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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, 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 EQ-Bank 12

Daniel earns $32 per hour as a delivery driver. He is also paid a $12 fuel allowance per shift.

How much will he earn from a 5-hour shift?   (2 marks)

Show Answers Only

\(\$172\)

Show Worked Solution

\(\text{Wages}=5\times 32=\$160\)

\(\text{Total earnings}= 160+ 12=\$172\)

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 11

A farmer in Western Australia observes that the rodent population on his property is doubling every two weeks.

A student claims that a linear model is NOT appropriate to predict the rodent population over time.

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

Show Answers Only

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

\(\text{Correct justification could include ONE of the following:}\)

    • \(\text{The rodent population is doubling every two weeks, which means it}\)
      \(\text{is increasing by a constant multiplier, not a constant amount — this}\)
      \(\text{is exponential growth, not linear.}\)
    • \(\text{A linear model increases by the same amount each time period, but}\)
      \(\text{the rodent population increases by a larger amount each fortnight as}\)
      \(\text{the population grows.}\)
    • \(\text{The graph of the rodent population over time would be an upsloping}\)
      \(\text{exponential curve, not a straight line.}\)
    • \(\text{A linear model would significantly underestimate the rodent population}\)
      \(\text{over time.}\)
Show Worked Solution

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

\(\text{Correct justification could include ONE of the following:}\)

    • \(\text{The rodent population is doubling every two weeks, which means it}\)
      \(\text{is increasing by a constant multiplier, not a constant amount — this}\)
      \(\text{is exponential growth, not linear.}\)
    • \(\text{A linear model increases by the same amount each time period, but}\)
      \(\text{the rodent population increases by a larger amount each fortnight as}\)
      \(\text{the population grows.}\)
    • \(\text{The graph of the rodent population over time would be an upsloping}\)
      \(\text{exponential curve, not a straight line.}\)
    • \(\text{A linear model would significantly underestimate the rodent population}\)
      \(\text{over time.}\)

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 15

A local council invests in a community solar farm that sells electricity back to the grid.

The solar farm is expected to operate for 12 years. Each year the farm incurs a maintenance cost of $3000.

The council uses a spreadsheet to model the costs and revenue of the project.   

  1. What are the total fixed costs for the solar farm project?   (1 mark)
  2. After how many years does the solar farm break even?   (1 mark)
  3. Calculate the profit the council makes over the full 12-year lifespan of the solar farm.   (2 marks)
Show Answers Only

a.    \($35\,000\)

b.    \(7\ \text{years}\)

c.    \($25\,000\)

Show Worked Solution

a.    \(\text{Total fixed costs}=$24\,000+$8000+$3000=$35\,000\)
    

b.    \(\text{From the spreadsheet, at year}\ 7:\)

\(\text{Total cost}=$56\,000,\ \text{Revenue}=$56\,000\ \checkmark\)

\(\therefore\ \text{Break-even}=7\ \text{years}\)
    

c.    \(\text{Project lifespan}=12\ \text{years}\)

\(\text{Variable cost} =12\times \$3000= $36\,000\)

\(\text{Total costs} =$35\,000+$36\,000= \$71\,000\)

\(\text{Revenue} =12\times \$8000 = \$96\,000\)

\(\therefore\ \text{Profit} = \$96\,000-\$71\,000 = \$25\,000\)

Algebra, STD1 EQ-Bank 23

Maya operates Sydney City Walking Tours, a city walking tour business.

The bus she hires holds up to 40 people. Each person on the tour pays $35 and receives a complimentary bottle of water.

Maya uses a spreadsheet to model the costs and revenue for each tour. 
  

  1. What are the total fixed costs for each tour?   (1 mark)

  2. How many people need to attend the tour for Maya to break-even?   (1 mark)

  3. Calculate the profit Maya makes if the tour is fully booked.   (2 marks)

Show Answers Only

a.    \($600\)

b.    \(20\ \text{people}\)

c.    \($600\)

Show Worked Solution

a.    \(\text{Total fixed costs}=$350+$250=$600\)
    

b.    \(\text{From the spreadsheet, Maya’s break-even is when }\)

\(\text{Total cost}=\text{Revenue}=$700\)

\(\therefore\ \text{People to break-even} = 20\)
  

c.    \(\text{Fully booked}=40\ \text{people}\)

\(\text{Variable cost} =40\times \$5= \$200\)

\(\text{Total costs} =$600+$200= \$800\)

\(\text{Revenue} =40\times \$35 = \$1400\)

\(\therefore\ \text{Profit} = \$1400-\$800 = \$600\)

Algebra, STD1 EQ-Bank 16

A cafe owner uses a spreadsheet to model the costs and revenue from selling cups of coffee.

The cafe has a fixed setup cost of $450 and a cost per cup 0f $1.50 to make each coffee. The owner sells each coffee for $4.50.

Part of the spreadsheet is shown.
  

  1. What does it mean for the cafe owner to break even?   (1 mark)

  2. Use the spreadsheet to identify the break-even point for the cafe owner.   (1 mark)

  3. If the cafe owner sells 250 cups of coffee, what profit is made?   (2 marks)

Show Answers Only

a.    \(\text{Break-even is when the total cost equals the revenue}\)

\(\text{i.e. the owner makes neither a profit nor a loss.}\)

b.    \(150\ \text{cups}\)

c.    \($300\)

Show Worked Solution

a.    \(\text{Break-even is when the total cost equals the revenue}\)

\(\text{i.e. the owner makes neither a profit nor a loss.}\)
  

b.    \(\text{From the spreadsheet, total cost equals revenue when }\)

\(\text{Cost}=\text{Revenue}=$675\)

\(\therefore\ \text{Cups sold to break-even} = 150\)

\(\text{Confirmed by cell C15:}\)

\(\text{Break-even (cups)}=\dfrac{450}{4.50-1.50}=\dfrac{450}{3}=150 \text{ cups}\)
  

c.    \(\text{At 250 cups (row 9): Revenue} = \$1125,\ \text{Total cost} = \$825\)

\(\therefore\ \text{Profit} = 1125-825 = \$300\)

Algebra, STD1 EQ-Bank 22

Gemstar Promotions is organising the annual Concert Under the Stars event to take place on the last weekend in January.

The outdoor venue holds up to 800 people. Each ticket holder receives a complimentary souvenir program valued at $15.

Garth from Gemstar Promotions uses a spreadsheet to model the costs and revenue for the concert. 
  

  1. What are the total fixed costs for the concert?   (1 mark)
  2. The promoter will only proceed with the concert if the loss is no more than $3000.   
  3. What is the minimum number of tickets that must be sold for the concert to go ahead?   (2 marks)
  4. Use the formula in cell C9 to calculate the number of tickets that must be sold for the promoter to break even?   (1 mark)
  5. Calculate the profit for Gemstar Promotions if the concert is fully booked.   (1 mark)
Show Answers Only

a.    \($30\,000\)

b.    \(450\ \text{people}\)

c.    \(500\ \text{people}\)

d.    \($18\,000\)

Show Worked Solution

a.    \(\text{Total fixed costs}=$8000+$6000+$16\,000=$30\,000\)
    

b.    \(\text{Maximum allowable loss}=$3000\)

\(\text{From the spreadsheet, at}\ 450\ \text{people}:\)

\(\text{Total cost}=$36\,750,\ \text{Revenue}=$33\,750\)

\(\text{Loss}=$36\,750-$33\,750=$3000\ \checkmark\)

\(\therefore\ \text{Minimum ticket sales}=450\)
  

c.    \(\text{Breakeven = C6/(C7-C8)}\)

\(\text{Breakeven}\ = \dfrac{30\,000}{75.00-15.00}=500\ \text{people}\)
  

d.    \(\text{Venue capacity}=800\ \text{people}\)

\(\text{Variable cost} =800\times \$15= $12\,000\)

\(\text{Total costs} =$30\,000+$12\,000= \$42\,000\)

\(\text{Revenue} =800\times \$75 = \$60\,000\)

\(\therefore\ \text{Profit} = \$60\,000-\$42\,000 = \$18\,000\)

Statistics, STD2 EQ-Bank 27

The shoe size and height of ten students were recorded.

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Shoe size} \rule[-1ex]{0pt}{0pt} & \text{6} & \text{7} & \text{7} & \text{8} & \text{8.5} & \text{9.5} & \text{10} & \text{11} & \text{12} & \text{12} \\
\hline \rule{0pt}{2.5ex} \text{Height} \rule[-1ex]{0pt}{0pt} & \text{155} & \text{150} & \text{165} & \text{175} & \text{170} & \text{170} & \text{190} & \text{185} & \text{200} & \text{195} \\
\hline
\end{array}

  1. Identify the dependent variable.   (1 mark)
  2. Complete the scatter plot AND draw a line of fit by eye.   (2 marks)


     
     

  3. Use the line of fit to estimate the height difference between a student who wears a size 7.5 shoe and one who wears a size 9 shoe.   (1 mark)

Show Answers Only

a.    `text{Height is the dependent variable (y-axis variable).}`

b.    `text(See Worked Solutions.)`

c.    `13\ text{cm  (or close given LOBF drawn)}`

Show Worked Solution

a.    `text{Height is the dependent variable (y-axis variable).}`

b.    
      2UG 2015 28e Answer

c.    `text{Shoe size 7½ gives a height estimate of 162 cm (see graph).}`

`text{Shoe size 9 gives a height estimate of 175 cm (see graph).}`

`text(Height difference)= 175-162= 13\ text{cm  (or close given LOBF)}`

Statistics, STD2 EQ-Bank 29

Each member of a group of males had his height and foot length measured and recorded. The results were graphed and a line of fit drawn.
 

  1. Identify the independent variable.   (1 mark)

  2. Why does the value of the `y`-intercept have no meaning in this situation?   (1 mark)

  3. George is 10 cm taller than his brother Harry. Use the line of fit to estimate the difference in their foot lengths.   (1 mark)

Show Answers Only

a.    `text{Height is the independent variable (x-axis variable).}`
 

b.    `text(The y-intercept occurs when)\ x = 0.\ text(It has no meaning to have)`

`text(a height of 0 cm.)`
  

c.    `text(A 20 cm height difference results in a foot length difference of 6 cm.)`

`text(A 10 cm height difference means George should have a 3 cm longer foot.)`

Show Worked Solution

a.    `text{Height is the independent variable (x-axis variable).}`
 

b.    `text(The y-intercept occurs when)\ x = 0.\ text(It has no meaning to have)`

`text(a height of 0 cm.)`
  

c.    `text(A 20 cm height difference results in a foot length difference of 6 cm.)`

`text(A 10 cm height difference means George should have a 3 cm longer foot.)`

ENGINEERING, TE 2025 HSC 27a

Security cameras are becoming more commonly used in society.

