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PHYSICS, M5 2025 HSC 29

A mass moves around a vertical circular path of radius \(r\), in Earth's gravitational field, without loss of mechanical energy. A string of length \(r\) maintains the circular motion of the mass.

When the mass is at its highest point \(B\), the tension in the string is zero.
 

  1. Show that the speed of the mass at the highest point, \(B\), is given by  \(v=\sqrt{r g}\).   (2 marks)

  2. Compare the speed of the mass at point \(A\) to that at point \(B\). Support your answer using appropriate mathematical relationships.   (3 marks)

Show Answers Only

a.    \(\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}\)

\(\dfrac{mv_B^2}{r}\) \(=T+mg\)
\(v_B^2\) \(=rg \ \ (T=0)\)
\(v_B\) \(=\sqrt{r g}\)

 
b.    \(\text {Total} \ ME=E_k+GPE\)

\(\text{At point} \ B :\)

 \(\text {Total} \ ME\) \(=\dfrac{1}{2} m v_B^2+mg(2r)\)
  \(=\dfrac{1}{2} m \times r g+2 mrg\)
  \(=\dfrac{5}{2} m r g\)

 
\(\text{At point} \ A :\)

\(\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg\)
 

\(\text{Since \(ME\) is conserved:}\)

\(\dfrac{5}{2} mrg\) \(=\dfrac{1}{2} m v_A^2+m r g\)
\(\dfrac{1}{2} mv_A^2\) \(=\dfrac{3}{2} m r g\)
\(v_A^2\) \(=3 rg\)
\(v_A\) \(=\sqrt{3rg}=\sqrt{3} \times v_B\)
Show Worked Solution

a.    \(\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}\)

\(\dfrac{mv_B^2}{r}\) \(=T+mg\)
\(v_B^2\) \(=rg \ \ (T=0)\)
\(v_B\) \(=\sqrt{r g}\)

 
b.    \(\text {Total} \ ME=E_k+GPE\)

\(\text{At point} \ B :\)

 \(\text {Total} \ ME\) \(=\dfrac{1}{2} m v_B^2+mg(2r)\)
  \(=\dfrac{1}{2} m \times r g+2 mrg\)
  \(=\dfrac{5}{2} m r g\)

 
\(\text{At point} \ A :\)

\(\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg\)
 

\(\text{Since \(ME\) is conserved:}\)

\(\dfrac{5}{2} mrg\) \(=\dfrac{1}{2} m v_A^2+m r g\)
\(\dfrac{1}{2} mv_A^2\) \(=\dfrac{3}{2} m r g\)
\(v_A^2\) \(=3 rg\)
\(v_A\) \(=\sqrt{3rg}=\sqrt{3} \times v_B\)

PHYSICS, M6 2025 HSC 26

The starting position of a simple AC generator is shown. It consists of a single rectangular loop of wire in a uniform magnetic field of 0.5 T. This loop is connected to two slip rings and the slip rings are connected via brushes to a voltmeter.
 

  1. The loop is rotated at a constant rate through an angle of 90 degrees from the starting position in the direction indicated, in 0.1 seconds.
  2. Calculate the magnitude of the average emf generated during this rotation.   (2 marks)
  1. The same coil was then rotated at 10 revolutions per second from the starting position. The voltage varies with time, as shown in the graph.
     

  1. On the same axes, sketch a graph that shows the variation of voltage with time if the rotational speed is 20 revolutions per second in the opposite direction, beginning at the original starting position.   (3 marks)

Show Answers Only

a.    \(\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}\)

\(\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}\)
 

b.    
           

Show Worked Solution

a.    \(\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}\)

\(\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}\)
 

b.    

BIOLOGY, M8 2025 HSC 24

The following flow chart represents the control of body temperature in humans.
 

  1. Complete the flow chart to give an example of mechanism A and an example of mechanism B.   (2 marks)
  2. Outline how mechanism B maintains homeostasis.   (2 marks)

Show Answers Only

a.    Mechanism A (Decreases temperature):

  • Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

  • Shivering → Vasoconstriction

b.    Maintaining homeostasis

  • When body temperature drops below normal range, thermoreceptors detect the change.
  • The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
  • Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
  • Body temperature increases back to normal range, restoring homeostasis.

Show Worked Solution

a.   Mechanism A (Decreases temperature):

  • Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

  • Shivering → Vasoconstriction

b.    Maintaining homeostasis

  • When body temperature drops below normal range, thermoreceptors detect the change.
  • The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
  • Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
  • Body temperature increases back to normal range, restoring homeostasis.

