { and } B t=9 \text { )}\)

\(sBt\ \ 9\ \text{(leaving an excess flow capacity of } s B=7 \text { )}\)

\(sCt\ \ 14\ \text{(leaving an excess flow capacity of } C t=3)\)

\(\text{Maximum Flow} = 11+11+9+14=45\)
 

c.    \(\text{Answers could include one of the following:}\)

\(\text{B} t \ \text{could be increased (by 7) leading to a maximum flow of 52.}\)

\(\text{A} t \ \text{could be increased (by 11) leading to a maximum flow of 52.}\)

Show Worked Solution

a.    \(\text{Inflow of \(C\) = Outflow of \(C\) = 14}\)

b.    \(\text{Method 1:}\)

\(\text{Minimum cut through}\ \ sC-Bt-At:\)

\(\text{Maximum flow} = 14+20+11=45\)
 

\(\text{Method 2:}\)
 

     

\(sAt\ \ 11\ \text{(leaving an excess flow capacity of } s A=11 \text { )}\)

\(sABt\ \ 11\ \text{(leaving an excess flow capacity of } AB=7 \text { and } B t=9 \text { )}\)

\(sBt\ \ 9\ \text{(leaving an excess flow capacity of } s B=7 \text { )}\)

\(sCt\ \ 14\ \text{(leaving an excess flow capacity of } C t=3)\)

\(\text{Maximum Flow} = 11+11+9+14=45\)
 

c.    \(\text{Answers could include one of the following:}\)

\(\text{B} t \ \text{could be increased (by 7) leading to a maximum flow of 52.}\)

\(\text{A} t \ \text{could be increased (by 11) leading to a maximum flow of 52.}\)

Algebra, STD2 EQ-Bank 29

The population ( \(P\) ) of a town is reducing. The population is modelled using

\(P=k(1.1)^{-t}\)  for  \( t \geq 0,\)

where \(t\) is time in years.

The population of the town today is 5000 .

  1. Use the model to estimate the population of the town 15 years from today.   (2 marks)

  2. The population of the town needs to always be above 500 .
  3. Explain why this model is NOT appropriate to use in this case.   (1 mark)

Show Answers Only

a.    \(P=5000(1.1)^{-15}=1196.96…=1197\ \text{(nearest person)}\)

b.    \(\text{When}\ \ t=30\ \ \Rightarrow\ \ P \approx 287\)

\(\text{Since the population needs to always be over 500, it is not appropriate.}\)

Show Worked Solution

a.    \(\text{When}\ \ t=0,\ \ P=5000:\)

\(5000=k(1.1)^{0}\ \ \Rightarrow\ \ k=5000\)

\(\text{Find}\ P\ \text{when}\ \ t=15:\)

\(P=5000(1.1)^{-15}=1196.96…=1197\ \text{(nearest person)}\)
 

b.    \(\text{When}\ \ t=30\ \ \Rightarrow\ \ P \approx 287\)

\(\text{Since the population needs to always be over 500, it is not appropriate.}\)

Measurement, STD2 EQ-Bank 20

The travel graph displays Jamie's trip which began at town `M` at 8 am and finished at town `N`.
 

   

  1. How far apart are the two towns?   (1 mark)

  2. At what time during the day did Jamie arrive back at town `M` ?   (1 mark)

  3. What was the total distance that Jamie travelled?   (1 mark)

  4. Between which times in the day was Jamie travelling at the fastest speed? Justify your answer, without calculations.   (1 mark)

Show Answers Only

a.   `140\ text{km}`

b.   `10\ text{am}`

c.    `220\ \text{km}`  

d.    `\text{Between 11 – 11:30 am}`

Show Worked Solution

a.   `140\ text{km}`
 

b.   `text{James arrives back at town when he is 140 km away (2nd time).}`

`=> 10\ text{am}`
 

c.    `\text{Total Distance} = 40 + 40 + 60+ 80=220\ text{km}` 
  

d.   `text{Fastest speed → graph is the steepest (either up or down)}`

`:.\ text{Fastest speed between 11 – 11:30 am}`

Measurement, STD2 EQ-Bank 19

The travel graph displays Nikau's car trip along a straight road from home and back again. The trip has been broken into four separate sections: `A`, `B`, `C`  and `D`.
 

   

  1. How far did Nikau travel in total?   (1 mark)

  2. In which section of the trip, `A`, `B`, `C` and `D`, did Nikau travel the fastest?   (1 mark)

Show Answers Only

a.    `400\ text(km)`

b.    `text(S)text(ection)\ D`

Show Worked Solution

a.    `text(Distance travelled)\ = 2 xx 200= 400\ text(km)`
 

b.    `text{Fastest section has the steepest slope (in either direction).}`

`:. text(S)text(ection)\ B\ text(was the fastest)`

`\ rightarrow text(50 km/30 mins) =100 text(km/hour)`

Measurement, STD2 EQ-Bank 27

Sue walks along a trail, starting at 7 am and finishing at 10 am. The travel graph shows Sue’s journey from the start to the finish. The journey has been broken into six sections, `A`, `B`, `C`, `D`, `E` and `F`.
 

     

  1. In which section of the journey did Sue travel fastest? Justify your answer.   (2 marks)

  2. Kim walked along the same trail, also starting at 7 am and finishing at 10 am. Kim walked at a constant speed for the entire journey.
  3. By showing Kim’s journey on the grid above, determine between what times Sue was ahead of Kim.   (3 marks)

Show Answers Only

a.    `text(S)text(ection)\ C\ \text{(Slope is the steepest)}.`

b.    `text(8:30 am – 9:15 am)`
 
           

Show Worked Solution

a.   `text{Fastest travel occurs when the slope is the steepest.}`

`=>\ text{Section}\ C`

b.

`text(Sue was ahead when her graph is higher than Kim’s.)`

`:.\ text(She was ahead between 8:30 am – 9:15 am)`

Measurement, STD2 EQ-Bank 35

The distance-time graph shows the first two stages of a car journey from home to a holiday house.
  

  1. At what speed, in metres per second, did the car travel during stage \(A\) of the journey? Give your answer correct to one decimal place.   (2 marks)
  2. After stage \(B\), the car continues to travel towards the holiday house at a constant speed of \(50\ \text{km/h}\) for 2 hours. Graph this part of the journey on the grid above.   (2 marks)

Show Answers Only

a.    \(27.8\ \text{m/s}\)

b.   