Describe the ethical concerns that should be considered when designing and developing telecommunications projects such as security cameras.   (3 marks)

Show Answers Only
  • Security cameras raise privacy concerns, as constant surveillance may infringe on individuals’ rights to move and act without being monitored.
  • Data security is a concern, as recorded footage may be vulnerable to unauthorised access, misuse, or data breaches.
  • Facial recognition technology used in camera systems can introduce bias and discrimination, raising concerns about fairness and individual rights.
Show Worked Solution
  • Security cameras raise privacy concerns, as constant surveillance may infringe on individuals’ rights to move and act without being monitored.
  • Data security is a concern, as recorded footage may be vulnerable to unauthorised access, misuse, or data breaches.
  • Facial recognition technology used in camera systems can introduce bias and discrimination, raising concerns about fairness and individual rights.

ENGINEERING, CS 2025 HSC 26b

Explain how the introduction of steel beams in bridge construction has affected people's lives.   (3 marks)

Show Answers Only
  • Steel beams have high tensile and compressive strength. This allows bridges to span greater distances and support heavier loads than traditional materials.
  • As a result, transport networks have expanded, enabling faster movement of people and goods across regions.
  • This has led to growth in trade and industry, lowering the cost of goods and improving economic opportunities for communities.
Show Worked Solution
  • Steel beams have high tensile and compressive strength. This allows bridges to span greater distances and support heavier loads than traditional materials.
  • As a result, transport networks have expanded, enabling faster movement of people and goods across regions.
  • This has led to growth in trade and industry, lowering the cost of goods and improving economic opportunities for communities.

ENGINEERING, CS 2025 HSC 22a

A building is being partially demolished and replaced. Part of the original structure will be retained.

Outline a method that could be used to recycle a material from the building.   (2 marks)

Show Answers Only

Answers could include any ONE of the following:

  • Steel: Steel framing members, such as beams and columns, are salvaged during demolition and sent to a steel mill. They are reprocessed in an electric arc furnace and reformed into new steel products for future construction use.
  • Concrete: Waste concrete from demolished slabs and footings is broken down by mechanical crushing equipment into smaller aggregate pieces. This recycled aggregate is then incorporated into new concrete mixes or used as compacted fill beneath pavements.
  • Glass: Glazing panels removed during demolition are sorted and processed into fine particles known as cullet. This material is fed back into a glass furnace and reformed into new products, or blended into bituminous pavement materials.
Show Worked Solution

Answers could include any ONE of the following:

  • Steel: Steel framing members, such as beams and columns, are salvaged during demolition and sent to a steel mill. They are reprocessed in an electric arc furnace and reformed into new steel products for future construction use.
  • Concrete: Waste concrete from demolished slabs and footings is broken down by mechanical crushing equipment into smaller aggregate pieces. This recycled aggregate is then incorporated into new concrete mixes or used as compacted fill beneath pavements.
  • Glass: Glazing panels removed during demolition are sorted and processed into fine particles known as cullet. This material is fed back into a glass furnace and reformed into new products, or blended into bituminous pavement materials.

ENGINEERING, TE 2025 HSC 21d

Discuss the effect of ONE telecommunications engineering innovation on people's lives.   (4 marks)

Show Answers Only

Innovation: Mobile phone technology

  • [P] Mobile phone technology has significantly transformed how people communicate and access information daily.
  • [E] Smartphones enable instant communication, remote working and access to banking, healthcare and education from anywhere.
  • [Ev] During the COVID-19 pandemic, mobile phones allowed millions to work and study from home, maintaining productivity.
  • [L] On one hand, mobile phones have improved quality of life by keeping people connected and informed at all times.
  • [P] However, mobile phone technology also presents significant challenges for individuals and society.
  • [E] Constant connectivity has contributed to rising rates of anxiety, sleep disruption and reduced face-to-face interaction.
  • [Ev] Studies show excessive screen time is linked to poorer mental health outcomes, particularly in young people.
  • [L] Despite the significant benefits, the negative impact on mental health and social connection must be carefully considered.
Show Worked Solution

Innovation: Mobile phone technology

  • [P] Mobile phone technology has significantly transformed how people communicate and access information daily.
  • [E] Smartphones enable instant communication, remote working and access to banking, healthcare and education from anywhere.
  • [Ev] During the COVID-19 pandemic, mobile phones allowed millions to work and study from home, maintaining productivity.
  • [L] On one hand, mobile phones have improved quality of life by keeping people connected and informed at all times.
  • [P] However, mobile phone technology also presents significant challenges for individuals and society.
  • [E] Constant connectivity has contributed to rising rates of anxiety, sleep disruption and reduced face-to-face interaction.
  • [Ev] Studies show excessive screen time is linked to poorer mental health outcomes, particularly in young people.
  • [L] Despite the significant benefits, the negative impact on mental health and social connection must be carefully considered.

Statistics, MET2 2025 VCAA 3

The time taken for a driver to travel to work each day, in minutes, is modelled by a continuous random variable \(T\) with probability density function

\(f(t)=\left\{\begin{array}{cl}
\dfrac{1}{1\,215\,000}(t-29)(59-t)^3 & 29 \leq t \leq 59 \\
0 & \text {otherwise}
\end{array}\right.\)

    1. Find the mean time taken, in minutes, for the driver to travel to work each day.   (1 mark)

    2. Find the standard deviation of the time taken, in minutes, for the driver to travel to work each day.   (2 marks)

  2. The driver allows \(k\) minutes to travel to work each day. If the journey takes longer than \(k\) minutes, the driver will be late. Whether the driver is late on a particular day is independent of whether they are late on any other day.
    1. If \(k=47\), write a definite integral to show that the probability of the driver being late is 0.08704    (1 mark)
    2. If \(k=47\), find the probability that the driver will be late on at least one day in a five-day working week.
    3. Give your answer correct to four decimal places.   (2 marks)

    4. For \(k=47\), let \(\hat{P}\) be the proportion of days the driver is late in any five-day working week. Find \(\operatorname{Pr}(0.4 \leq \hat{P} \leq 0.6)\) correct to four decimal places.   (2 marks)

    5. Find the integer \(k\) such that the probability, correct to one decimal place, of the driver being late at least once in any five-day working week is 0.2    (2 marks)

  3. At a given traffic light, the wait time is modelled by a normal distribution with a mean of 2.5 minutes and a standard deviation of \(\sigma\) minutes.
    1. If \(\sigma=0.6\), find the probability that the wait time will be less than 3.5 minutes.
    2. Give your answer correct to two decimal places.   (1 mark)

    3. Find the value of \(\sigma\) such that there is a 2% chance of a wait time longer than 3.5 minutes.
    4. Give your answer correct to two decimal places.   (1 mark)

  4. The driver passes through three traffic lights \((A, B\) and \(C)\) on their journey to work. The probability of each traffic light being red is shown in the table below.
  5. \begin{array}{|l|c|c|c|}
    \hline \rule{0pt}{2.5ex}\text {Traffic light} \rule[-1ex]{0pt}{0pt}& \quad A \quad & \quad B \quad & \quad C  \quad\\
    \hline \rule{0pt}{2.5ex} \text {Probability that the traffic light is red} \quad  \rule[-1ex]{0pt}{0pt}& 0.2 & 0.3 & 0.1 \\
    \hline
    \end{array}
  6. Let \(Y\) be the random variable representing the number of traffic lights that are red on the driver's journey to work. Assume that each traffic light being red is independent of any other traffic light being red.
  7. Complete the following table for the probability distribution of \(Y\).    (2 marks)

  8. \begin{array}{|c|c|c|c|c|}
    \hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 3 \ \ \quad \\
    \hline \rule{0pt}{2.5ex}\operatorname{Pr}(Y=y) \rule[-1ex]{0pt}{0pt}& & & & \\
    \hline
    \end{array}
Show Answers Only

a.i.   \(\displaystyle \int_{29}^{59}(t \times f(t)) d t=39\)
 

a.ii. \(\text{Strategy 1}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59} t^2 f(t) d t-39^2}=\dfrac{10 \sqrt{14}}{7}\)

\(\text{Strategy 2}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59}(t-39)^2 f(t)\, d t}=\dfrac{10 \sqrt{14}}{7}\)
 

b.i.  \(P\text{(driver being late)}=\displaystyle\int_{47}^{59} f(t)\, d t=0.08704\)
 

b.ii.  \(P(\text{(driver late at least 1 day in week)}=0.3658 \ \ \text{(4 d.p.)}\)
 

b.iii. \(0.0631 \ \text{(4 d.p.)}\)
 

b.iv. \(k=49\)
 

c.i.  \(P(W<3.5)=0.95\)
 

c.ii   \(\sigma=0.49\)
 

d.

\begin{array}{|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad \\
\hline \rule{0pt}{2.5ex}\operatorname{Pr}(Y=y) \rule[-1ex]{0pt}{0pt}& \frac{63}{125}=0.504& \frac{199}{500}=0.398 & \frac{23}{250}=0.092 & \frac{3}{500}=0.006 \\
\hline
\end{array}

Show Worked Solution

a.i.   \(\text{Calculate (by CAS):}\)

\(\displaystyle \int_{29}^{59}(t \times f(t)) d t=39\)
 

a.ii. \(\text{Strategy 1}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59} t^2 f(t) d t-39^2}=\dfrac{10 \sqrt{14}}{7}\)

\(\text{Strategy 2}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59}(t-39)^2 f(t)\, d t}=\dfrac{10 \sqrt{14}}{7}\)
 

b.i.  \(P\text{(driver being late)}\)

\(=\displaystyle\int_{47}^{59} f(t)\, d t=0.08704\)
 

b.ii.  \(P(\text{(driver late at least 1 day in week)}\)

\(=1-P(\text{never late in 5 days})\)

\(=1-(0.08704)^5\)

\(=0.3658 \ \ \text{(4 d.p.)}\)
 

b.iii. \(\text{Let} \ \ Y \sim \text{Bi}(5,0.08704)\)

\(\operatorname{Pr}(0.4 \leqslant \hat{P} \leqslant 0.6)=\operatorname{Pr}(2 \leqslant Y \leqslant 3)=0.0631 \ \text{(4 d.p.)}\)
 

b.iv. \(\text{Solve for} \ k:\)

\(1-\left(1-\displaystyle \int_k^{59} f(t) d t\right)^5=0.2\)

\(k=49\)
 

c.i.  \(W \sim N\left(\mu, \sigma^2\right) \sim\left(2.5,0.6^2\right)\)

\(\text{Solve (by CAS):}\)

\(P(W<3.5)=0.95\)
 

c.ii   \(\text{Find \(z\)-score when \(P(W>3.5)=0.02\)}\)

\(z \text {-score }=2.0537 \ldots\)

\(\text{Solve for} \ \sigma :\)

\(\dfrac{3.5-2.5}{\sigma}=2.0537 \ldots \ \Rightarrow \ \sigma=0.49\ \text{(2 d.p.)}\)
 

d.

\begin{array}{|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad \\
\hline \rule{0pt}{2.5ex}\operatorname{Pr}(Y=y) \rule[-1ex]{0pt}{0pt}& \frac{63}{125}=0.504& \frac{199}{500}=0.398 & \frac{23}{250}=0.092 & \frac{3}{500}=0.006 \\
\hline
\end{array}

Functions, MET2 2025 VCAA 2

Let  \(f: R \rightarrow R, \ f(x)=\dfrac{x}{2}+7\)  and

\(g: R \rightarrow R, \ g(x)=A e^{k x}\)  where  \(A, k \in R\).