BIOLOGY, M5 2025 HSC 31

Congenital amegakaryocytic thrombocytopenia (CAMT) is a rare, inherited disorder where bone marrow no longer makes platelets that are important for clotting and preventing bleeding. The pedigree below shows the inheritance of CAMT in a family.
 

  1. What type of inheritance is shown in the pedigree above? Justify your answer?   (3 marks)
Type of Inheritance:  
  1. A CAMT mutation was found to produce the following amino acid sequence:
    1. Glutamine – Tyrosine – Isoleucine – Aspartic acid.
  2. The same DNA fragment has been sequenced from an unaffected individual.
  3. Template strand           GTC ATA CAG CTG.
  4. The following codon chart displays all the codons and corresponding amino acids. The chart translates mRNA sequences into amino acids.
      

  5. Use the codon chart shown to explain the type of mutation which causes CAMT.   (3 marks)
Show Answers Only

a.    Type of inheritance: Autosomal recessive

  • Both males and females are affected equally, ruling out sex-linked inheritance.
  • Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
  • The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
  • Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b.   Type of mutation causing CAMT

  • The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
  • In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
  • This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.
Show Worked Solution

a.    Type of inheritance: Autosomal recessive

  • Both males and females are affected equally, ruling out sex-linked inheritance.
  • Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
  • The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
  • Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b.   Type of mutation causing CAMT

  • The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
  • In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
  • This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.

BIOLOGY, M5 2025 HSC 33

The following diagram shows the cell division processes occurring in two related individuals.
 

  1. Compare the cell division processes carried out by cells \(R\) and \(S\) in Individual 1.   (3 marks)
  2. Explain the relationship between Individuals 1 and 2.   (2 marks)
  3. \(A\) and \(B\) are two separate mutations. Analyse how mutations \(A\) and \(B\) affect the genetic information present in cells  \(U\), \(V\), \(W\) and \(X\).   (4 marks)
Show Answers Only

a.    Similarities:

  • Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

  • Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
  • Cell S undergoes meiosis producing four genetically different haploid gametes.
  • Mitosis in R maintains chromosome number for growth and repair.
  • Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b.    Relationship between Individuals 1 and 2

  • Individual 2 is the offspring of Individual 1.
  • This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

c.    Mutation’s affect on genetic information

  • Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
  • Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
  • Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
  • The significance is that only mutation A can be inherited by offspring, while mutation B cannot.
Show Worked Solution

a.    Similarities:

  • Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

  • Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
  • Cell S undergoes meiosis producing four genetically different haploid gametes.
  • Mitosis in R maintains chromosome number for growth and repair.
  • Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b.    Relationship between Individuals 1 and 2

  • Individual 2 is the offspring of Individual 1.
  • This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

c.    Mutation’s affect on genetic information

  • Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
  • Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
  • Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
  • The significance is that only mutation A can be inherited by offspring, while mutation B cannot.

BIOLOGY, M6 2025 HSC 34

The following graph shows the changes in allele frequencies in two separate populations of the same species. Each line represents an introduced allele.

  1. Explain why the fluctuations of the allele frequencies are more pronounced in the small population, compared to the larger population.   (2 marks)
  2. Evaluate the effects of gene flow on the gene pools of the two populations.   (4 marks)
Show Answers Only

a.  Genetic Drift and Population Size

  • Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
  • Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b.    Evaluation Statement

  • Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

  • The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
  • This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
  • The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

  • The small population (20) shows limited benefit from gene flow.
  • Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
  • Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

  • Overall, gene flow proves highly effective in large populations for maintaining diversity,.
  • However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.
Show Worked Solution

a.  Genetic Drift and Population Size

  • Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
  • Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b.    Evaluation Statement

  • Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

  • The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
  • This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
  • The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

  • The small population (20) shows limited benefit from gene flow.
  • Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
  • Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

  • Overall, gene flow proves highly effective in large populations for maintaining diversity,.
  • However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.

BIOLOGY, M8 2025 HSC 26

The diagram shows the steps in LASIK (Laser-Assisted In Situ Keratomileusis) surgery.
 

Compare the LASIK technology shown with ONE other technology that can be used to treat a named visual disorder.   (4 marks)

Visual disorder:  

Show Answers Only

Visual disorder: Myopia (short-sightedness)

Similarities:

  • Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
  • Both technologies enable clear vision at distance for myopia patients.

Differences:

  • LASIK permanently reshapes the cornea using laser ablation to correct vision.
  • Spectacles use external convex or concave lenses to refract light without altering eye structure.
  • LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
  • LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

Show Worked Solution

Visual disorder: Myopia (short-sightedness)

Similarities:

  • Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
  • Both technologies enable clear vision at distance for myopia patients.