Show Worked Solution

a.    \(\text{Distance travelled} = 150\ \text{km}\ =150\,000\ \text{m}\)

\(\text{Stage A duration = 1.5 hours}\)

\(\text{Express 1.5 hours in minutes:}\)

\(\text{Minutes} = 1.5 \times 60 \times 60 = 5400\ \text{seconds}\)

\(\text{Speed} = \dfrac{150\,000}{5400}=27.77…=27.8\ \text{m/s}\)

 

b.    \(\text{Position after 2 hours at 50km/h}=(150+2\times 50 , 2 + 2) = (250 , 4)\)
 

   

Networks, STD2 EQ-Bank 26

A project requires the completion of 9 activities, \(A\) to \(I\). The project is due to be completed in 13 days.

The directed network diagram shows these activities with their completion times in days.
 

   

  1. Use the information from the network diagram to complete the Gantt chart, including the critical path and all other activities required for the project.   (3 marks)

       

  2. List all activities, not on the critical path, which could be occurring at midday on day 8.   (1  mark)

Show Answers Only

a.    
     

b.    \(C\ \text{and}\ E\)

Show Worked Solution

a.    
         

 
b.    \(\text{By inspection of the Gantt chart}\)

\(\text{Activities which could be occurring ~7.5 on chart (midday day 8):}

\(C\ \text{and}\ E\)

Probability, STD2 EQ-Bank 32

In Year 11 there are 80 students. Of these, 50 play a sport \((S), 25\) are involved in debating ( \(D\) ), and 20 do neither.

  1. Using this information, complete the Venn diagram.   (2 marks)

 
               

  1. A Year 11 student is selected at random.
  2. What is the probability that the student plays a sport and is involved in debating?   (1 mark)

  3. Two Year 11 students are selected at random.
  4. What is the probability that both students are involved in debating only?   (2 marks)

Show Answers Only

a.    \(\text{Venn diagram}\)

 

b.    \(\dfrac{3}{16}\)

c.    \(\dfrac{9}{632} \)

Show Worked Solution

a.    \(\text{Venn diagram}\)

 

 

b.    \(\text{Using the Venn diagram:}\)

\(P\text{(plays both)} = \dfrac{15}{80}=\dfrac{3}{16}\)
 

c.    \(P\text{(both students involved in debating only)}\)

\(=\dfrac{10}{80} \times\ \dfrac{9}{79}=\dfrac{9}{632} (\approx 0.0142)\)

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.
 

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

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 F1 EQ-Bank 30

Aanya is buying a used car with a sale price of \(\$15\,800\). 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.

Aanya is considering two loan options to finance the total amount payable for the car.

Loan A 

 - A bank loan with simple interest at 6.5% per annum.

 - Repaid in equal monthly repayments over 4 years.

Loan B

 - A dealership loan with simple interest at 5.8% per annum

 - Repaid in equal monthly repayments over 5 years.
 

  1. Calculate the total amount Aanya needs to borrow.   (1 mark)
  2. Calculate the monthly repayment for each loan option, correct to the nearest cent and identify which loan has the lowest monthly repayment.   (3 marks)
  3. Identify ONE disadvantage of choosing this loan compared to the other.   (1 mark)
Show Answers Only

a.    \(\$16\,324\)

b.  \(\text{Loan B has the lowest monthly repayment (\$350.97 < \$428.51)}\)

c.  \(\text{Loan B costs}\ \ 21\,057.96-20\,568.24=\$489.72\ \ \text{more in total repayments.}\)

Show Worked Solution

a.    \(\text{Calculate total amount to borrow:}\)

\(\text{Stamp duty}=\dfrac{15\,800}{100}\times 3=158\times 3=\$474\)

\(\text{Total borrowed}=15\,800+50+474=\$16\,324\)
 

b.    \(\text{Loan A:}\)

\(I=Prn=16\,324\times 0.065\times 4=\$4244.24\)

\(\text{Total to repay}=16\,324+4244.24=\$20\,568.24\)

\(\text{Monthly repayment}=\dfrac{20\,568.24}{48}=428.505\approx \$428.51\)
 

\(\text{Loan B:}\)

\(I=Prn=16\,324\times 0.058\times 5=\$4733.96\)

\(\text{Total to repay}=16\,324+4733.96=\$21\,057.96\)

\(\text{Monthly repayment}=\dfrac{21\,057.96}{60}=350.966\approx \$350.97\)
 

\(\text{Loan B has the lowest monthly repayment (\$350.97 < \$428.51)}\)
 

c.    \(\text{Consider the total repayments of each loan (see part b):}\)

\(\text{Loan B costs}\ \ 21\,057.96-20\,568.24=\$489.72\ \ \text{more in total repayments.}\)

Financial Maths, STD2 F1 EQ-Bank 29

A second-hand motorbike has a sale price of $12 400. In addition to the sale price, the following costs are charged:

  • transfer of registration $85
  • stamp duty which is calculated at $4.50 for every $100, or part thereof, of the sale price.

Jordan borrows the total amount to be paid for the motorbike, including transfer of registration and stamp duty. Simple interest at the rate of 8.5% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 2 years.

Calculate Jordan's monthly repayment, correct to the nearest cent.   (5 marks)

Show Answers Only

\(\$635.85\)

Show Worked Solution

\(\text{Stamp duty}=\dfrac{12\,400}{100}\times 4.5=\$558\)

\(\text{Total borrowed}=12\,400+85+558=\$13\,043\)

\(\text{Interest} =Prn=13\,043\times 0.085\times 2=\$2217.31\)

\(\text{Total to repay}=13\,043+2217.31=\$15\,260.31\)

 
\(\therefore\ \text{Monthly repayment}=\dfrac{15\,260.31}{24}=635.846\approx \$635.85\)

Financial Maths, STD1 F3 2025 HSC 24 (Adapted)

A used car has a sale price of \(\$18\,600\). 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.

Tahlia borrows the total amount to be paid for the car, including transfer of registration and stamp duty. Simple interest at the rate of 7.2% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 4 years.

Calculate Tahlia's monthly repayment, correct to the nearest cent.   (5 marks)

Show Answers Only

\(\$515.41\)

Show Worked Solution

\(\text{Stamp duty}=\dfrac{18\,600}{100}\times 3=\$558\)

\(\text{Total borrowed}=18\,600+50+558=\$19\,208\)

\(\text{Interest}=Prn=19\,208\times 0.072\times 4=\$5531.904\)

\(\text{Total to repay}=19\,208+5531.904=\$24\,739.904\)

 
\(\therefore\ \text{Monthly repayment}=\dfrac{24\,739.904}{48}=515.4146…\approx \$515.41\)

Financial Maths, STD2 F1 EQ-Bank 28

At the end of the 2024-2025 financial year, Marcus had a taxable income of $168 000.