The graphs of  \(y=f(x)\)  and  \(y=g(x)\)  intersect at the points \((-12,1)\) and \((2,8)\), as shown below.
 

   

  1. Write down two simultaneous equations in terms of \(A\) and \(k\).
  2. Solve them, using algebra, to show that  \(A=2^{\tfrac{18}{7}}\)  and  \(k=\dfrac{3}{14} \log _e(2)\).   (3 marks)

  3. Find the value of \(b\), where  \(b \in R\), such that \(g(x)\) can be expressed in the form  \(g(x)=A \times 2^{b x}\).   (1 mark)

  4. Use a definite integral to evaluate the area bounded by the graphs of  \(y=f(x)\)  and  \(y=g(x)\), where  \(x \in[-12,2]\).
  5. Give the area correct to two decimal places.   (2 marks)

  6. Let  \(h(x)=f(x)-g(x)\).
    1. Write down an expression for the derivative of \(h(x)\).   (1 mark)

    2. Find the maximum value of \(h(x)\), where  \(x \in[-12,2]\).   (1 mark)
    3. Give your answer correct to two decimal places.
  7. Let \(g^{-1}\) be the inverse of \(g\).
  8. Find the points where the graph of  \(y=g^{-1}(x)\)  intersects with the graph of  \(y=2(x-7)\).   (2 marks)

  9. Let \(F\) be an anti-derivative of \(f\) that passes through \((0, c)\), where \(c \in R\).
    1. Show that it is not possible for the graph of  \(y=F(x)\)  to pass through both \((-12,1)\) and \((2,8)\).   (2 marks)

    2. The graph of  \(y=F(x)\) can be dilated by a factor of \(m\) from the \(x\)-axis such that its image passes through both \((-12,1)\) and \((2,8)\).
    3. Find the values of \(m\) and \(c\).   (2 marks)

Show Answers Only

a.    \(f(x)=\dfrac{x}{2}+7, \ g(x)=A e^{k x}\)

\(\text{Intersection occurs at }(-12,1) \text { and }(2,8):\)

\(A e^{-12 k}\) \(=1\ \ldots\ (1)\)
\(A e^{2 k}\) \(=18\ \ldots\ (2)\)

 
\(\text{Divide:}\ \ (2) ÷ (1)\)

\(e^{2 k-(-12 k)}\) \(=8\)
\(e^{14 k}\) \(=8\)
\(14 k\) \(=\log _e 8\)
\(14 k\) \(=3\log _e 2\)
\(14 k\) \(=\dfrac{3}{14} \log _e 2\)

 

\(\text{Substitute \(k\) into (1):}\)

\(A e^{-12\left(\tfrac{3}{14} \log _e 2\right)}\) \(=1\)
\(A e^{-\tfrac{18}{7} \log_e2}\) \(=1\)
\(A \times 2^{-\tfrac{18}{7}}\) \(=1\)
\(A\) \(=2^{\tfrac{18}{7}}\)

 

b.    \(g(x)=2^{\tfrac{18}{7}} \times 2^{\tfrac{3 x}{14}}\)
 

c.    \(\text{Area}=\displaystyle \int_{-12}^2 f(x)-g(x)=15.87 \ \text{u}^2\)
 

d.i.  \(h^{\prime}(x)=\dfrac{1}{2}-\dfrac{6 \log _e 2}{7} \times 2^{\tfrac{3 x}{14}+\tfrac{4}{7}}\)
 

d.ii. \(h(x)_{\text{max}}=1.72\)
 

e.   \(\text{Intersection at}\ (1,-12) \ \text{and} \ (8,2).\)
 

f.i.  \(f(x)=\dfrac{x}{2}+7\)

\(F(x)=\displaystyle \int f(x)\ d x=\dfrac{1}{4} x^2+7 x+c\)
 

\(\text{If} \ F(x) \ \text{passes through} \ (-12,1):\)

\(F(-12)=\dfrac{1}{4}(-12)^2+7(-12)+c=1 \ \ \Rightarrow\ \ c=49\)
 

\(\text{If \(F(x)\) passes through \((2,8)\):}\)

\(F(2)=\dfrac{1}{4}(2)^2+7(2)+c=8 \ \ \Rightarrow \ \ c=-7\)

\(\text{Since \(c\) cannot have 2 values, it cannot pass through both points.}\)
 

f.ii.  \(m=\dfrac{1}{9}, \ c=57\)

Show Worked Solution

a.    \(f(x)=\dfrac{x}{2}+7, \ g(x)=A e^{k x}\)

\(\text{Intersection occurs at }(-12,1) \text { and }(2,8):\)

\(A e^{-12 k}\) \(=1\ \ldots\ (1)\)
\(A e^{2 k}\) \(=18\ \ldots\ (2)\)
Mean mark (a) 51%.

\(\text{Divide:}\ \ (2) ÷ (1)\)

\(e^{2 k-(-12 k)}\) \(=8\)
\(e^{14 k}\) \(=8\)
\(14 k\) \(=\log _e 8\)
\(14 k\) \(=3\log _e 2\)
\(14 k\) \(=\dfrac{3}{14} \log _e 2\)

 

\(\text{Substitute \(k\) into (1):}\)

\(A e^{-12\left(\tfrac{3}{14} \log _e 2\right)}\) \(=1\)
\(A e^{-\tfrac{18}{7} \log_e2}\) \(=1\)
\(A \times 2^{-\tfrac{18}{7}}\) \(=1\)
\(A\) \(=2^{\tfrac{18}{7}}\)

 

b.    \(g(x)\) \(=2^{\tfrac{18}{7}} \times e^{\left(\tfrac{3}{14} \log _e 2\right) x}\)
    \(=2^{\tfrac{18}{7}} \times e^{\left(\tfrac{3 x}{14} \log _e 2\right)}\)
    \(=2^{\tfrac{18}{7}} \times e^{\left(\log _e 2^{\tfrac{3x}{14}}\right)}\)
    \(=2^{\tfrac{18}{7}} \times 2^{\tfrac{3 x}{14}}\)

Mean mark (b) 55%.
 

c.    \(\text{Area}=\displaystyle \int_{-12}^2 f(x)-g(x)=15.87 \ \text{u}^2\)
 

d.i.  \(h(x)=f(x)-g(x)\)

\(h(x)=\dfrac{x}{2}+7-2^{\tfrac{18}{7}} \times e^{k x}\)

\(h^{\prime}(x)=\dfrac{1}{2}-\dfrac{6 \log _e 2}{7} \times 2^{\tfrac{3 x}{14}+\tfrac{4}{7}}\)
 

d.ii. \(\text{Solve}\ \ h^{\prime}(x)=0\ \ \text{for}\ x\ \text{(by CAS):}\) 

\(x=-3.829\)

\(\text{Substitute into} \ \ h(x):\)

\(h(x)_{\text{max}}=1.72\)
 

e.   \(\text{Strategy 1}\)

\(\text{By CAS, find} \ \ g^{-1}(x):\)

\(g^{-1}(x)=\dfrac{2\left(7 \log _e x-7 \log _e 8+3 \log _e 2\right)}{3 \log _e 2}\)

\(\text{Solve} \ \ g^{-1}(x)=2(x-7) \ \ \text{for} \ x:\)

\(x=1,8\)

\(\text{Intersection at}\ (1,-12) \ \text{and} \ (8,2).\)
 

\(\text{Strategy 2}\)

\(y=2(x-7) \ \ \text{is the inverse of}\ \  y=\dfrac{x}{2}+7\ \ \text{(i.e.}\ f(x)).\)

\(\text{Two inverse functions will intersect at}\ (1,-12) \ \text{and} \ (8,2).\)
 

f.i.  \(f(x)=\dfrac{x}{2}+7\)

\(F(x)=\displaystyle \int f(x)\ d x=\dfrac{1}{4} x^2+7 x+c\)
 

\(\text{If} \ F(x) \ \text{passes through} \ (-12,1):\)

\(F(-12)=\dfrac{1}{4}(-12)^2+7(-12)+c=1 \ \ \Rightarrow\ \ c=49\)
 

\(\text{If \(F(x)\) passes through \((2,8)\):}\)

\(F(2)=\dfrac{1}{4}(2)^2+7(2)+c=8 \ \ \Rightarrow \ \ c=-7\)

\(\text{Since \(c\) cannot have 2 values, it cannot pass through both points.}\)

♦ Mean mark (f.i) 50%.

f.ii.  \(F(x)=\dfrac{1}{4} x^2+7 x+c\)

\(\text{Dilation of \(m\) from \(x\)-axis passes through \((-12,1)\) and \((2,8)\).}\)

\(\text{Solve simultaneously: }\)

\(m \times F(-12)=36 m-84 m+m c=1\ \ldots\ (1)\)

\(m \times F(2)=m+14 m+m c=8\ \ldots\ (2)\)

\(m=\dfrac{1}{9}, \ c=57\)

♦♦ Mean mark (f.ii) 34%.

Calculus, MET2 2025 VCAA 1

Let  \(g: R \rightarrow R\)  be defined by  \(g(x)=4 x^3-3 x^4\).

  1. Find the coordinates of both stationary points of \(g\).   (2 marks)

  2. Sketch the graph of \(y=g(x)\) on the axes below, labelling the stationary points and axial intercepts with their coordinates.   (2 marks)

   

  1. Complete the following gradient table with appropriate values of \(x\) and \(g^{\prime}(x)\) to show that \(g\) has a stationary point of inflection.   (2 marks)

\begin{array}{|c|c|c|c|}
\hline \rule{0pt}{2.5ex}x  \rule[-1ex]{0pt}{0pt}& \quad \quad \quad \quad& \quad \quad \quad \quad & \quad \quad \quad \quad\\
\hline\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& & & \\
\hline
\end{array}

  1. Find the average value of \(g\) between  \(x=0\)  and  \(x=2\).   (2 marks)
  2. Let \(h\) be the result after applying a sequence of transformations to \(g\), such that \(h\) has a stationary point of inflection at  \((1,0)\) and a local maximum at \((-1,1)\).
  3. Write down a possible sequence of three transformations to map from \(g\) to \(h\).   (3 marks)

  4. Let  \(X \sim \operatorname{Bi}(4, p)\) be a binomial random variable.
  5. Show that  \(\operatorname{Pr}(X \geq 3)=g(p)\) for all \(p \in[0,1]\).   (2 marks)

Show Answers Only

a.    \(\text{SP’s at} \ (0,0) \ \text{and}\ (1,1)\).
 

b.   

c.