Differences:

  • LASIK permanently reshapes the cornea using laser ablation to correct vision.
  • Spectacles use external convex or concave lenses to refract light without altering eye structure.
  • LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
  • LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

BIOLOGY, M7 2025 HSC 28

Alpha-gal syndrome (AGS) is a tick-borne allergy to red meat caused by tick bites. Alpha-gal is a sugar molecule found in most mammals but not humans, and can also be found in the saliva of ticks. The diagram shows how a tick bite might cause a person to develop an allergic reaction to red meat.
 

 

  1. The flow chart shows the process of antibody production following exposure to alpha-gal. 
  2.   
  3. Describe the role of X, Y and Z in the process of antibody production.   (4 marks)

  4. An allergic reaction to alpha-gal sugar is similar to a secondary immune response.
    1.    
  5. Describe the features of antibody production shown in the graph.   (2 marks)

  6. Explain the role of memory cells in the immune response.   (3 marks)
Show Answers Only

a.    Antibody Production Process

  • X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
  • Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
  • Y is a B-cell that has receptors specific to the alpha-gal antigen.
  • B-cells are activated by Helper T-cells and undergo clonal expansion.
  • Some B-cells differentiate into memory cells for long-term immunity.
  • Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
  • Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b.    Features of Antibody Production

  • Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
  • Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c.    Role of Memory Cells

  • Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
  • Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
  • This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.
Show Worked Solution

a.    Antibody Production Process

  • X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
  • Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
  • Y is a B-cell that has receptors specific to the alpha-gal antigen.
  • B-cells are activated by Helper T-cells and undergo clonal expansion.
  • Some B-cells differentiate into memory cells for long-term immunity.
  • Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
  • Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b.    Features of Antibody Production

  • Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
  • Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c.    Role of Memory Cells

  • Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
  • Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
  • This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.

Financial Maths, STD2 EQ-Bank 01 MC

Jacinta buys several items at the supermarket. The docket for her purchases is shown below.

What is the amount of GST included in the total? 

  1. $1.15
  2. $1.27
  3. $1.55
  4. $1.71
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Total price of taxable items}\ +\ 10\%\ \text{ GST}\ =1.29+7.23+4.13=$12.65\)

\(\therefore\ 110\%\ \text{of original price}\) \(=$12.65\)
\(\therefore\ \text{GST}\) \(=$12.65\times\dfrac{100}{110}\)
  \(=$1.15\)

\(\Rightarrow A\)

Measurement, STD2 EQ-Bank 01 MC

The sheets of paper Jenny uses in her photocopier are 21 cm by 30 cm. The paper is 80 gsm, which means that one square metre of this paper has a mass of 80 grams. Jenny has a pile of this paper weighing 25.2 kg.

How many sheets of paper are in the pile?

  1. 500
  2. 2000
  3. 2500
  4. 5000
Show Answers Only

\(D\)

Show Worked Solution

\(1\ \text{square metre} = 100\ \text{cm}\times 100\ \text{cm}=10\,000\ \text{cm}^2\)

\(\text{Area of paper sheet}\ = \ 21\times 30=630\ \text{cm}^2\)

\(\text{Number of 80 gsm sheets in 25.2 kg}\ =\dfrac{25.2\times 1000}{80}=315\)

\(\therefore\ \text{Sheets in pile}\) \(=315\times\dfrac{10\,000}{630}\)
  \(=5000\)

  

CHEMISTRY, M6 2025 HSC 31

Hydrazine is a compound of hydrogen and nitrogen. The complete combustion of 1.0 L of gaseous hydrazine requires 3.0 L of oxygen, producing 2.0 L of nitrogen dioxide gas and 2.0 L of water vapour. All volumes are measured at 400°C.

  1. Use the chemical equation for the combustion of hydrazine to show that the molecular formula for hydrazine is \(\ce{N2H4}\).   (2 marks)

  2. The relationship between the acid equilibrium constant \(\left(K_a\right)\) and the corresponding conjugate base equilibrium constant \(\left(K_b\right)\) is shown.
      1. \(K_a \times K_b=K_w\)
  3. Use a relevant chemical equation to calculate the pH of a  0.20 mol L\(^{-1}\) solution of \(\ce{N2H5+}\) using the following data:
    • the \(K_b\) of hydrazine is \(1.7 \times 10^{-6}\) at 25°C
    • \(\ce{N2H5+}\) is the conjugate acid of \(\ce{N2H4}\).   (4 marks)
Show Answers Only

a.    Using Avogadro’s law:

  • At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
  •    \(\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}\)
  • Mole ratio  \(\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ \) matches the volume ratios given.
  • Therefore \(\ce{N2H4}\) is the correct molecular formula for hydrazine.