  1. The table shows the income tax rate for Australian residents for the 2024-2025 financial year.

\(\begin{array} {|l|l|}\hline \rule{0pt}{2.5ex}\text{ Taxable income}\rule[-1ex]{0pt}{0pt} & \text{ Tax payable}\\\hline \rule{0pt}{2.5ex}\text{\$0 – \$18 200}\rule[-1ex]{0pt}{0pt} & \text{Nil}\\\hline \rule{0pt}{2.5ex}\text{\$18 201 – \$45 000}\rule[-1ex]{0pt}{0pt} & \text{16 cents for each \$1 over \$18 200}\\\hline\rule{0pt}{2.5ex}\text{\$45 001 – \$135 000}\rule[-1ex]{0pt}{0pt} & \text{\$4288 plus 30 cents for each \$1 over \$45 000}\\\hline\rule{0pt}{2.5ex}\text{\$135 001 – \$190 000}\rule[-1ex]{0pt}{0pt} & \text{\$31 288 plus 37 cents for each \$1 over \$135 000}\\\hline\rule{0pt}{2.5ex}\text{\$190 001 and over}\rule[-1ex]{0pt}{0pt} & \text{\$51 638 plus 45 cents for each \$1 over \$190 000}\\\hline\end{array}\)

  1. Using the table, calculate Marcus's tax payable.   (3 marks)

  2. The Medicare levy is 2% of taxable income.
  3. Calculate the Medicare levy payable by Marcus.   (1 mark)

Show Answers Only

a.    \(\$43\,498\)

b.    \(\$3360\)

Show Worked Solution

a.    \(\text{Calculate tax payable:}\)

\(\text{Tax payable}\) \(=31\,288+0.37\times (168\,000-135\,000)\)
  \(=31\,288+0.37\times 33\,000\)
  \(=31\,288+12\,210\)
  \(=\$43\,498\)

 

b.    \(\text{Calculate Medicare levy:}\)

\(\text{Medicare levy}=0.02\times 168\,000=\$3360\)

Financial Maths, STD1 F1 2025 HSC 19 (Adapted)

At the end of the 2024-2025 financial year, Priya's taxable income was $92 400.

  1. The table shows the income tax rate for Australian residents for the 2024-2025 financial year.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\text{    Taxable income}\rule[-1ex]{0pt}{0pt} & \text{    Tax payable}\\
\hline
\rule{0pt}{2.5ex}\text{\$0 – \$18 200}\rule[-1ex]{0pt}{0pt} & \text{Nil}\\
\hline
\rule{0pt}{2.5ex}\text{\$18 201 – \$45 000}\rule[-1ex]{0pt}{0pt} & \text{16 cents for each \$1 over \$18 200}\\
\hline
\rule{0pt}{2.5ex}\text{\$45 001 – \$135 000}\rule[-1ex]{0pt}{0pt} & \text{\$4288 plus 30 cents for each \$1 over \$45 000}\\
\hline
\rule{0pt}{2.5ex}\text{\$135 001 – \$190 000}\rule[-1ex]{0pt}{0pt} & \text{\$31 288 plus 37 cents for each \$1 over \$135 000}\\
\hline
\rule{0pt}{2.5ex}\text{\$190 001 and over}\rule[-1ex]{0pt}{0pt} & \text{\$51 638 plus 45 cents for each \$1 over \$190 000}\\
\hline
\end{array}

  1. Using the table, calculate Priya's tax payable.   (3 marks)

  2. The Medicare levy is 2% of taxable income.
  3. Calculate the Medicare levy payable by Priya.   (1 mark)

Show Answers Only

a.    \(\$18\,508\)

b.    \(\$1848\)

Show Worked Solution

a.    \(\text{Calculate tax payable:}\)

\(\text{Tax payable}\) \(=4288+0.30\times (92\,400-45\,000)\)
  \(=4288+0.30\times 47\,400\)
  \(=4288+14\,220\)
  \(=\$18\,508\)

 

b.    \(\text{Calculate Medicare levy:}\)

\(\text{Medicare levy}=0.02\times 92\,400=\$1848\)

Financial Maths, STD1 F1 EQ-Bank 7 MC

Mia earns $32 per hour working as a swim instructor in Wollongong. Her hourly pay rate increases by 4%.

How much will she earn for a 6-hour shift with this increase?

  1. $7.68
  2. $33.28
  3. $192.00
  4. $199.68
Show Answers Only

\(D\)

Show Worked Solution

\(\text{New hourly rate}=32\times 1.04=\$33.28\)

\(\therefore\ \text{Shift earnings}=33.28\times 6=\$199.68\)

\(\Rightarrow D\)

Financial Maths, STD1 F1 2024 HSC 4 MC (Adapted)

Liam works at a cafe in Newcastle and earns $24 per hour. His hourly pay rate increases by 3%.

How much will he earn for a 5-hour shift with this increase?

  1. $3.60
  2. $24.72
  3. $120.00
  4. $123.60
Show Answers Only

\(D\)

Show Worked Solution

\(\text{New hourly rate}=24\times 1.03=\$24.72\)

\(\therefore\ \text{Shift earnings}=24.72\times 5=\$123.60\)

\(\Rightarrow D\)

Financial Maths, STD1 F1 2022 HSC 21 (Adapted)

A real estate agent's commission for selling houses is 3% for the first \(\$600\,000\) of the sale price and 1.8% for any amount over \(\$600\,000\).

Calculate the commission earned in selling a house for \(\$950\,000\).   (2 marks)

Show Answers Only

\(\$24\,300\)

Show Worked Solution

\(\text{Commission on first}\ \$600\,000=3\%\times 600\,000=\$18\,000\)

\(\text{Amount over}\ \$600\,000= 950\,000- 600\,000=\$350\,000\)

\(\text{Commission on remainder}=1.8\%\times 350\,000=\$6300\)

\(\text{Total commission}= 18\,000+ 6300=\$24\,300\)

Financial Maths, STD1 F1 EQ-Bank 20

A real estate agent's commission for selling apartments is 2.5% for the first \(\$400\,000\) of the sale price and 1.5% for any amount over \(\$400\,000\).