\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}&  \quad -1 \quad & \quad \quad 0 \quad \quad & \quad \quad \frac{1}{2} \quad \quad\\
\hline
\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& 24 & 0 & \frac{3}{2} \\
\hline
\end{array}

 
d.    \(I=-\dfrac{8}{5}\)
 

e.    \(\text{Possible sequences include:}\)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 1}\rule[-1ex]{0pt}{0pt}&\text{Sequence 2} \\
\hline
\rule{0pt}{2.5ex}\text {1. Reflect in the } y \text {-axis.}\rule[-1ex]{0pt}{0pt}&\text{1. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}& \text{2. Translate 1 unit left.} \\
\hline
\rule{0pt}{2.5ex}\text {3. Translate 1 unit right.}\quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.}  \\
\hline
\end{array}

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 3}\rule[-1ex]{0pt}{0pt}&\text{Sequence 4} \\
\hline
\rule{0pt}{2.5ex}\text{1. Reflect in the \(y\)-axis.} \rule[-1ex]{0pt}{0pt}&\text{1. Translate 0.5 units left.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Translate 0.5 units right.}\rule[-1ex]{0pt}{0pt}&\text{2. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{3. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.} \\
\hline
\end{array}

 
f.    \(X \sim \operatorname{Bi}(4, p)\)

\(\operatorname{Pr}(X \geqslant 3)\) \(=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)\)
  \(=4 p^3(1-p)+p^4\)
  \(=4 p^3-4 p^4+p^4\)
  \(=4 p^3-3 p^4\)
  \(=g(p)\)
Show Worked Solution

a.    \(g(x)=4 x^3-3 x^4\)

\(g^{\prime}(x)=12 x^2-12 x^3=12 x^2(1-x)\)

\(\text{Solve} \ \ g^{\prime}(x)=0:\)

\(x=0,1\)

\(\text{SP’s at} \ (0,0) \ \text{and}\ (1,1)\).
 

b.    \(\text{Find} \ x \text {-intercepts:}\)

\(\text{Solve} \ \ g(x)=0 \ \ \Rightarrow\ \ x=0, \dfrac{4}{3}\)
 

c.

\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}&  \quad -1 \quad & \quad \quad 0 \quad \quad & \quad \quad \frac{1}{2} \quad \quad\\
\hline
\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& 24 & 0 & \frac{3}{2} \\
\hline
\end{array}

 
d.    \(\text{Solve integral (by CAS):}\)

\(I=\dfrac{1}{2-0} \displaystyle \int_0^2 g(x)\ d x=-\dfrac{8}{5}\)
 

e.    \(\text{Possible sequences include:}\)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 1}\rule[-1ex]{0pt}{0pt}&\text{Sequence 2} \\
\hline
\rule{0pt}{2.5ex}\text {1. Reflect in the } y \text {-axis.}\rule[-1ex]{0pt}{0pt}&\text{1. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}& \text{2. Translate 1 unit left.} \\
\hline
\rule{0pt}{2.5ex}\text {3. Translate 1 unit right.}\quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.}  \\
\hline
\end{array}

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 3}\rule[-1ex]{0pt}{0pt}&\text{Sequence 4} \\
\hline
\rule{0pt}{2.5ex}\text{1. Reflect in the \(y\)-axis.} \rule[-1ex]{0pt}{0pt}&\text{1. Translate 0.5 units left.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Translate 0.5 units right.}\rule[-1ex]{0pt}{0pt}&\text{2. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{3. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.} \\
\hline
\end{array}

 
f.    \(X \sim \operatorname{Bi}(4, p)\)

\(\operatorname{Pr}(X \geqslant 3)\) \(=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)\)
  \(=4 p^3(1-p)+p^4\)
  \(=4 p^3-4 p^4+p^4\)
  \(=4 p^3-3 p^4\)
  \(=g(p)\)
♦ Mean mark (e) 40%.
Mean mark (f) 51%

ENGINEERING, TE 2025 HSC 7 MC

Which of the pictorial drawings is an example of isometric drawing?
 

 

Show Answers Only

\(A\)

Show Worked Solution
  • In an isometric drawing, all three axes are drawn at 120° to each other, with vertical lines remaining vertical.
  • Drawing A shows all receding axes at 30° to the horizontal — the defining feature of isometric projection.
  • Drawings B and C show one face drawn true shape with receding lines at an oblique angle — characteristic of oblique projection.
  • Drawing D shows receding lines at an angle inconsistent with isometric construction.

\(\Rightarrow A\)

Functions, EXT1 F1 2025 MET2 5 MC

Which of the following sets represents a function that has an inverse function?

  1. \(\{(1,3),(2,0),(2,1)\}\)
  2. \(\{(-1,3),(2,2),(3,1)\}\)
  3. \(\{(-1,3),(0,1),(1,3)\}\)
  4. \(\{(1,0),(2,3),(1,3)\}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Option A and D are not functions (one x-value has two y-values).}\)

\(\text{Consider option B:}\)

\(\{(-1,3),(2,2),(3,1)\} \ \ \text{is a} \  1:1 \ \text {function.}\)

\(\therefore \ \text {Inverse exists}\)

\(\text{Note that option C is not one-to-one because two different inputs (-1 and 1) map to}\)

\(\text{the same output (3), so its inverse would not be a function.}\)

\(\Rightarrow B\)

Trigonometry, 2ADV T3 2025 MET2 1 MC

A function that has a range of \([6,12]\) is

  1. \(f(x)=6+3 \cos (9 x)\)
  2. \(f(x)=6+6 \cos (3 x)\)
  3. \(f(x)=9-3 \cos (6 x)\)
  4. \(f(x)=9-6 \cos (3 x)\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{By trial and error,}\)

\(\text{Consider option C:}\ \  f(x)=9-3 \cos (6 x)\)

\(\text{Since}\ \ -1 \leqslant \cos (6 x) \leqslant 1\)

\(\text{Range is} \ \ [-3+9,3+9]=[6,12]\)

\(\Rightarrow C\)

Statistics, SPEC2 2025 VCAA 6

The volume of water, \(V\) mL, consumed by a student during a school day may be assumed to be normally distributed with a mean of 1000 mL and a standard deviation of 80 mL .

    1. Write down the mean and standard deviation of the sampling distribution for the average volume of water consumed by randomly selected samples of 25 students.
    2. Give your answers in millilitres.   (1 mark)

    3. What is the probability, correct to four decimal places, that the average volume of water consumed by a random sample of 25 students on a particular school day is more than 970 mL?   (1 mark)

The canteen at a particular school stocks two brands of water in bottles, Wasser and Apa.

The manufacturer of Wasser bottled water knows that the volume of water dispensed into bottles may be assumed to be normally distributed with a standard deviation of 5 mL. Engineers at the company take a random sample of 30 bottles and measure the volume of water in each bottle. The sample mean is found to be 750 mL.

  1. Find a 95% confidence interval for the mean volume of water dispensed into each Wasser bottle.
  2. Give your values in millilitres, correct to one decimal place.   (1 mark)

  3. The engineers decide to take 300 random samples, each containing 30 bottles, and calculate the respective 95% confidence intervals. All samples are independent.
  4. In how many of these confidence intervals would the engineers expect the value of the true mean volume dispensed to be included?   (1 mark)

  5. What is the minimum size of the sample required to ensure that the difference between the sample mean and the mean volume dispensed is no more than 1 mL at the 95% confidence level?   (1 mark)

The volume of water dispensed into Apa water bottles may be assumed to be normally distributed with a mean of 750 mL and a standard deviation of 5 mL. After a service, a random sample of 50 bottles gave a sample mean of 748 mL. The company now claims that the mean volume of water dispensed is less than the stated mean of 750 mL.

A one-tailed statistical test at the 1% level of significance is proposed.

  1. Write down the null and alternative hypotheses that will be used in testing the company's claim.   (1 mark)

    1. Determine the \(p\) value for this test.
    2. Give your answer correct to four decimal places.   (1 mark)

    3. Is the company's claim correct?
    4. Explain your conclusion in terms of the \(p\) value.   (1 mark)

  3. At the 1% level of significance for a sample size of 50 bottles, find the critical value of the sample mean, below which a sample mean value would support the conclusion that the mean volume of water dispensed is now less than 750 mL.
  4. Give your answer correct to three decimal places.   (1 mark)

  5. Assume that, after the service, the true mean volume of water in the Apa bottles was found to be 747.5 mL and that the population standard deviation, \(\sigma\), is 5 mL.
  6. At the 1% level of significance, for a sample size of 50 , find the probability that the company will conclude that the service has not reduced the mean volume of water in an Apa bottle.
  7. Give your answer correct to three decimal places.   (1 mark)

Show Answers Only

a.i.  \(\mu=1000,\ \ \sigma=16\)

a.ii.  \(P(V>970)=0.9696\)

b.    \(\text{95% CI}=(748.2,751.8)\) 

c.    \(95 \% \times 300=285\)

d.    \(n=97\) 

e.    \(H_0: \ \mu=750, \ H_1: \ \mu<750\)

f.i.  \(p=0.0023\) 

f.ii.   \(\text{Since \(0.0023<0.01\) (significance level), claim is correct.}\)

g.    \(\overline{X}=748.355 \ \text{mL}\)

h.    \(p=0.113\)

Show Worked Solution
a.i.   \(\mu\) \(=1000\)
  \(\sigma\) \(=\dfrac{80}{\sqrt{16}}=16\)

 

a.ii.  \(P(V>970)=0.9696\)
 

b.    \(\overline{X} \sim N\left(750, \dfrac{5}{\sqrt{30}}\right)\)

\(\text{95% CI}\) \(=\left(750-1.96 \times \dfrac{5}{\sqrt{30}}, 750+1.96\times\dfrac{5}{\sqrt{30}}\right)\)
  \(=(748.2,751.8)\)

 

c.    \(95 \% \times 300=285\)
 

d.    \(\text {Solve for} \ n:\)

\(1\) \(\geqslant 1.96 \times \dfrac{5}{\sqrt{n}}\)
\(n\) \(\geqslant 96.04\)
\(n\) \(=97\)

 

e.    \(H_0: \ \mu=750\)

\(H_1: \ \mu<750\)
 

f.i.    \(p\) \(=\operatorname{Pr}\left(z<\dfrac{748-750}{\frac{5}{\sqrt{50}}}\right)\)
    \(=\operatorname{Pr}\left(z<-\dfrac{2 \sqrt{50}}{5}\right)\)
    \(=0.0023\)

 

f.ii.   \(\text{Since \(0.0023<0.01\) (significance level), claim is correct.}\)
 

g.    \(\text{Solve for} \ c:\ \ \operatorname{Pr}(\overline{X}<c)=0.01\)

\(c=748.355 \ \text{mL}\)
 

h.    \(p=0.113\)

Vectors, SPEC2 2025 VCAA 5

Consider three planes defined by the equations  \(\Pi_1: \ 2 x+9 z=8, \Pi_2: \ 3 x+6 y+5 z=7\)  and  \(\Pi_3: \ x+9 y-3 z=7\).