b.    \(\text{pH} = 4.46\)

Show Worked Solution

a.    Using Avogadro’s law:

  • At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
  •    \(\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}\)
  • Mole ratio  \(\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ \) matches the volume ratios given.
  • Therefore \(\ce{N2H4}\) is the correct molecular formula for hydrazine.

b.    \(K_a(\ce{N2H5+}) = \dfrac{K_w}{K_b(\ce{N2H4})} = \dfrac{1 \times 10^{-14}}{1.7 \times 10^{-6}} = 5.88235 \times 10^{-9}\)

  • The ionisation of \(\ce{N2H5+}\) is given the chemical equation below:
  •    \(\ce{N2H5+(aq) + H2O(l) \leftrightharpoons N2H4(aq) + H3O+(aq)}\)
  •    \(K_a = \dfrac{\ce{[H3O+][N2H4]}}{\ce{[N2H5+]}}\)
     
  • Using an Ice Table where all numbers are in mol L\(^{-1}\).

\begin{array} {|c|c|c|c|}
\hline  & \ce{[N2H5+]} & \ce{[N2H4]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.20 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.20 -x & x & x \\
\hline \end{array}

 

  • Substituting into the \(K_a\) expression:
   \(\dfrac{x^2}{0.20-x}\) \(=5.88235 \times 10^{-9}\)  
\(\dfrac{x^2}{0.20}\) \(=5.88235 \times 10^{-9}\), as \(x\) is really small  
\(x\) \(=3.42997 \times 10^{-5}\)  

 

   \(\text{pH}\) \(=-\log_{10}(\ce{[H3O+]})\)  
  \(=-\log_{10}(3.42997 \times 10^{-5})\)  
  \(=4.46\)  

CHEMISTRY, M5 2025 HSC 29

Consider the following reaction.

\(\ce{2NO2(g) \rightleftharpoons N2O4(g) \quad \quad \Delta H= -57.2 \ \text{kJ mol}^{-1}}\)

A sealed reaction vessel of fixed volume contains a mixture of \(\ce{NO2}\) and \(\ce{N2O4}\) gases at equilibrium.

Explain the impact of the addition of argon, an inert gas, on the temperature of the system.   (4 marks)

Show Answers Only
  • The equilibrium expression for the reaction is:
  •    \(Q = \dfrac{\ce{[N2O4]}}{\ce{[NO2]^2}} = K_{eq}​\)
  • When argon is added at constant volume, the total pressure increases, but the concentrations of \(\ce{NO2}\) and \(\ce{N2O4}\) remain unchanged because the amount of each gas and the volume remain constant. Since the concentrations in the equilibrium expression do not change, the value of \(Q\) remains equal to \(K_{eq}\), so the equilibrium position does not shift.
  • Because the equilibrium does not shift, no extra forward or reverse reaction occurs, meaning there is no absorption or release of heat. Even though the reaction is exothermic, the system does not produce or consume heat if equilibrium is unchanged.
Show Worked Solution
  • The equilibrium expression for the reaction is:
  •    \(Q = \dfrac{\ce{[N2O4]}}{\ce{[NO2]^2}} = K_{eq}​\)
  • When argon is added at constant volume, the total pressure increases, but the concentrations of \(\ce{NO2}\) and \(\ce{N2O4}\) remain unchanged because the amount of each gas and the volume remain constant. Since the concentrations in the equilibrium expression do not change, the value of \(Q\) remains equal to \(K_{eq}\), so the equilibrium position does not shift.
  • Because the equilibrium does not shift, no extra forward or reverse reaction occurs, meaning there is no absorption or release of heat. Even though the reaction is exothermic, the system does not produce or consume heat if equilibrium is unchanged.

CHEMISTRY, M7 2025 HSC 28

Kevlar and polystyrene are two common polymers.

A section of their structures is shown.
 

     

  1. Kevlar is produced through a reaction of two different monomers, one of which is shown. Draw the missing monomer in the box provided.   (1 mark)

  1. Kevlar chains are hard to pull apart, whereas polystyrene chains are not.
  2. With reference to intermolecular forces, explain the difference in the physical properties of the two polymers.   (3 marks)

Show Answers Only

a.    
           

b.    The physical differences between the two polymers are:

  • Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
  • The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
  • Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
  • These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.
Show Worked Solution

a.    
           

b.    The physical differences between the two polymers are:

  • Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
  • The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
  • Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
  • These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.

CHEMISTRY, M7 2025 HSC 27

Mixtures of hydrocarbons can be obtained from crude oil by the process of fractional distillation. Examples include petrol, diesel and natural gas.