Calculate the commission earned in selling an apartment for \(\$720\,000\).   (2 marks)

Show Answers Only

\(\$14\,800\)

Show Worked Solution

\(\text{Commission on first}\ \$400\,000 =2.5\%\times 400\,000=\$10\,000\)

\(\text{Amount over}\ \$400\,000= 720\,000- 400\,000=\$320\,000\)

\(\text{Commission on remainder}=1.5\%\times 320\,000=\$4800\)

\(\text{Total commission}= 10\,000+ 4800=\$14\,800\)

Financial Maths, STD1 F1 EQ-Bank 29

A tutoring business charges $44 per hour for one-on-one lessons. The price includes 10% GST.

  1. Calculate the amount of GST included in a single one-hour lesson.   (1 mark)

  2. In one week, the tutor delivers 25 one-hour lessons. The tutor's costs for the week are $200 for printing, equipment and travel. Calculate the tutor's profit for the week.   (2 marks)

Show Answers Only

a.    \(\$4.00\)

b.    \(\$900\)

Show Worked Solution

a.    \(\text{Calculate GST component:}\)

\(\text{GST is included in the price, so divide by 11:}\)

\(\text{GST}=\dfrac{44}{11}=\$4.00\)

 

b.    \(\text{Calculate profit:}\)

\(\text{Revenue}=25\times 44=\$1100\)

\(\text{Profit}= 1100- 200=\$900\)

Financial Maths, STD1 F1 2021 HSC 12 (Adapted)

A coding bootcamp runs a school holiday course and charges $200 per student.

The costs of running the course are:

  • Instructor: $130 per hour
  • Venue hire: $60 per hour plus 10% GST.

The course runs for 3 hours a day for 5 days.

What profit does the coding bootcamp make if 18 students pay for the course?   (3 marks)

Show Answers Only

\(\$660\)

Show Worked Solution

\(\text{Total hours}=3\times 5=15\ \text{hours}\)

\(\text{Revenue}=18\times 200=\$3600\)

\(\text{Instructor cost}=15\times 130=\$1950\)

\(\text{Venue hire}=15\times 60=\$900\)

\(\text{GST on venue hire}=10\%\times 900=\$90\)

\(\text{Total venue hire}= 900+ 90=\$990\)

\(\text{Total costs}= 1950+ 990=\$2940\)

\(\text{Profit}= 3600- 2940=\$660\)

Financial Maths, STD1 F1 EQ-Bank 26

A photography studio runs a school holiday workshop and charges $180 per student.

The costs of running the workshop are:

  • Photography instructor: $140 per hour
  • Studio hire: $50 per hour plus 10% GST.

The workshop runs for 3 hours a day for 4 days.

What profit does the photography studio make if 15 students pay for the workshop?   (3 marks)

Show Answers Only

\(\$360\)

Show Worked Solution

\(\text{Total hours}=3\times 4=12\ \text{hours}\)

\(\text{Revenue}=15\times 180=\$2700\)

\(\text{Instructor cost}=12\times 140=\$1680\)

\(\text{Studio hire}=12\times 50=\$600\)

\(\text{GST on studio hire}=10\%\times 600=\$60\)

\(\text{Total studio hire}= 600+ 60=\$660\)

\(\text{Total costs}= 1680+ 660=\$2340\)

\(\text{Profit}= 2700- 2340=\$360\)

Financial Maths, STD1 F1 EQ-Bank 12

Ethan has a weekly net income of $580 from his part-time job at JB Hi-Fi. He has created the following budget:

Item Weekly amount
Rent $220
Food $120
Transport $60
Other expenses $50
Savings $130

 
Ethan is saving for an overseas trip that will cost $4680.

  1. How many weeks will it take Ethan to save enough for the trip if he sticks to this budget?   (1 mark)

  2. Ethan decides to save more by reducing his "Food" by $20 per week and reducing his "Other expenses" to $20 per week.
  3. Determine how many fewer weeks it will now take Ethan to save for the trip.   (2 marks)

Show Answers Only

a.    \(\text{36 weeks}\)

b.    \(\text{10 fewer weeks}\)

Show Worked Solution

a.    \(\text{Calculate weeks to save:}\)

\(\text{Weeks}=\dfrac{4680}{130}=36\ \text{weeks}\)

 

b.    \(\text{Calculate new weekly savings:}\)

\(\text{Food reduction}=\$20\)

\(\text{Other expenses reduction}= 50- 20=\$30\)

\(\text{Extra savings per week}= 20+ 30=\$50\)

\(\text{New weekly savings}= 130+ 50=\$180\)

\(\text{New weeks}=\dfrac{4680}{180}=26\ \text{weeks}\)

\(\therefore\ \text{Fewer weeks}= 36- 26=10\ \text{fewer weeks}\)

Financial Maths, STD1 F1 2019 HSC 26 (Adapted)

Harper has a weekly net income of $720. She has created a budget where she allocates this income to rent, food, phone, subscriptions and the rest to savings.

Her budget is shown below, with some details missing.

Item Weekly amount
Rent $300
Food ?
Phone $25
Subscriptions $35
Savings ?

 
Harper allocates 25% of her weekly net income to food.

How many weeks will it take Harper to save $4860?   (3 marks)

Show Answers Only

\(\text{27 weeks}\)

Show Worked Solution

\(\text{Food}=25\%\times 720=0.25\times 720=\$180\)

\(\text{Savings}= 720- 300- 180- 25- 35=\$180\)

\(\text{Weeks}=\dfrac{4860}{180}=27\ \text{weeks}\)

Financial Maths, STD1 F1 EQ-Bank 24

Riley has a weekly net income of $640. He has created a budget where he allocates this income to rent, groceries, transport, entertainment and the rest to savings.

His budget is shown below, with some details missing.

Item Weekly amount
Rent $250
Groceries ?
Transport $55
Entertainment $45
Savings ?

 
Riley allocates 15% of his weekly net income to groceries.

How many weeks will it take Riley to save $5820?   (3 marks)

Show Answers Only

\(\text{30 weeks}\)

Show Worked Solution

\(\text{Groceries}=15\%\times 640=0.15\times 640=\$96\)

\(\text{Savings}= 640- 250- 96- 55- 45=\$194\)

\(\text{Weeks}=\dfrac{5820}{194}=30\ \text{weeks}\)

Financial Maths, STD1 F1 2019 HSC 11 (Adapted)

Jack works as a casual lifeguard and earns $30 per hour. He is also paid a $20 wet-weather allowance for each shift in which he works in the rain.