  1. Find the point of intersection of the three planes.   (1 mark)

    1. Find a vector that gives the direction of the line of intersection of the planes \(\Pi_2\) and \(\Pi_3\).   (2 marks)

    2. Find a set of parametric equations that give the coordinates of the points that lie on this line of intersection.   (1 mark)

  3. Find the shortest distance from the point \((1,1,2)\) to the plane \(\Pi_3\).   (2 marks)

  4. Consider a family of planes, \(\Psi\), with equation  \(6 x+27 z=m\), where \(m \in N\).
    1. Show that the plane \(\Pi_1\) is parallel to each member of \(\Psi\).   (1 mark)

    2. Find all values of \(m\) for which the shortest distance between plane \(\Pi_1\) and the plane of the form  \(6 x+27 z=m\)  is  \(\dfrac{23}{3 \sqrt{85}}\).   (3 marks)

Show Answers Only

a.    \(\text{Intersection at}\ (-5,2,2)\)
 

b.i.  \(\underset{\sim}{d}=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)\)
 

b.ii. \(x=-5-63 \lambda\)

\(y=2 + 14 \lambda\)

\(z=2+21 \lambda\)
 

c.    \(\dfrac{3}{\sqrt{91}}\)
 

d.i.  \(\Psi: \ 6 x+27 z=m\)

\(\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}\)

\(\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}\)

\(\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}\)

d.ii.  \(m=1,47\)

Show Worked Solution

a.    \(\Pi_1: \ 2 x+9 z=8\)

\(\Pi_2: \ 3 x+6 y+5 z=7\)

\(\Pi_3: \ x+9 y-3 z=7\)

\(\text {Solve simultaneously (by CAS):}\)

\(\text{Intersection at}\ (-5,2,2)\)
 

b.i.  \(\underset{\sim}{d} \ \text{is} \perp \text{to normals of} \ \Pi_2 \ \text{and}\  \Pi_3.\)

\(\underset{\sim}{d}=n_2 \times n_3=\left|\begin{array}{ccc}i & j & k \\ 3 & 6 & 5 \\ 1 & 9 & -3\end{array}\right|=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)\)
 

b.ii. \((-5,2,2)\ \text{lies on both planes (from part a.)}\)

\(x=-5-63 \lambda\)

\(y=2 + 14 \lambda\)

\(z=2+21 \lambda\)

♦ Mean mark (b.ii) 51%.

c.    \(\text{Find shortest distance from (1, 1, 2) to \(\Pi_3\).}\)

\(\Pi_3: \ x+9 y-3 z=7 \ \ \Rightarrow \ \ (7,0,0)\ \text{lies on plane.}\)

\(\text {Spanning vector to} \ (1,1,2) \ \text{is} \ (-6 \underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})\)

\(\text{Distance}=\dfrac{1}{\sqrt{91}} \times \abs{\left(\begin{array}{c}-6 \\ 1 \\ 2\end{array}\right)\left(\begin{array}{c}1 \\ 9 \\ -3\end{array}\right)}=\dfrac{3}{\sqrt{91}}\)
 

d.i.  \(\Psi: \ 6 x+27 z=m\)

\(\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}\)

\(\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}\)

\(\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}\)
  

d.ii. \(\text{Point on}\ \Pi_1: (4,0,0)\)

\(\text{Point on} \ \Psi:\left(\dfrac{m}{6}, 0,0\right)\)

\(\text{Spanning vector}=\left(4-\dfrac{m}{6}\right) \underset{\sim}{i}\)

\(\text{Distance}=\abs{\left(4-\dfrac{m}{6}\right) \underset{\sim}{i} \cdot \dfrac{(2 \underset{\sim}{i}+9\underset{\sim}{k})}{\sqrt{85}}}=\dfrac{23}{3 \sqrt{85}}\)

\(\text{Solve}\ \ \abs{8-\dfrac{m}{3}}=\dfrac{23}{3} \ \ \text{for} \ m:\)

\(m=1,47\)

♦ Mean mark (d.ii) 45%.

Vectors, SPEC2 2025 VCAA 4

The path of a moving particle with position vector

\(\underset{\sim}{r}(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

is shown below for time \(t \geq 0\).

All lengths are in metres and time is measured in seconds.
 

  1. Write down the coordinates of the particle's starting point.   (1 mark)

  2. On the graph above, draw an arrow from the point \((9,0)\) to indicate the direction of motion of the particle.   (1 mark)

  3. Find the value of \(t\) for which the particle will first return to its starting point.   (1 mark)

  4. Show that the speed of the particle, in m s\(^{-1}\), at time \(t\) can be expressed as  \(\sqrt{125-100 \cos \left(\dfrac{3 t}{2}\right)}\).   (3 marks)

  5. What is the maximum speed of the particle in m s\(^{-1}\)?   (1 mark)

  6. On the graph above, trace the path of the particle for  \(t \in[0, \pi]\).   (1 mark)

  7. Find the length of the path traced in part f, giving your answer in metres, correct to one decimal place.   (2 marks)

Show Answers Only

a.    \(\text{Initial coordinates:}\ (1,0)\)
 

b.    
     

c.   \(t=4 \pi \ \text{seconds}\)

d.    \(r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}\)

\(\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

\(\text {Speed}=\abs{\dot{r}(t)}:\)

\(\text{Speed}^2\) \(=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2\)
  \(=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)\)
       \(\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)\)
  \(=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)\)
  \(=125-100 \cos \left(\dfrac{5 t}{2}-t\right)\)
\(\text{speed}\) \(=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}\)

 

e.    \(\text{Speed}_{\text {max}}=15 \ \text{ms}^{-1}\)
 

f.    \(\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).\)
 


 

g.    \(\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}\)

Show Worked Solution

a.    \(\text{At} \ \ t=0:\)

\(r(t)=(5 \cos 0-4 \cos 0)\underset{\sim}{i} + (5 \sin 0-4 \sin 0) {\underset{\sim}{j}}=\underset{\sim}{i}\)

\(\text{Initial coordinates:}\ (1,0)\)
 

b.    
     

c.   \(\text{Solve for}\ t \ \text{(by CAS):}\)

\(5 \cos (t)-4 \cos \left(\dfrac{5t}{2}\right)=1\)

\(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)=0\)

\(t=4 \pi \ \text{seconds}\)

Mean mark (c) 53%.
 

d.    \(r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}\)

\(\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

\(\text {Speed}=\abs{\dot{r}(t)}:\)

\(\text{Speed}^2\) \(=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2\)
  \(=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)\)
       \(\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)\)
  \(=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)\)
  \(=125-100 \cos \left(\dfrac{5 t}{2}-t\right)\)
\(\text{speed}\) \(=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}\)

 

e.    \(\text{Speed}_{\text {max }} \ \text {occurs when} \ \ \cos \left(\frac{3 t}{2}\right)=-1\)

\(\text{Speed}_{\text {max}}=\sqrt{125+100}=15 \ \text{ms}^{-1}\)
 

f.    \(\text{At} \ \ t=\pi, \quad r(\pi)=-5\underset{\sim}{i}-4\underset{\sim}{j}\)

\(\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).\)
 


 

g.    \(\text{Length of curve (by CAS):}\)

\(\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}\)

Functions, MET2 2025 VCAA 5 MC

Which of the following sets represents a function that has an inverse function?

  1. \(\{(1,3),(2,0),(2,1)\}\)
  2. \(\{(-1,3),(2,2),(3,1)\}\)
  3. \(\{(-1,3),(0,1),(1,3)\}\)
  4. \(\{(1,0),(2,3),(1,3)\}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Option A and D are not functions (x-values have two y-values).}\)

\(\text{Consider option B:}\)

\(\{(-1,3),(2,2),(3,1)\} \ \text{is a one-to-one function.}\)

\(\therefore \ \text {Inverse exists}\)

\(\text{Note that option C is not one-to-one because two different inputs (-1 and 1) map to}\)

\(\text{the same output (3), so its inverse would not be a function.}\)

\(\Rightarrow B\)

Graphs, MET2 2025 VCAA 1 MC

A function that has a range of \([6,12]\) is

  1. \(f: R \rightarrow R, \ f(x)=6+3 \cos (9 x)\)
  2. \(f: R \rightarrow R, \ f(x)=6+6 \cos (3 x)\)
  3. \(f: R \rightarrow R, \ f(x)=9-3 \cos (6 x)\)
  4. \(f: R \rightarrow R, \ f(x)=9-6 \cos (3 x)\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{By trial and error,}\)

\(\text{Consider option C:}\ \  f(x)=9-3 \cos (6 x)\)

\(\text{Since}\ \ -1 \leqslant \cos (6 x) \leqslant 1\)

\(\text{Range is} \ \ [-3+9,3+9]=[6,12]\)

\(\Rightarrow C\)

Complex Numbers, SPEC2 2025 VCAA 2

  1. Sketch \(\{z: z \bar{z}=4, z \in C\}\) on the Argand plane below.   (1 mark)
     

    1. Show that \(\{z:|z-2 i|=|z-\sqrt{3}-i|, z \in C\}\) may be expressed as  \(y=\sqrt{3} x\).   (2 marks)

    2. Sketch \(\{z:|z-2 i|=|z-\sqrt{3}-i|, z \in C\}\) on the Argand plane in part a.   (1 mark)

    1. Find the points of intersection of the curves defined in part a and in part b.i, expressing your answers in the form  \(a+i b\), where  \(a, b \in R\).   (2 marks)

    2. Label these points on the Argand plane in part a.   (1 mark)

Consider the points \(P\) and \(Q\) labelled on the Argand plane below.
 

  1. A ray originating at point \(P\) and passing through point \(Q\) has the equation  \(\operatorname{Arg}\left(z-z_0\right)=\theta\), where \(\theta\) is a radian measure.
  2. Write down the values of \(z_0\) and \(\theta\).   (1 mark)

  3. Find the area of the minor segment bounded by the chord connecting the points \(P\) and \(Q\) and the circle given by  \(|z|=3\).
  4. Give your answer in the form \(c \pi+d\), where \(c, d \in R\).   (2 marks)

Show Answers Only

a.    \(z \bar{z}=4\ \ \Rightarrow\ \ \text{Circle, centre (0, 0), radius = 2}\)
 

b.i  \(\abs{z-2 i}=\abs{x+(y-2) i}=x^2+(y-2)^2\)

\(\abs{z-\sqrt{3}-i}=\abs{x-\sqrt{3}+(y-1)^2}=(x-\sqrt{3})^2+(y-1)^2\)

\(\text{Equating expressions:}\)

\(x^2+(y-2)^2=(x-\sqrt{3})^2+(y-1)^2\)

\(x^2+y^2-4 y+4=x^2-2 \sqrt{3} x+3+y^2-2 y+1\)

\(-2 y\) \(=-2 \sqrt{3} x\)
\(y\) \(=\sqrt{3} x\)

 

b.ii    \(\text {See image in part (a).}\)

  \(\abs{z-2 i}=\abs{z-\sqrt{3}-i} \ \Rightarrow \ \text{see straight line in image.}\)
 

c.i   \(\text{Find intersection:}\)

\(x^2+y^2=4 \ \ \text{and} \ \ y=\sqrt{3} x\)

\(x^2+3 x^2=4 \ \ \Rightarrow \ \ x^2=1 \ \ \Rightarrow \ \ x= \pm 1\)

\(\text{Intersection co-ordinates:}\ \ (1, \sqrt{3}),(-1,-\sqrt{3})\)