  1. Outline an environmental implication for a use of a named hydrocarbon mixture that is obtained from crude oil.   (2 marks)

  2. Ethene is a simple hydrocarbon obtained from crude oil.
  3. Using structural formulae, write the chemical equation for the conversion of ethene to ethanol, including any other necessary reagents.   (3 marks)

  1. When ethanol is reacted with ethanoic acid, ethyl ethanoate is formed, as shown by the equation.
      1. \(\text{ethanol} \ + \ \text{ethanoic acid } \ \rightleftharpoons \ \text{ethyl ethanoate} \ +\ \text{water}\)
  2. The graph below shows the concentration of ethanol from the start of the reaction, \(t_0\), up to a time \(t_1\).
  3. At time \(t_1\), an additional amount of ethanol is added to the system.
  4. Sketch on the graph the changes that occur in the concentration of ethanol between time \(t_1\), and when the system reaches a new equilibrium before time \(t_2\).   (3 marks)

Show Answers Only

a.    Environmental implication:

  • The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
  • Increased \(\ce{CO2}\) levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
  • A typical component of petrol, octane, burns according to:
  •    \(\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}\)

b.    
           

c.    
       

Show Worked Solution

a.    Environmental implication:

  • The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
  • Increased \(\ce{CO2}\) levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
  • A typical component of petrol, octane, burns according to:
  •    \(\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}\)

b.    
           

c.    
       

  • There will be a sudden increase at \(t_1\) and then the concentration of ethanol will decrease smoothly until a new equlibrium concentration (greater than the original equilibrium concentration) is reached.

CHEMISTRY, M5 2025 HSC 20 MC

The solubility constant for silver\(\text{(I)}\) oxalate \(\ce{(Ag2C₂O4)}\) was determined using the following method.

  • 2.0 g of solid \(\ce{Ag2C2O4}\) was added to 100 mL of distilled water.
  • A sample of the saturated solution above the undissolved \(\ce{Ag₂C₂O}\) was diluted by a factor of 2000, using distilled water.
  • This diluted solution was analysed using atomic absorption spectroscopy (AAS).

The calibration curve for the AAS is provided below.
 

The absorbance of the diluted sample was 0.055.

What is the \(K_{s p}\) for silver oxalate?

  1. \(8.8 \times 10^{-14}\)
  2. \(5.3 \times 10^{-12}\)
  3. \(1.1 \times 10^{-11}\)
  4. \(2.1 \times 10^{-11}\)
Show Answers Only

\(B\)

Show Worked Solution
  • By interpolation, the observed concentration of silver ions for an absorbance of \(0.055\)  is  \(0.11 \times 10^{-6}\ \text{mol L}^{-1}\).
  • Since this sample was diluted by a factor of 2000:
  •    \(\ce{[Ag+]_{\text{saturated}} = 2000 \times [Ag+]_{\text{diluted}}} = 2000 \times 0.11 \times 10^{-6} = 2.2 \times 10^{-4}\ \text{mol L}^{-1}\)
  • The equation for the dissolution of \(\ce{Ag2C₂O4}\) is
  •    \(\ce{Ag2C₂O4 \leftrightharpoons 2Ag+(aq) + C2O4^{2-}(aq)}\)
  • Hence \(\ce{[C2O4^{2-}]_{\text{saturated}} = 0.5 \times [Ag+]_{\text{saturated}}} = 0.5 \times 2.2 \times 10^{-4} = 1.1 \times 10^{-4}\)
  •    \(K_{sp} = \ce{[Ag+]^2[C2O4^{2-}]} = (2.2 \times 10^{-4})^2(1.1 \times 10^{-4}) = 5.3 \times 10^{-12}\)

\(\Rightarrow B\)

CHEMISTRY, M8 2025 HSC 18 MC

The concentration of silver ions in a solution is determined by titrating it with aqueous sodium chloride, using yellow potassium chromate as the indicator.