In one week, Jack worked three shifts of 5 hours each. It rained during two of these shifts.

How much did Jack earn in total this week?   (2 marks)

Show Answers Only

\(\$490\)

Show Worked Solution

\(\text{Total hours}=3\times5=15\ \text{hours}\)

\(\text{Wages}=15\times 30=\$450\)

\(\text{Wet-weather allowances}=2\times 20=\$40\)

\(\text{Total earnings}=450+40=\$490\)

Financial Maths, STD1 F1 EQ-Bank 3 MC

Mason works a 36-hour week at a local cafe in Manly and is paid at an hourly rate of $24. Hours worked on Saturday are paid at time-and-a-half, and hours worked on Sunday are paid at double time.

In a particular week, Mason worked his regular 36 hours plus 4 hours on Saturday and 3 hours on Sunday.

How much did Mason earn in total this week?

  1. $864
  2. $1032
  3. $1152
  4. $1164
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Normal pay}=36\times 24=\$864\)

\(\text{Saturday rate}= 24\times1.5=\$36\ \text{per hour}\)

\(\text{Saturday pay}=4\times 36=\$144\)

\(\text{Sunday rate}= 24\times2=\$48\ \text{per hour}\)

\(\text{Sunday pay}=3\times 48=\$144\)

\(\text{Total pay}= 864+ 144+ 144=\$1152\)

\(\Rightarrow C\)

Financial Maths, STD1 F1 EQ-Bank 4 MC

Lachlan works a 38-hour week at his local Coles supermarket and is paid at an hourly rate of $20. Any overtime hours worked are paid at time-and-a-half.

In a particular week, he worked his regular 38 hours plus 6 hours of overtime.

How much did Lachlan earn in this week?

  1. $760
  2. $880
  3. $940
  4. $1140
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Normal pay}=38\times 20=\$760\)

\(\text{Overtime rate}= 20\times1.5=\$30\ \text{per hour}\)

\(\text{Overtime pay}=6\times 30=\$180\)

\(\text{Total pay}=760+ 180=\$940\)

\(\Rightarrow C\)

Financial Maths, STD1 F1 EQ-Bank 30

Hugo purchased a jet ski for $25 000. The value of the jet ski decreases according to a linear model. The graph shows the value of the jet ski, $\(V\), against the time, \(t\) months, since it was purchased.
 

  1. By how much does the value of the jet ski decrease every 10 months?   (1 mark)

  2. Find the value of the jet ski after 6 years.   (1 mark)

  3. Identify ONE problem with using this model to determine the value of Hugo's jet ski over time.   (1 mark)

Show Answers Only

a.    \(\$2000\)

b.    \(\$10\,600\)

c.    \(\text{Jet ski will have a negative value after 125 months.}\)

Show Worked Solution

a.    \(\text{Total decrease over 100 months}= 25\,000-5000= \$20\,000\)

\(\text{Decrease per 10 months}= \dfrac{10}{100} \times 20\,000 = \$2000\)
 

b.    \(\text{6 years} = 6 \times 12 = 72\ \text{months}\)

\(\text{Depreciation rate}= \$200\ \text{per month}\)

\(V = 25\,000-(200 \times 72)=\$10\,600\)
 

c.    \(\text{Model limitations:}\)

\(\text{The linear model predicts the jet ski’s value reaches \$0 at 125 months and negative}\)

\(\text{values beyond that, which is unrealistic.}\)

Financial Maths, STD1 F1 2021 HSC 19 (Adapted)

Maya purchased a motorbike for $12 000. The value of the motorbike decreases according to a linear model. The graph shows the value of the motorbike, $\(V\), against the time, \(t\) months, since it was purchased.
 

  1. By how much does the value of the motorbike decrease every 10 months?   (1 mark)

  2. Find the value of the motorbike after 4 years.   (1 mark)

  3. Identify ONE problem with using this model to determine the value of Maya's motorbike over time.   (1 mark)

Show Answers Only

a.     \(\$1000\)

b.     \(\$7200\)

c.     \(\text{Motorbike will have a negative value after 120 months.}\)

Show Worked Solution

a.    \(\text{Total decrease over 100 months}= 12\,000-2000= \$10\,000\)

\(\text{Decrease per 10 months}= \dfrac{10\,000}{10}= \$1000\)
 

b.    \(\text{4 years} = 4 \times 12 = 48\ \text{months}\)

\(\text{Depreciation rate}= \$100\ \text{per month}\)

\(V = 12\,000-(100 \times 48)=\$7200\)
 

c.    \(\text{Model limitations:}\)

\(\text{The linear model will eventually predict a value of \$0 (at 120 months) and negative}\)

\(\text{values beyond that, which is unrealistic.}\)

Financial Maths, STD1 F1 EQ-Bank 25

Riley borrowed $4000 at a simple interest rate of 6% per annum. The loan is to be repaid as a single lump sum at the end of 8 months.

Calculate the total amount Riley must repay.   (2 marks)

Show Answers Only

\(\$4160\)

Show Worked Solution
\(\text{Interest}\) \(=Prn\)
  \(=4000\times 0.06\times \dfrac{8}{12}=\$160\)

 
\(\therefore\ \text{Total repayment} =4000+160=\$4160\)

Financial Maths, STD2 F1 EQ-Bank 2 MC

A tradesperson's ute was valued at $54 000 when new. The value of the ute depreciates at a rate of 25 cents per kilometre travelled.

What is the value of the ute after it has travelled a total distance of 86 400 km?

  1. $21 600
  2. $32 400
  3. $53 784
  4. $54 216
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Depreciation} = 86\,400\times\dfrac{25}{100}=\$21\,600\)

\(\text{Value} = 54\,000-21\,600=\$32\,400\)

\(\Rightarrow B\)

Financial Maths, STD1 F1 2023 HSC 6 MC (Adapted)

A courier van was valued at $48 000 when new. The value of the van depreciates at a rate of 18 cents per kilometre travelled.

What is the value of the van after it has travelled a total distance of 95 400 km?

  1. $17 172
  2. $30 828
  3. $37 440
  4. $65 172
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Depreciation} = 95\,400\times\dfrac{18}{100}=\$17\,172\)

\(\text{Value} = 48\,000-17\,172=\$30\,828\)

\(\Rightarrow B\)

Financial Maths, STD2 F1 2010 HSC 5 MC (Adapted)

Noah saw the following advertisement at his local bank in Newcastle:

Noah invests $6000 for a term of 9 months.