\(z=1+\sqrt{3} i\)

\(z=-1-\sqrt{3} i\)
 

c.ii  \(\text{See image in part (a)}\)
 

d.    \(z_0=\dfrac{3}{2}+\dfrac{3 \sqrt{3}}{2} i\)

\(\theta=-\dfrac{\pi}{3}\)
 

e.    \(A\) \(=\dfrac{1}{2} \times 3^2 \times\left(\dfrac{\pi}{3}-\sin \dfrac{\pi}{3}\right)\)
    \(=\dfrac{3}{2} \pi-\dfrac{9 \sqrt{3}}{4}\)
Show Worked Solution

a.    \(z \bar{z}=4\ \ \Rightarrow\ \ \text{Circle, centre (0, 0), radius = 2}\)
 

b.i  \(\abs{z-2 i}=\abs{x+(y-2) i}=x^2+(y-2)^2\)

\(\abs{z-\sqrt{3}-i}=\abs{x-\sqrt{3}+(y-1)^2}=(x-\sqrt{3})^2+(y-1)^2\)

\(\text{Equating expressions:}\)

\(x^2+(y-2)^2=(x-\sqrt{3})^2+(y-1)^2\)

\(x^2+y^2-4 y+4=x^2-2 \sqrt{3} x+3+y^2-2 y+1\)

\(-2 y\) \(=-2 \sqrt{3} x\)
\(y\) \(=\sqrt{3} x\)

 

b.ii    \(\text {See image in part (a).}\)

  \(\abs{z-2 i}=\abs{z-\sqrt{3}-i} \ \Rightarrow \ \text{see straight line in image.}\)
 

c.i   \(\text{Find intersection:}\)

\(x^2+y^2=4 \ \ \text{and} \ \ y=\sqrt{3} x\)

\(x^2+3 x^2=4 \ \ \Rightarrow \ \ x^2=1 \ \ \Rightarrow \ \ x= \pm 1\)

\(\text{Intersection co-ordinates:}\ \ (1, \sqrt{3}),(-1,-\sqrt{3})\)

\(z=1+\sqrt{3} i\)

\(z=-1-\sqrt{3} i\)
 

c.ii  \(\text{See image in part (a)}\)
 

d.    \(z_0=\dfrac{3}{2}+\dfrac{3 \sqrt{3}}{2} i\)

\(\theta=-\dfrac{\pi}{3}\)

Mean mark (d) 51%.
e.    \(A\) \(=\dfrac{1}{2} \times 3^2 \times\left(\dfrac{\pi}{3}-\sin \dfrac{\pi}{3}\right)\)
    \(=\dfrac{3}{2} \pi-\dfrac{9 \sqrt{3}}{4}\)

Calculus, SPEC2 2025 VCAA 1

  1. Sketch the graph of  \(y(x)=\dfrac{3 x}{x^3+x+2}\)  on the axes below.
  2. Label the asymptotes with their equations, and label the turning point and the point of inflection with their coordinates. Give the coordinates of the point of inflection correct to one decimal place.   (3 marks)

  1. The region bounded by the graph of  \(y=\dfrac{3 x}{x^3+x+2}\), the coordinate axes and the line  \(x=2\)  is rotated about the \(x\)-axis to form a solid of revolution.
    1. Write down a definite integral that, when evaluated, will give the volume of the solid of revolution.   (1 mark)

    2. Find the volume of the solid of revolution correct to two decimal places.   (1 mark)

  2. Find the equations of the vertical asymptotes of the curve given by  \(y=\dfrac{3 x}{x^3-5 x+2}\).   (1 mark)

  3. A family of curves is given by  \(y(x)=\dfrac{3 x}{x^3+a x+2}\), where  \(a \in R\).
    1. Consider the case where the graph has a stationary point \(P\).
    2. Find the \(y\)-coordinate of \(P\) in terms of \(a\).   (1 mark)

    3. For a given value of \(a\), the graph has no stationary points.
    4. Find the equations of the vertical asymptotes of the graph in this case.   (1 mark)

    5. For a given value of \(a\), the graph will have a point of inflection at  \(x=2\).
    6. Find the value of \(a\).   (2 marks)

Show Answers Only

a.    

b.i.    \(V=\displaystyle \int_0^2 \pi\left(\frac{3 x}{x^3+x+2}\right)^2 \, d x\)

b.ii.  \(V=2.29\ \text{u}^3\)

c.   \(x=-\sqrt{2}-1, \ x=\sqrt{2}-1, \ x=2\)

d.i.  \(\text{SP at} \ \left(1, \dfrac{3}{3+a}\right)\)
 

d.ii. \(\text{No SP’s exist when}\ \ 3+a=0 \ \ \Rightarrow \ \ a=-3\)

\(x=-2,1\)

d.iii.  \(a=\dfrac{24}{5}\)

Show Worked Solution

a.    

b.i.    \(V=\displaystyle \int_0^2 \pi\left(\frac{3 x}{x^3+x+2}\right)^2 \, d x\)
 

b.ii.  \(V=2.29\ \text{u}^3\)
 

c.   \(\text{Find vertical asymptotes.}\)

\(\text {Solve:} \ \ x^3-5 x+\dfrac{1}{2}=0\)

\(x=-\sqrt{2}-1, \ x=\sqrt{2}-1, \ x=2\)
 

d.i.  \(y=\dfrac{3 x}{x^3+a x+2}\)

\(\text{Find \(x\) when \(y^{\prime}=0 \ \ \Rightarrow \ \ x=1\)}\)

\(\text{SP at} \ \ \left(1, \dfrac{3}{3+a}\right)\)
 

d.ii. \(\text{No SP’s exist when}\ \ 3+a=0 \ \ \Rightarrow \ \ a=-3\)

\(y=\dfrac{3 x}{x^3-3 x+2}\)

\(\text{Vertical asymptotes occur when}\ \ x^3-3 x+2=0\)

\(\text{(By CAS):} \ \ x=-2,1\)

♦ Mean mark (d.ii) 46%.

d.iii.  \(\text{If POI exists at} \ \ x=2:\)

\(y^{\prime \prime}(2)=0 \ \Rightarrow \ \dfrac{-3(5 a-24)}{2(a+5)^3}=0\)

\(a=\dfrac{24}{5}\)

Calculus, EXT1 C3 2025 SPEC2 8 MC

Consider the direction field below.
 

The direction field best represents the differential equation

  1. \(\dfrac{d y}{d x}=x^2-y\)
  2. \(\dfrac{d y}{d x}=x-y^2\)
  3. \(\dfrac{d y}{d x}=y-x\)
  4. \(\dfrac{d y}{d x}=x-y\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{At \((-1,0)\), gradient is positive (eliminate B and D)}\).

\(\text{At \((1,0)\), gradient is positive (eliminate C)}\).

\(\Rightarrow A\)

Proof, EXT2 P1 2025 SPEC2 1 MC

A tiger is a type of cat.

Consider the following statement.

'If I have a tiger, then I have a cat.'

The contrapositive of this statement is

  1. if I do not have a tiger, then I do not have a cat.
  2. if I have a cat, then I have a tiger.
  3. if I do not have a cat, then I do not have a tiger.
  4. if I do not have a tiger, then I have a different type of cat.
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Statement: If} \ \ X \Rightarrow Y\)

\(\text{Contrapositive: If}\ \ \neg \ Y\ \Rightarrow \neg \ X\)

\(\text{If I don’t have a cat} \ \ \Rightarrow \ \ \text{I don’t have a tiger.}\)

\(\Rightarrow C\)

Probability, 2ADV EQ-Bank 19

A new test has been developed for determining whether or not people are carriers of the Gaussian virus.

Two hundred people are tested. A two-way table is being used to record the results.
 

  1.  What is the value of `A`?   (1 mark)

  2.  A person randomly selected from the tested group is a carrier of the virus.
  3. What is the probability that the test results would show this?   (1 mark) 
  4. If the person randomly selected has a negative result from their test, what is the probability they are not a carrier of the virus?   (2 marks)

Show Answers Only

a.    `98`

b.    `37/43`

c.    `49/55`

Show Worked Solution

a.    `A= 200-(74 + 12 + 16)= 98`
 

b.    `P` `= text(#Positive carriers)/text(Total carriers)`
  `= 74/86`
  `= 37/43`

 

c.    `text(Total number with negative results) = 110`

`text{Total non-carriers with negative results}\ = 98`

`P\text{(not a carrier)|negative result}\ = 98/110=49/55`

Probability, 2ADV EQ-Bank 16

Lie detector tests are not always accurate. A lie detector test was administered to 200 people.

The results were:

• 50 people lied. Of these, the test indicated that 40 had lied;
• 150 people did NOT lie. Of these, the test indicated that 20 had lied.

  1. Complete the table using the information above   (1 mark)
      
        

  2. For what percentage of the people tested was the test accurate?   (1 mark)

  3. What is the probability that the test indicated a lie for a person who did NOT lie?   (1 mark)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `text(85%)`

c.    `2/15`

Show Worked Solution

a.

b.  `text(Percentage of people with accurate readings)`

`= text(# Accurate readings)/text(Total readings) xx 100`

`= (40+130)/200`

`= 85 text(%)`
 

c.  `text{P(lie detected when NOT a lie)}`

`= 20/150`

`= 2/15`

Statistics, 2ADV EQ-Bank 4 MC

A random variable \(X\) is defined as the volume of fuel (in litres) dispensed by a bowser per use. Which statement best describes \(X\) ?

  1. \(X\) is discrete because the machine dispenses a fixed amount each time.
  2. \(X\) is discrete because volume can only take positive values.
  3. \(X\) is continuous because the machine is used many times per day.
  4. \(X\) is continuous because volume can take any value within an interval.
Show Answers Only

\(D\)

Show Worked Solution
  • Volume is a measurement variable that can take any value within a range, so \(X\) is a continuous random variable.

\(\Rightarrow D\)

Statistics, 2ADV EQ-Bank 3 MC

Which of the following is an example of a continuous random variable?

  1. The number of text messages sent by a student in a day
  2. The time taken for a swimmer to complete 100 m race
  3. The number of cars in a school car park at 9 am
  4. The number of siblings a student has
Show Answers Only

\(B\)

Show Worked Solution
  • Time is a measurement that can take any value within an interval and is therefore continuous.
  • Options A, B and D are all counts that can only take whole number values, making them discrete.

\(\Rightarrow B\)

Statistics, 2ADV EQ-Bank 2 MC

A hospital records the number of patients admitted to the emergency department each day. How would this data best be classified?

  1. Discrete, because the number of patients changes every day
  2. Discrete, because the count can only take whole number values
  3. Continuous, because the number of patients can be very large
  4. Continuous, because patient admission is an ongoing process
Show Answers Only

\(B\)

Show Worked Solution
  • The number of patients is a count and can only take whole number values, making it a discrete numerical variable.

\(\Rightarrow B\)

Statistics, 2ADV EQ-Bank 1 MC

A scientist measures the wingspan of butterflies in a study. How would this data best be classified?