Which row of the table correctly identifies the colour change at the endpoint and the more soluble salt?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
 \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \text{Colour change} \ \ & \text{More soluble salt} \\
\text{at endpoint}\rule[-1ex]{0pt}{0pt}& \text{} \\
\hline
\rule{0pt}{2.5ex}\text{Red to yellow}\rule[-1ex]{0pt}{0pt}& \ce{Ag2CrO4}\\
\hline
\rule{0pt}{2.5ex}\text{Yellow to red}\rule[-1ex]{0pt}{0pt}& \ce{Ag2CrO4}\\
\hline
\rule{0pt}{2.5ex}\text{Red to yellow}\rule[-1ex]{0pt}{0pt}& \ce{AgCl} \\
\hline
\rule{0pt}{2.5ex}\text{Yellow to red}\rule[-1ex]{0pt}{0pt}& \ce{AgCl} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • The two key reactions that take place in this titration are
  •    \(\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}\)
  •    \(\ce{2Ag+(aq) + CrO4^{2-}(aq) -> Ag2CrO4(s)}\)
  • For the titration to occur effectively all of the \(\ce{Cl-}\) ions must react with \(\ce{Ag+}\) ion and be precipitated out of the solution as the sodium chloride is Titrant (known volume and known concentration).
  • Only once all of the \(\ce{Cl-}\) ions are used up will silver ions begin to react the chromate ions signally the endpoint of the titration.
  • Hence, \(\ce{Ag2CrO4}\) is the more soluble salt and as the potassium chromate is initially yellow, the colour change must be from yellow to red.

\(\Rightarrow B\)

CHEMISTRY, M8 2025 HSC 17 MC

The chemical environment of an atom depends on the species surrounding that atom within a molecule.

In which of the following compounds does the number of carbon chemical environments equal the number of proton chemical environments?
 

 

Show Answers Only

\(A\)

Show Worked Solution
  • In the diagrams below, each colour represents a different carbon or proton environment (ignoring the oxygen atoms).

  • Only \(A\) has equal number of carbon and proton environments with two of each.

\(\Rightarrow A\)

CHEMISTRY, M7 2025 HSC 16 MC

A single straight strand of polyester was produced through a condensation reaction of 1000 molecules of 3-hydroxypropanoic acid, \(\ce{HOCH2CH2COOH}\).

What is the approximate molar mass of the strand (in g mol\(^{-1}\) )?

  1. 72 062
  2. 72 080
  3. 90 060
  4. 90 078
Show Answers Only

\(B\)

Show Worked Solution
  • The molar mass of 3-hydroxypropanoic acid \(= 6(1.008) +3(16.00) + 3(12.01) = 90.078\ \text{g mol}^{-1}\).
  • When 1000 molecules of 3-hydroxypropanoic acid undergoes condensation reactions, 999 molecules of water are formed as by-products.
  • Hence the molar mass of the polymer strand can be calculated by:
\(MM_{\text{strand}}\) \(=MM_{\text{monomers}}-MM_{\text{water}}\)  
  \(=90.078 \times 1000 -18.016 \times 999\)  
  \( = 72\ 080\ \text{g mol}^{-1}\)  

\(\Rightarrow B\)

CHEMISTRY, M7 2025 HSC 15 MC

Consider the following sequence of reactions.

  • Prop-2-en-1-ol was reacted with hydrogen gas to form liquid \(X\).
  • \(X\) was oxidised, producing liquid \(Y\) that formed bubbles of a gas when reacted with aqueous sodium carbonate.
  • \(Y\) was heated under reflux with methanol and a drop of concentrated sulfuric acid, producing an organic liquid, \(Z\).

This process has been presented in the flow chart below.
 

  

Which option correctly identifies the structures for \(X\), \(Y\) and \(Z\)?
 

Show Answers Only

\(D\)

Show Worked Solution
  • The first reaction that occurs is a hydrogenation addition reaction in which prop-2-en-1-ol reacts under a Pd catalyst to produce propan-1-ol.
  • The primary alcohol propan-1-ol then undergoes oxidation to produce the carboxylic acid propanoic acid. This is confirmed as when it is reacted with sodium carbonate, it undergoes an acid-carbonate reaction to produce carbon dixoide which is observed through the bubbles.
  • The third reaction is an estification reaction in which propanic acid reacts with methanol under relfex to produce methyl-propanoate.
  • All three of the correctly drawn compounds can be observed in \(D\)

\(\Rightarrow D\)

CHEMISTRY, M5 2025 HSC 14 MC

The equation for the decomposition of hydrogen iodide is shown.

\(\ce{2HI(g)\rightleftharpoons I2(g) + H2(g)} \quad \quad \Delta H=+52 \ \text{kJ mol}^{-1}\)

The equilibrium formed during this reaction was investigated in two experiments carried out at different temperatures. The initial and equilibrium concentrations for both experiments are shown in the table, with only the \(K_{e q}\) for Experiment 1 shown.
 