How much interest will Noah earn at the end of the term?

  1. $216
  2. $259
  3. $288
  4. $298
Show Answers Only

\(A\)

Show Worked Solution

\(P=\$6000,\ r=4.8\%=0.048,\ n=\dfrac{9}{12}\ \text{years}\)

\(I=Prn=6000 \times 0.048 \times \dfrac{9}{12}=\$216\)

\(\Rightarrow A\)

Financial Maths, STD1 F1 EQ-Bank 5 MC

Chloe invests $5000 for 2 years and 3 months. Simple interest is paid on the investment at a rate of 4% per annum.

What is the total value of the investment at the end of this period?

  1. $450
  2. $5200
  3. $5400
  4. $5450
Show Answers Only

\(D\)

Show Worked Solution

\(\text{2 years and 3 months = 27 months}\)

\(I=Prn=5000 \times 0.04 \times \dfrac{27}{12}=\$450\)

\(\therefore\ \text{Value of Investment} =5000+450=\$5450\)

\(\Rightarrow D\)

Financial Maths, STD2 F1 2010 HSC 5 MC (Adapted)

Hayden invests $3000 for 1 year and 8 months. Simple interest is paid on the investment at a rate of 5% per annum.

What is the total value of the investment at the end of this period?

  1. $3150
  2. $3250
  3. $3270
  4. $3500
Show Answers Only

\(B\)

Show Worked Solution

\(\text{1 year and 8 months = 20 months}\)

\(I=Prn=3000 \times 0.05 \times \dfrac{20}{12}=\$250\)

\(\therefore\ \text{Value of Investment} =3000+250=\$3250\)

\(\Rightarrow B\)

Financial Maths, STD1 F1 2023 HSC 2 MC (Adapted)

Mia deposits $4000 into a savings account at the Commonwealth Bank that pays simple interest at a rate of 4% per annum.

How much interest will she earn in the first three years?

  1. $120
  2. $160
  3. $480
  4. $4480
Show Answers Only

\(C\)

Show Worked Solution

\(P=\$4000,\ r=4\%=0.04,\ n=3\ \text{years}\)

\(I=Prn=4000 \times 0.04 \times 3=\$480\)

\(\Rightarrow C\)

Algebra, STD1 EQ-Bank 25

A scientist is studying a colony of bacteria in a laboratory. At the start of the experiment there are 500 bacteria.

Each hour the number of bacteria is modelled to be 1.25 times the number of the previous hour.

Let  \(t=0\)  be the start of the experiment.

  1. By first completing the table of values below, draw a graph showing how the bacteria population is modelled from  \(t=0\)  to  \(t=8\) hours.   (3 marks)

\(\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & & & & \\ \hline \end{array}\)
 

 

  1. Using your graph from (a), or otherwise, determine after how many hours the bacteria population first exceeds 2000.   (1 mark)

Show Answers Only

a.    \(\text{Table of values:}\)

\(\text{Number of bacteria} = 500 \times 1.25^t\)

\(\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & \textbf{625} & \textbf{781} & \textbf{1221} & \textbf{2980} \\ \hline \end{array}\)
 

b.    \(\text{From the graph, the bacteria population reaches 2000 at approximately }\ t \approx 6.25 \ \text{hours.}\)

Show Worked Solution

a.    \(\text{Table of values:}\)

\(\text{Number of bacteria} = 500 \times 1.25^t\)

\(\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & \textbf{625} & \textbf{781} & \textbf{1221} & \textbf{2980} \\ \hline \end{array}\)
 

b.    \(\text{From the graph, the bacteria population reaches 2000 at approximately }\ t \approx 6.25 \ \text{hours.}\)

Algebra, STD1 EQ-Bank 22

The population of a certain town on 1 January 2024 was 2000 people.

Each year the population of this town is modelled to be 1.18 times the population of the previous year.

Let  \(t=0\)  be 1 January 2024 .

By first completing the table for the indicated years, draw a graph showing how the population is modelled from 1 January 2024 to 1 January 2034.   (4 marks)

\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 & & & & \\
\hline
\end{array}

Show Answers Only

\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 &2360 &2785 &4576 & 10468 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 &2360 &2785 &4576 & 10468 \\
\hline
\end{array}

Algebra, STD1 EQ-Bank 13

A boat is purchased for $15 000. It depreciates in value by $3000 per year.

Let  \(V\) = Value of the boat in dollars, and  \(t\) = time in years.

  1. Complete the table of values below that models the relationship between the value of the boat and time in years.   (1 mark)
      
    \(\begin{array}{|c|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}V & 15\,000 &  & 9000 &  & 3000 &  \\[12pt] \hline \end{array}\)

  2. Using the table of values from (a), neatly graph the value of the boat from 0 to 5 years on the grid below.   (2 marks)

      

     

  3. Identify ONE limitation of this linear model.   (1 mark)

Show Answers Only

a.    \(\text{Table of values}\)

\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline V & 15\ 000 & \textbf{12 000} & 9000 & \textbf{6000} & 3000 & \ \ \ \ \textbf{0}\ \ \ \  \\ \hline \end{array}

b.  

     

c.     \(\text{Limitations could include ONE of the following:}\)

    • \(\text{The model predicts the boat has zero value after 5 years, which is}\)
      \(\text{unrealistic as most boats retain some value.}\)
    • \(\text{Beyond 5 years the model would predict a negative value, which is}\)
      \(\text{not possible.}\)
    • \(\text{The model assumes a constant rate of depreciation, but in reality}\)
      \(\text{a boat may depreciate more quickly in early years.}\)
Show Worked Solution

a.    \(\text{Table of values}\)

\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline V & 15\ 000 & \textbf{12 000} & 9000 & \textbf{6000} & 3000 & \ \ \ \ \textbf{0}\ \ \ \  \\ \hline \end{array}

 
b.  

     

c.     \(\text{Limitations could include ONE of the following:}\)

    • \(\text{The model predicts the boat has zero value after 5 years, which is}\)
      \(\text{unrealistic as most boats retain some value.}\)
    • \(\text{Beyond 5 years the model would predict a negative value, which is}\)
      \(\text{not possible.}\)
    • \(\text{The model assumes a constant rate of depreciation, but in reality}\)
      \(\text{a boat may depreciate more quickly in early years.}\)

Algebra, STD1 EQ-Bank 20

Sunny SUPs Pty Ltd charges a $2.50 online booking fee plus $25 per hour for stand-up paddle board hire.