  1. Discrete, because wingspan can be measured precisely
  2. Discrete, because each butterfly has a unique wingspan
  3. Continuous, because wingspan can take any value within a range
  4. Continuous, because there are many butterflies in the study
Show Answers Only

\(C\)

Show Worked Solution
  • Wingspan is a measurement that can take any value within a range, making it a continuous numerical variable.

\(\Rightarrow C\)

Complex Numbers, EXT2 N2 2025 SPEC1 8

Consider the function with rule  \(f(z)=z^4+6 z^2+25\), where \(z \in C\).

  1. Consider  \(z_1=1+2 i\).
  2. Plot and label \(z_1\) and \(\overline{z}_1\) on the Argand plane below.   (1 mark)  

     
  3. Given that  \(1+2 i\)  is a solution of  \(f(z)=0\), find a quadratic factor of \(f(z)\).   (2 marks)  

  4. Hence, find all remaining solutions of  \(f(z)=0\).   (2 marks)

Show Answers Only

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(z^2-2 z+5\)

c.    \(z=-1+2 i, z=-1-2 i\)

Show Worked Solution

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)

\(\Rightarrow 1-2 i \ \ \text{is also a solution (conjugate factor theorem).}\)

\(\text {Express as a quadratic factor:}\)

\((z-(1+2 i))(z-(1-2 i))\) \(=((z-1)-2 i)((z-1)+2 i)\)
  \(=(z-1)^2-4 i^2\)
  \(=z^2-2 z+5\)

 

c.    \(f(z)=z^4+6 z^2+25, \quad z \in C\)

\(\text{Since \(\ z^2-2 z+5\ \) is a factor:}\)

\(f(z)\) \(=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)\)
  \(=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)\)
  \(=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b\)
  \(=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b\)
Mean mark (c) 53%.

\(\text{Equating co-efficients:}\)

\(a-2=0 \ \Rightarrow \ a=2\)

\(5 b=25 \ \Rightarrow \ b=5\)

\(f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)\)
 

\(\text{Solve:} \ \ z^2+2 z+5=0\)

\(z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i\)

\(\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i\)

Calculus, SPEC2 2025 VCAA 8 MC

Consider the direction field below.
 

The direction field best represents the differential equation

  1. \(\dfrac{d y}{d x}=x^2-y\)
  2. \(\dfrac{d y}{d x}=x-y^2\)
  3. \(\dfrac{d y}{d x}=y-x\)
  4. \(\dfrac{d y}{d x}=x-y\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{At \((-1,0)\), gradient is positive (eliminate B and D)}\).

\(\text{At \((1,0)\), gradient is positive (eliminate C)}\).

\(\Rightarrow A\)

Calculus, SPEC2 2025 VCAA 7 MC

Using the substitution  \(u=\cos (\theta), \dfrac{1}{2} \displaystyle \int_0^{\tfrac{\pi}{2}} \dfrac{\sin (2 \theta)}{1+\cos (\theta)} \, d \theta\)  can be expressed as

  1. \(\displaystyle\int_0^{\tfrac{\pi}{2}} u \sqrt{\frac{1-u}{1+u}}\ d u\)
  2. \(\displaystyle\int_0^1\left(1+\frac{1}{1+u}\right) d u\)
  3. \(\displaystyle\int_0^{\tfrac{\pi}{2}}\left(1-\frac{1}{1+u}\right) d u\)
  4. \(\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) d u\)
Show Answers Only

\(D\)

Show Worked Solution

\(u=\cos (\theta) \ \Rightarrow \ du=-\sin (\theta)\ d \theta\)

\(\text{When} \ \ \theta=\dfrac{\pi}{2}, u=0\)

\(\text{When}\ \  \theta=0, u=1\)

\(I\) \(=\displaystyle\frac{1}{2} \int_0^{\tfrac{\pi}{2}} \frac{\sin (2 \theta)}{1+\cos \theta} \, d \theta\)
  \(=\displaystyle\int_0^{\tfrac{\pi}{2}} \frac{\sin (\theta) \cos (\theta)}{1+\cos (\theta)} \, d \theta\)
  \(=\displaystyle \int_1^0-\frac{u}{1+u} \, d u\)
  \(=\displaystyle\int_0^1 \frac{1+u-1}{1+u} \, d u\)
  \(=\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) \, d u\)

 

\(\Rightarrow D\)

Statistics, STD2 S1 2010 HSC 26b*

A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.

The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.
 

2010 26b

  1. Find the value of  Χ  in the table.   (1 mark)

  2. On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data.   (1 mark)
  3. Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer.   (1 mark)
Show Answers Only

a.    `15`

b. & c.
               

`text(Median)\ ~~155`

Show Worked Solution

a.    `X= 25-10= 15`

b. & c.
               

`text(Median)\ ~~155`

Complex Numbers, SPEC1 2025 VCAA 8

Consider the function with rule  \(f(z)=z^4+6 z^2+25\), where \(z \in C\).

  1. Consider  \(z_1=1+2 i\).
  2. Plot and label \(z_1\) and \(\overline{z}_1\) on the Argand plane below.   (1 mark)  

     
  3. Given that  \(1+2 i\)  is a solution of  \(f(z)=0\), find a quadratic factor of \(f(z)\).   (2 marks)  

  4. Hence, find all remaining solutions of  \(f(z)=0\).   (2 marks)

Show Answers Only

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(z^2-2 z+5\)

c.    \(z=-1+2 i, z=-1-2 i\)

Show Worked Solution

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)

\(\Rightarrow 1-2 i \ \ \text{is also a solution.}\)

\(\text {Express as a quadratic factor:}\)

\((z-(1+2 i))(z-(1-2 i))\) \(=((z-1)-2 i)((z-1)+2 i)\)
  \(=(z-1)^2-4 i^2\)
  \(=z^2-2 z+5\)

 

c.    \(f(z)=z^4+6 z^2+25, \quad z \in C\)

\(\text{Since \(\ z^2-2 z+5\ \) is a factor:}\)

\(f(z)\) \(=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)\)
  \(=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)\)
  \(=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b\)
  \(=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b\)
Mean mark (c) 53%.

\(\text{Equating co-efficients:}\)

\(a-2=0 \ \Rightarrow \ a=2\)

\(5 b=25 \ \Rightarrow \ b=5\)

\(f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)\)
 

\(\text{Solve:} \ \ z^2+2 z+5=0\)

\(z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i\)

\(\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i\)

Trigonometry, 2ADV T3 2025 MET1 3

Let  \(f(x)=2 \cos (2 x)+1\)  over the domain \(x \in\left[0, 2 \pi \right]\).

  1. State the range of \(f(x)\).   (1 mark)

  2. Solve  \(f(x)=0\)  for \(x\).   (3 marks)

  3. Sketch the graph of  \(y=f(x)\)  for  \(x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]\) on the axes below.
  4. Label the endpoints with their coordinates.   (2 marks)

Show Answers Only

a.    \(\text{Range of } f(x):-1 \leqslant y \leqslant 3\)

b.    \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)

c.   

   

Show Worked Solution

a.    \(\text{Amplitude}=2 \ \ \text{about} \ \ y=1.\)

\(\text{Range of } f(x):\ -1 \leqslant y \leqslant 3\)
 

b.     \(2 \cos (2 x)+1\) \(=0\)
  \(\cos (2 x)\) \(=-\dfrac{1}{2}\)

 
\(\text{Base angle}=\dfrac{\pi}{3}\)

\(2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots\)

\(2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}\)

  \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)
  

c.   

   

♦ Mean mark (c) 49%.

Calculus, MET1 2025 VCAA 7

Let \(f: R \rightarrow R, f(x)=x^3-x^2-16 x-20\).

  1. Verify that  \(x=5\)  is a solution of  \(f(x)=0\).   (1 mark)

  2. Express \(f(x)\) in the form  \((x+d)^2(x-5)\), where \(d \in R\).   (2 marks)

  3. Consider the graph of  \(y=f(x)\), as shown below.
  4. Complete the coordinate pairs of all axial intercepts of  \(y=f(x)\).   (1 mark)

     

   

  1. Let \(g: R \rightarrow R, g(x)=x+2\).
    1. State the coordinates of the stationary point of inflection for the graph of  \(y=f(x) g(x)\).   (1 mark)

    2. Write down the values of \(x\) for which  \(f(x) g(x) \geq 0\).   (1 mark)

Show Answers Only

a.    \(f(5)=5^3-5^2-16 \times 5-20 =0\)

b.    \(f(x)=(x+2)^2(x-5)\)

c.   
       

d.i.   \((-2,0)\)

d.ii.  \(x \in(-\infty,-2] \cup[5, \infty)\)

Show Worked Solution

a.    \(f(x)=x^3-x^2-16 x-20\)

\(f(5)=5^3-5^2-16 \times 5-20=125-25-80-20=0\) 

\(\therefore x=5\ \ \text{is a solution of}\ f(x)\).
 

b.    \(\text{By long division:}\)
 
           

\(f(x)=\left(x^2+4 x+4\right)(x-5)=(x+2)^2(x-5)\)
 

c.   
       
 

d.i.    \(y\) \(=f(x) \cdot g(x)\)
    \(=(x+2)^2(x-5)(x+2)\)
    \(=(x+2)^3(x-5)\)

 

\((x+2)^3\ \text{factor}\ \Rightarrow\ \text{SP of inflection at} \ (-2,0)\)

♦ Mean mark (d.i) 44%.
♦♦ Mean mark (d.ii) 27%.
 

d.ii.
     

\(\text{By inspection of graph:}\)

\(f(x) g(x) \geqslant 0 \ \ \text{when}\ \  x \leqslant-2\ \cup\  x \geqslant 5\)

\(x \in(-\infty,-2] \cup[5, \infty) \ \text{also correct.}\)

Probability, MET1 2025 VCAA 4

The probability distribution for the discrete random variable \(X\) is given in the table below, where \(k\) is a positive real number.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad \quad 0 \quad \quad & \quad \quad 1 \quad \quad & \quad \quad 2 \quad \quad & \quad \quad 3 \quad \quad \\
\hline
\rule{0pt}{2.5ex}\operatorname{Pr}(X=x) \rule[-1ex]{0pt}{0pt}& \dfrac{4}{k} &\dfrac{2 k}{75} &\dfrac{k}{75} & \dfrac{2}{k} \\
\hline
\end{array}

  1. Show that  \(k=10\)  or  \(k=15\).   (2 marks)

  2. Let  \(k=15\).
    1. Find \(\operatorname{Pr}(X>1)\).   (1 mark)

    2. Find \(E (X)\).   (1 mark)

Show Answers Only

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

  
\(\therefore k=10 \ \ \text{or 15}\)

b.i.   \(\operatorname{Pr}(X>1)=\dfrac{1}{3}\) 

b.ii.  \(E(X)=\dfrac{90}{75}\)

Show Worked Solution

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

 

\(\therefore k=10 \ \ \text{or 15}\)
 

b.i.   \(\operatorname{Pr}(X>1)\) \(=\operatorname{Pr}(X=2)+\operatorname{Pr}(X=3)\)
    \(=\dfrac{15}{75}+\dfrac{2}{15}\)
    \(=\dfrac{1}{3}\)

 

b.ii.   \(E(X)\) \(=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}\)
    \(=\dfrac{30}{75}+\dfrac{30}{75}+\dfrac{30}{75}\)
    \(=\dfrac{90}{75}\)

BIOLOGY, M8 EQ-Bank 5 MC

Which coordination system detects changes in blood glucose and sends signals to effector organs to restore homeostasis?