Which row in the table correctly compares features of the two experiments?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
 \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \quad K_{e q}\quad \quad\quad& \ \ \text{Temperature of}\ \  \\
\ \rule[-1ex]{0pt}{0pt}& \textit{experiment} \\
\hline
\rule{0pt}{2.5ex}\text{Lower in 1}\rule[-1ex]{0pt}{0pt}&\text{Lower in 2}\\
\hline
\rule{0pt}{2.5ex}\text{Lower in 1}\rule[-1ex]{0pt}{0pt}& \text{Higher in 2}\\
\hline
\rule{0pt}{2.5ex}\text{Higher in 1}\rule[-1ex]{0pt}{0pt}& \text{Lower in 2} \\
\hline
\rule{0pt}{2.5ex}\text{Higher in 1}\rule[-1ex]{0pt}{0pt}& \text{Higher in 2} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • The equilibrium constant for experiment two:
  •    \(K_{eq} = \dfrac{\ce{[H2][I2]}}{\ce{[HI]^2}} = \dfrac{0.02 \times 0.02}{0.04^2} = 0.25\)
  • Therefore the \(K_{eq}\) for experiment 1 is lower than the \(K_{eq}\) for experiment 2.
  • For an endothermic reaction (thinking of heat as a reactant), increasing the temperature of the reaction will shift the equilibrium to the products and increase \(K_{eq}\) which is what occured in experiment 2. 
  • Hence, the temperature of the experiment is higher in 2.

\(\Rightarrow B\)

Vectors, EXT2 V1 2025 HSC 16c

Consider the point \(B\) with three-dimensional position vector \(\underset{\sim}{b}\) and the line  \(\ell: \underset{\sim}{a}+\lambda \underset{\sim}{d}\), where \(\underset{\sim}{a}\) and \(\underset{\sim}{d}\) are three-dimensional vectors, \(\abs{\underset{\sim}{d}}=1\) and \(\lambda\) is a parameter.

Let \(f(\lambda)\) be the distance between a point on the line \(\ell\) and the point \(B\).

  1. Find \(\lambda_0\), the value of \(\lambda\) that minimises \(f\), in terms of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{d}\).   (2 marks)

  2. Let \(P\) be the point with position vector  \(\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\).
  3. Show that \(PB\) is perpendicular to the direction of the line \(\ell\).   (1 mark)

  4. Hence, or otherwise, find the shortest distance between the line \(\ell\) and the sphere of radius 1 unit, centred at the origin \(O\), in terms of \(\underset{\sim}{d}\) and \(\underset{\sim}{a}\).
  5. You may assume that if \(B\) is the point on the sphere closest to \(\ell\), then \(O B P\) is a straight line.   (3 marks)

Show Answers Only

i.    \(\lambda_0=\underset{\sim}{d}(\underset{\sim}{b}-\underset{\sim}{a})\)

ii.   \(\text{See Worked Solutions.}\)

iii.  \(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Show Worked Solution

i.    \(\ell=\underset{\sim}{a}+\lambda \underset{\sim}{d}, \quad\abs{\underset{\sim}{d}}=1\)

\(\text{Vector from point \(B\) to a point on \(\ell\)}:\ \underset{\sim}{a}+\lambda \underset{\sim}{d}-\underset{\sim}{b}\)

\(f(\lambda)=\text{distance between \(\ell\) and \(B\)}\)

\(f(\lambda)=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}\)

\(\text{At} \ \ \lambda_0, f(\lambda) \ \ \text{is a min}\ \Rightarrow \ f(\lambda)^2 \ \ \text {is also a min}\)

\(f(\lambda)^2\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot  \underset{\sim}{d}\)
  \(=\lambda^2|\underset{\sim}{d}|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)
  \(=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)

 

\(f(\lambda)^2 \ \ \text{is a concave up quadratic.}\)

\(f(\lambda)_{\text {min}}^2 \ \ \text{occurs at the vertex.}\)

\(\lambda_0=-\dfrac{b}{2 a}=-\dfrac{2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b})}{2}=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})\)
 

ii.    \(P \ \text{has position vector} \ \ \underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\)

\(\text{Show} \ \ \overrightarrow{PB} \perp \ell:\)

\(\overrightarrow{PB}=\underset{\sim}{b}-\underset{\sim}{p}=\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\)

\(\overrightarrow{P B} \cdot \underset{\sim}{d}\) \(=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}\)
  \(=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}\)
  \(=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2\)
  \(=0\)

  

\(\therefore \overrightarrow{PB}\ \text{is perpendicular to the direction of the line}\ \ell. \)
 

iii.   \(\text{Shortest distance between} \ \ell \ \text{and sphere (radius\(=1\))}\)

\(=\ \text{(shortest distance \(\ell\) to \(O\))}-1\)
 

\(f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to point \(B\)}\)

\(\text{Set} \ \ \underset{\sim}{b}=0 \ \Rightarrow \ f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to \(0\)}\)

\(\Rightarrow \lambda_0=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})=-\underset{\sim}{d} \cdot \underset{\sim}{a}\)

\(f\left(\lambda_0\right)\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}\)
\(f\left(\lambda_0\right)^2\) \(=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2\)
  \(=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2\)
\(f\left(\lambda_0\right)\) \(=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}\)

 

\(\text {Shortest distance of \(\ell\) to sphere \(\left(d_{\min }\right)\):}\)

\(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

CHEMISTRY, M6 2025 HSC 11 MC

The structures of two substances, \(\text{X}\) and \(\text{Y}\), are shown.
 