A student claims that Sunny SUPs' income can be modelled using a linear equation.

Is the student correct? Justify your answer.   (2 marks)

Show Answers Only

\(\text{Yes, the student is correct.}\)

\(\text{Correct answers could include any ONE of the following justifications:}\)

    • \(\text{The income increases by a constant amount of \$25 for each additional}\)
      \(\text{hour hired.}\)
    • \(\text{The income can be written in the form }\  I=25h+2.50,\text{which is a linear}\)
      \(\text{equation.}\)
    • \(\text{The graph of income against hours hired would be a straight line.}\)
    • \(\text{The rate of change of income is constant at \$25 per hour.}\)
Show Worked Solution

\(\text{Yes, the student is correct.}\)

\(\text{Correct answers could include any ONE of the following justifications:}\)

    • \(\text{The income increases by a constant amount of \$25 for each additional}\)
      \(\text{hour hired.}\)
    • \(\text{The income can be written in the form }\  I=25h+2.50,\text{which is a linear}\)
      \(\text{equation.}\)
    • \(\text{The graph of income against hours hired would be a straight line.}\)
    • \(\text{The rate of change of income is constant at \$25 per hour.}\)

Algebra, STD1 EQ-Bank 26

Rangers at a nature reserve are monitoring the spread of an invasive weed. At the start of monitoring there are 100 weeds. The number of weeds is growing at a rate of 50% per month.

Let  \(N\) = number of weeds, and  \(t\) = time in months.

  1. Complete the table of values below that models the growth of the weeds over an 8 month period. Round all answers to the nearest whole number.   (2 marks)
      
    \(\begin{array}{|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}N & 100 & & & & \\[12pt] \hline \end{array}\)

  2. Using the table of values from (a), neatly plot the points and join with a smooth curve.   (2 marks)

      

Show Answers Only

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|} \hline \quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad  \\[6pt] \hline \ N  & 100 & 150 & 225 & 506 & 2563 \\[6pt] \hline \end{array} 

b.      

   

Show Worked Solution

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|} \hline \quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad  \\[6pt] \hline \ N  & 100 & 150 & 225 & 506 & 2563 \\[6pt] \hline \end{array}

\(\text{Algebraic method}\)

\(\text{Formula:  }\ N=100\times1.5^t\)

\(t=0:\ N=100\times1.5^{0}=100\)

\(t=1:\ N=100\times1.5^{1}=150\)

\(t=2:\ N=100\times1.5^{2}=225\)

\(t=4:\ N=100\times1.5^{4}=506.25\approx506\)

\(t=8:\ N=100\times1.5^{8}=2562.89\ldots\approx2563\)
  

\(\text{Using CASIO calculator with constant multiplier}\)

\begin{array} {|c|c|c|c|}
\hline t & \text{Input} & \text{Output}\ (N) & \text{Rounded}\ (N) \\
\hline {0} & 100= & 100 & 100 \\
\hline {1} & \text{Ans}\times 1.5= & 150  & 150 \\
\hline {2} & = & 225 & 225 \\
\hline {4} & = & 506.25 & 506 \\
\hline {8} & = & 2562.890625 & {2563} \\
\hline \end{array}  

b.      

   

Algebra, STD1 EQ-Bank 27

A conservation program is tracking the recovery of a native wildflower species in a national park. At the start of the program there are 200 plants. The number of plants is predicted to grow at a rate of 25% per year.

Let  \(N\) = number of plants, and  \(t\) = time in years.

  1. Complete the table of values below that models the growth of the plants. Round all answers to the nearest whole number.   (2 marks)
    \begin{array}{|c|c|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
    \hline
    \rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
    \hline
    \end{array}

  2. Using the table of values from (a), neatly plot the points and join with a smooth curve.   (2 marks)

      

Show Answers Only

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
\hline
\end{array}

b.     

Show Worked Solution

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
\hline
\end{array}

\(\text{Algebraic method}\)

\(\text{Formula:   }\ N=200\times1.25^t\)

\(t=0:\ N=200\times1.25^{0}=200\)

\(t=1:\ N=200\times1.25^{1}=250\)

\(t=2:\ N=200\times1.25^{2}=312.5\approx313\)

\(t=3:\ N=200\times1.25^{3}=390.625\approx391\)

\(t=4:\ N=200\times1.25^{4}=488.28\ldots\approx488\)
  

\(\text{Using CASIO calculator with constant multiplier}\)

\begin{array} {|c|c|c|c|}
\hline t & \text{Input} & \text{Output}\ (N) & \text{Rounded}\ (N) \\
\hline \colorbox{lightblue}{0} & 200= & 200 & \colorbox{lightblue}{200} \\
\hline \colorbox{lightblue}{1} & \text{Ans}\times 1.25= & 250  & \colorbox{lightblue}{250} \\
\hline \colorbox{lightblue}{2} & = & 312.5 & \colorbox{lightblue}{313} \\
\hline \colorbox{lightblue}{3} & = & 390.625 & \colorbox{lightblue}{391} \\
\hline \colorbox{lightblue}{4} & = & 488.28\ldots & \colorbox{lightblue}{488} \\
\hline \end{array}  

b.     

   

Algebra, STD1 EQ-Bank 15

A household's monthly water bill consists of a fixed service charge of $45 plus $3 per kilolitre of water used.

Let  \(C\) = monthly cost in dollars, and  \(k\) = water usage in kilolitres.

  1. Complete the table of values below that models the relationship between water usage and monthly cost.   (1 mark)
      
    \(\begin{array}{|c|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad k \quad & \quad 0 \quad & \quad 10 \quad & \quad 20 \quad & \quad 30 \quad & \quad 40 \quad & \quad 50 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}C &  & 75 & 105 &  & 165 & 195 \\[12pt] \hline \end{array}\)

  2. Using the table of values from (a), neatly graph the monthly cost for water usage from 0 to 50 kL on the grid below.   (1 mark)

      

  3. Using your graph from (b), or otherwise, find the monthly cost when the household uses 35 kL of water.   (1 mark)

Show Answers Only

a.    \(\text{Table of values}\)

\begin{array}{|c|c|c|c|c|c|c|} \hline k & 0 & 10 & 20 & 30 & 40 & 50 \\ \hline C & \textbf{45} & 75 & 105 & \textbf{135} & 165 & 195 \\ \hline \end{array}

b.     

c.    \($120\)

Show Worked Solution

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|c|} \hline k & 0 & 10 & 20 & 30 & 40 & 50 \\ \hline C & \textbf{45} & 75 & 105 & \textbf{135} & 165 & 195 \\ \hline \end{array}

 
b.  
              

c.    \(\text{From the graph, when } k=35,\ \ C=\$120\)

Algebra, STD1 EQ-Bank 14

GreenCut Lawn Services charges a fixed call-out fee of $25 plus $40 per hour of work.