  1. Circulatory
  2. Digestive
  3. Hormonal
  4. Neural
Show Answers Only

`D`

Show Worked Solution
  • D is correct: Neural pathways detect changes and send signals to effectors to restore homeostasis.

Other Options:

  • A is incorrect: Circulatory system transports substances; does not detect or signal.
  • B is incorrect: Digestive system processes nutrients; not a coordination system.
  • C is incorrect: Hormonal is a valid coordination system but uses chemical messengers via blood, not neural signals.

Graphs, MET1 2025 VCAA 3

Let  \(f:[0,2 \pi] \rightarrow R, f(x)=2 \cos (2 x)+1\).

  1. State the range of \(f\).   (1 mark)

  2. Solve  \(f(x)=0\)  for \(x\).   (3 marks)

  3. Sketch the graph of  \(y=f(x)\)  for  \(x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]\) on the axes below.
  4. Label the endpoints with their coordinates.   (2 marks)

Show Answers Only

a.    \(\text{Range of } f(x):-1 \leqslant y \leqslant 3\)

b.    \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)

c.   

   

Show Worked Solution

a.    \(\text{Amplitude}=2 \ \ \text{about} \ \ y=1.\)

\(\text{Range of } f(x):\ -1 \leqslant y \leqslant 3\)
 

b.     \(2 \cos (2 x)+1\) \(=0\)
  \(\cos (2 x)\) \(=-\dfrac{1}{2}\)

 
\(\text{Base angle}=\dfrac{\pi}{3}\)

\(2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots\)

\(2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}\)

  \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)
  

c.   

   

♦ Mean mark (c) 49%.

Statistics, STD2 S1 EQ-Bank 18

A boxplot for the sale prices of a sample of 203 homes is shown.
 

  1. Calculate the range of the sale price data in the boxplot.   (1 mark)

  2. Calculate the upper fence for any outliers within the sale price data of the boxplot.   (2 marks)

Show Answers Only

a.    \(\text{Range}=900\,000\)

b.    \(1\,350\,000\)

Show Worked Solution

a.    \(\text{Range}=1\,300\,000-400\,000=900\,000\)
 

b.    \(IQR=900\,000-600\,000=300\,000\)

\(Q_3=900\,000\)

\(\text{Upper Fence}\) \(=Q_3+1.5 \times IQR\)
  \(=900\,000+1.5 \times 300\,000\)
  \(=1\,350\,000\)

Networks, GEN2 2025 VCAA 15

Frances lives in a housing estate.

On the graph below the vertices represent her favourite locations, and the edges represent the roads between them.
 

  1. Calculate the sum of the degrees of all the vertices in this graph.   (1 mark)
  2. Euler's formula, \(v+f=e+2\), holds for this graph.
  3. Complete the formula by writing the appropriate numbers in the boxes below.   (1 mark)

  1. Frances is at the gym. She would like to visit each of the other locations once and end at her home.
  2. What is the mathematical term used to describe this route?   (1 mark)
  3. Using edges from the original graph, construct a spanning tree below.   (1 mark)

     


Show Answers Only

a.    \(\text{Sum of degrees}=2+4+2+3+3=14\)

b.    \(5+4=7+2\)

c.    \(\text{Hamiltonian path}\)

d.    \(\text{Spanning tree (one of many possibilities):}\)
 

Show Worked Solution

a.    \(\text{Sum of degrees}=2+4+2+3+3=14\)
 

b.    \(v+f=e+2 \ \ \Rightarrow \ \ 5+4=7+2\)
 

c.    \(\text{Hamiltonian path}\)

♦ Mean mark (c) 53%.

d.    \(\text{Spanning tree (one of many possibilities):}\)
 

Data Analysis, GEN2 2025 VCAA 2

A boxplot for the sale prices of a sample of 203 homes is shown.
 

  1. Calculate the range of the sale price data in the boxplot.   (1 mark)

  2. Calculate the upper fence for the sale price data in the boxplot.   (1 mark)

Show Answers Only

a.    \(\text{Range}=900\,000\)

b.    \(1\,350\,000\)

Show Worked Solution

a.    \(\text{Range}=1\,300\,000-400\,000=900\,000\)
 

b.    \(IQR=900\,000-600\,000=300\,000\)

\(Q_3=900\,000\)

\(\text{Upper Fence}\) \(=Q_3+1.5 \times IQR\)
  \(=900\,000+1.5 \times 300\,000\)
  \(=1\,350\,000\)

Data Analysis, GEN2 2025 VCAA 1

Table 1, below, shows the prices in dollars, price, for a sample of 20 homes sold in an inner Melbourne suburb during 2017.

The type of home sold is either an apartment or a house.

Table 1

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \ \textit{Price(\$)}\quad \ \rule[-1ex]{0pt}{0pt}& \textit{Type} \\
\hline \rule{0pt}{2.5ex}350\,000 \rule[-1ex]{0pt}{0pt}& \ \ \text{apartment}\ \ \\
\hline \rule{0pt}{2.5ex}490\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}500\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}620\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}720\,000 \rule[-1ex]{0pt}{0pt}\rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}830\,000 & \text{apartment}\\
\hline \rule{0pt}{2.5ex}875\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}995\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,100\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,520\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}800\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}840\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,010\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,263\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,398\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,460\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline
\end{array}

  1. Find the median, in dollars, of the variable price.   (1 mark)

  2. State whether the variable type is numerical, nominal or ordinal.   (1 mark)

    1. Complete the table below by finding the standard deviation, to the nearest whole number, for the sale price of apartments in the sample.   (1 mark)

    2. Table 2
      \begin{array}{|c|c|}
      \hline \rule{0pt}{2.5ex} \quad \ \ \textbf{Type of home} \quad \ \ & \quad \textbf{Standard deviation of} \quad \\
      & \rule[-1ex]{0pt}{0pt}\textbf{sale price (\$)}\\
      \hline \rule{0pt}{2.5ex}\text{house} \rule[-1ex]{0pt}{0pt}& 300\,911 \\
      \hline \rule{0pt}{2.5ex}\text{apartment} \rule[-1ex]{0pt}{0pt}& \\
      \hline
      \end{array}
    3. Using the information in Table 2, comment on the relative spread in the distribution of the sale prices of houses compared with apartments in this sample.   (1 mark)

  4. Table 3, below, shows the percentage of houses and apartments with prices in the given ranges. Some information is missing.
  5. Use the data from Table 1 to complete Table 3.
  6. Table 3 
Show Answers Only

a.    \(\text{Median}=920\,000\)

b.    \(\text{Variable is nominal}\)

c.i.  \(\text{Std deviation = \$346 466}\)

c.ii.  \(\text {Apartment sale prices have a higher spread (standard deviation) than the}\)

\(\text{spread of house sale prices.}\)

d.

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Median}=\dfrac{10^{\text{th}}+11^{\text{th}}}{2}=920\,000\)
 

b.    \(\text{Type is qualitative and cannot be ordered}\)

\(\Rightarrow \ \text{Variable is nominal}\)
 

c.i.  \(\text{By calculator,}\)

\(\text{Std deviation = \$346 466}\)
 

c.ii.  \(\text {Apartment sale prices have a higher spread (standard deviation) than the}\)

\(\text{spread of house sale prices.}\)

♦ Mean mark (c.ii) 47%.

d.

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Networks, GEN1 2025 VCAA 34 MC

Consider the following graph.
 

The number of bridges in this graph is

  1. 1
  2. 2
  3. 3
  4. 4
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Bridge – an edge, if removed, will disconnect the network.}\)

\(\text{The top 3 edges satisfy this test.}\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 29 MC

The female population of an animal species has been divided into four age groups (1, 2, 3 and 4), with age group 1 being the youngest.

The age groups, birth rates and survival rates are presented in the life-cycle diagram below.
 

 
Which one of the following is a Leslie matrix that corresponds with this life-cycle diagram?
 

  1. \(\begin{bmatrix}0 & 2.1 & 4.6 & 1.8 \\ 0.9 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.2 & 0\end{bmatrix}\)
     
  2. \(\begin{bmatrix}0 & 0.9 & 0.7 & 0.2 \\ 2.1 & 0 & 0 & 0 \\ 0 & 4.6 & 0 & 0 \\ 0 & 0 & 1.8 & 0\end{bmatrix}\)
     
  3. \(\begin{bmatrix}2.1 & 4.6 & 1.8 & 0 \\ 0.9 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.2 & 0\end{bmatrix}\)
     
  4. \(\begin{bmatrix}2.1 & 4.6 & 1.8 & 0.8 \\ 0 & 0.9 & 0 & 0 \\ 0 & 0 & 0.7 & 0 \\ 0 & 0 & 0 & 0.2\end{bmatrix}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Age group 1 has no birthrate \(\Rightarrow  \ e_{11}=0\) (eliminate C, D)}\)

\(\text{Only option A has birth/death rates in the correct place.}\)

\(\Rightarrow A\)

Matrices, GEN1 2025 VCAA 27 MC

Consider the matrix \(E\) where

\begin{align*}
E=\begin{bmatrix}
m & -9 \\
4 & n
\end{bmatrix}
\end{align*}

For the inverse of \(E\) to exist, the values of \(m\) and \(n\), respectively, cannot be

  1. \(3\) and \(12\)
  2. \(12\) and \(3\)
  3. \(3\) and \(-12\)
  4. \(-3\) and \(-12\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Inverse exists}\ \ \Rightarrow\ \ \operatorname{det}\,E \neq 0\)

\(\operatorname{det}\,E=m \times n-(-9 \times 4)=mn+36\)

\(\text {Only option C gives} \ \ mn=-36\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 26 MC

Consider the matrices \(A, B\) and \(C\) where

\begin{align*}
A=\begin{bmatrix}
2 & 0 \\
1 & 6
\end{bmatrix}, B=\begin{bmatrix}
3 \\
5
\end{bmatrix} \ \text{and}\ \  C=AB.
\end{align*}

The calculation that correctly determines element \(c_{21}\) is

  1. \(2 \times 3+0 \times 5\)
  2. \(2 \times 5+3 \times 0\)
  3. \(1 \times 3+6 \times 5\)
  4. \(1 \times 5+6 \times 3\)
Show Answers Only

\(C\)

Show Worked Solution

\(AB=\begin{bmatrix}2 & 0 \\ 1 & 6\end{bmatrix}\begin{bmatrix}3 \\ 5\end{bmatrix}=\begin{bmatrix} 6 \\ 33\end{bmatrix}\)

\(c_{12}=1 \times 3+6 \times 5=33\)

\(\Rightarrow C\)