Let \(C\) = total charge in dollars, and  \(h\) = number of hours worked.

  1. Complete the table of values below that models the relationship between GreenCut's Lawn Services hours worked and total charge.   (1 mark)

    \begin{array}{|c|c|c|c|c|c|c|}
    \hline
    \quad \rule{0pt}{2.5ex}h\quad  \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad &\quad  2\quad  & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad\\
    \hline
    \rule{0pt}{2.5ex}C & & 65 & 105 \rule[-1ex]{0pt}{0pt}& 145 & & 225 \\
    \hline
    \end{array}

  2. Using the table of values from (a), neatly graph the total charge for work completed from 0 to 5 hours on the grid below.   (1 mark)

      

  3. Using your graph from (b), or otherwise, find the total charge for 2.5 hours of work.   (1 mark)

  4. A customer has a budget of $160. Using your graph from (b), or otherwise, determine the maximum number of complete hours GreenCut can work within this budget.   (1 mark)

Show Answers Only

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|c|}
\hline
\quad \rule{0pt}{2.5ex}h\quad  \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad &\quad  2\quad  & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad\\
\hline
\rule{0pt}{2.5ex}C & \textbf{25}& 65 & 105 \rule[-1ex]{0pt}{0pt}& 145 &\textbf{185} & 225 \\
\hline
\end{array}

 
b.    

   

c.    \($125\)

d.    \(3\ \text{hours}\)

Show Worked Solution

a.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|c|}
\hline
\quad \rule{0pt}{2.5ex}h\quad  \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad &\quad  2\quad  & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad\\
\hline
\rule{0pt}{2.5ex}C & \textbf{25}& 65 & 105 \rule[-1ex]{0pt}{0pt}& 145 &\textbf{185} & 225 \\
\hline
\end{array}

 
b.  

     

c.    \(\text{From the graph, when }\ h=2.5, \ C=\$125\)
 

d.    \(\text{From the graph, \$160 lies between } h=3 \text{ and } h=4.\)

\(\therefore\ \text{Maximum complete hours} = 3\ \text{hours}\)

Algebra, STD1 EQ-Bank 24

Zara is looking for a new mobile phone plan. She has found two plans that suit her needs and wants to work out which plan is cheaper depending on how many minutes she uses.

Plan \(\text{A}\):   $20 per month fixed charge plus $0.10 per minute

Plan \(\text{B}\):  $0.30 per minute, no fixed charge

Let  \(C\) = total monthly cost in dollars, and  \(m\) = number of minutes used.

  1. Plan \(\text{B}\) can be modelled by the equation  \(C=0.30m\).
  2. Write an equation for the monthly cost of Plan \(\text{A}\).   (1 mark)

  3. The graph of Plan \(\text{B}\) is provided on the grid below. Use the equation from (a) to add the graph of Plan \(\text{A}\) to the grid.   (2 marks)

      
  5. For how many minutes per month do both plans cost the same amount?   (1 mark)

  6. Zara uses an average of 140 minutes per month.
  7. Which plan she should choose and how much does she save compared to the other plan.   (1 mark)

Show Answers Only

a.    \(\text{Plan A: }C=20+0.10m\)

b.    

c.    \(100\ \text{minutes}\)

d.    \(\text{Plan A, cheaper by } \$8\)

Show Worked Solution

a.    \(\text{Plan A: }\ C=20+0.10m\)
  

b.    \(\text{Table of values}\)

\(\begin{array}{|c|c|c|c|c|c|} \hline m & 0 & 50 & 100 & 150 & 200 \\ \hline \text{Plan A} & 20 & 25 & 30 & 35 & 40 \\ \hline \end{array}\)
  


  

c.    \(\text{From the graph, the lines intersect at }\ m=100.\)

\(\therefore\ \text{Both plans cost the same at } 100\ \text{minutes}\)
  

d.    \(\text{Plan A:   }\ C=20+0.10\times140=\$34\)

\(\text{Plan B:   }\ C=0.30\times140=\$42\)

\(\text{Difference} = 42-34=\$8\)

\(\therefore\ \text{Zara should choose Plan A, which is \$8 cheaper than Plan B.}\)

Algebra, STD1 EQ-Bank 36

Two cyclists, Aiko and Ben, are riding along the same straight track in the same direction.

Aiko starts 2000 m ahead of Ben. Aiko rides at a constant speed of 250 metres/minute and Ben rides at a constant speed of 500 metres/minute.

Let    \(d\) = distance from the starting point in metres, and 

   \(t\) = time in minutes.

  1. The equation  \(d=2000+250t\)  models Aiko's distance from the starting point.
  2. Write an equation to model Ben's distance.   (1 mark)

  3. Use the equations from (a) to graph Aiko and Ben's journeys on the grid below.   (2 marks)

      
  4. After how many minutes does Ben catch Aiko?   (1 mark)

  5. How far has each cyclist travelled from their own starting point when they meet?   (2 marks)

Show Answers Only

a.    \(d=500t\)

b.    

c.    \(8\ \text{minutes}\)

d.    \(\text{Aiko travelled 2000 m, Ben travelled 4000 m.}\)

Show Worked Solution

a.    \(d=500t\)

b.    \(\text{Table of values:}\)

\(\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 \\ \hline \text{Aiko} & 2000 & 2500 & 3000 & 3500 & 4000 & 4500 \\ \hline \text{Ben} & 0 & 1000 & 2000 & 3000 & 4000 & 5000 \\ \hline \end{array}\)
 

c.    \(\text{From the graph, the lines intersect at }\ t=8.\)

\(\therefore\ \text{Ben catches Aiko after 8 minutes}\)
 

d.    \(\text{When}\ \ t=8, d=4000:\)

\(\text{Aiko started 2000 m ahead, so the distance travelled}\)

\(=4000-2000=2000\ \text{m}\)

\(\text{Ben started at the origin, so the distance travelled =4000 m}\)