Which row of the table correctly classifies these substances as a Brønsted-Lowry acid or a Brønsted-Lowry base?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
 \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Brønsted-Lowry} & \textit{Brønsted-Lowry} \\
\textit{acid}\rule[-1ex]{0pt}{0pt}& \textit{base} \\
\hline
\rule{0pt}{2.5ex}\text{-}\rule[-1ex]{0pt}{0pt}&\text{X and Y}\\
\hline
\rule{0pt}{2.5ex}\text{X and Y}\rule[-1ex]{0pt}{0pt}& \text{-}\\
\hline
\rule{0pt}{2.5ex}\text{Y}\rule[-1ex]{0pt}{0pt}& \text{X} \\
\hline
\rule{0pt}{2.5ex}\text{X}\rule[-1ex]{0pt}{0pt}& \text{Y} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • A Brønsted-Lowry acid is a proton donor and a Brønsted-Lowry base is a proton acceptor.
  • \(X\) is propanoate (the conjugate base of propanoic acid) and is therefore a proton accepter making it a Brønsted-Lowry base.
  • \(Y\) is ethanamine and is considered a weak base where the \(\ce{NH2}\) group can accept a proton to become \(\ce{NH3+}\).

\(\Rightarrow A\)

Mechanics, EXT2 M1 2025 HSC 16b

A particle of mass 1 kg is projected from the origin with a speed of 50 ms\(^{-1}\), at an angle of \(\theta\) below the horizontal into a resistive medium.
 

The position of the particle \(t\) seconds after projection is \((x, y)\), and the velocity of the particle at that time is  \(\underset{\sim}{v}=\displaystyle \binom{\dot{x}}{\dot{y}}\).

The resistive force, \(\underset{\sim}{R}\), is proportional to the velocity of the particle, so that  \(\underset{\sim}{R}=-k \underset{\sim}{v}\), where \(k\) is a positive constant.

Taking the acceleration due to gravity to be 10 ms\(^{-2}\), and the upwards vertical direction to be positive, the acceleration of the particle at time \(t\) is given by:

\(\underset{\sim}{a}=\displaystyle \binom{-k \dot{x}}{-k \dot{y}-10}\).    (Do NOT prove this.) 

Derive the Cartesian equation of the motion of the particle, given  \(\sin \theta=\dfrac{3}{5}\).   (5 marks)

Show Answers Only

 

Show Worked Solution

\(\sin \theta=\dfrac{3}{5} \ \Rightarrow \ \cos \theta=\dfrac{4}{5}\)

\(\text{Components of initial velocity:}\)

\(\dot{x}(0)=50\, \cos \theta=50 \times \dfrac{4}{5}=40 \ \text{ms}^{-1}\)

\(\dot{y}(0)=50\, \sin \theta=50 \times \dfrac{3}{5}=-30\ \text{ms}^{-1}\)

\(\text{Horizontal motion:}\)

  \(\dfrac{d \dot{x}}{dt}\) \(=-k \dot{x} \ \ \text{(given)}\)  
\(\dfrac{dt}{d \dot{x}}\) \(=-\dfrac{1}{k \dot{x}}\)  
\(\displaystyle \int dt\) \(=-\dfrac{1}{k} \int \dfrac{1}{\dot{x}}\, d x\)  
\(t\) \(=-\dfrac{1}{k} \ln \dot{x}+c\)  

 
\(\text{When} \ \ t=0, \ \dot{x}=40 \ \Rightarrow \ c=\dfrac{1}{k} \ln 40\)

\(t\) \(=\dfrac{1}{k} \ln 40-\dfrac{1}{k} \ln \abs{\dot{x}}=\dfrac{1}{k} \ln \abs{\dfrac{40}{\dot{x}}}\)
    \(k t\) \(=\ln \abs{\dfrac{40}{\dot{x}}}\)
  \(e^{k t}\) \(=\dfrac{40}{\dot{x}}\)
\(\dot{x}\) \(=40 e^{-k t}\)
\(x\) \(\displaystyle=\int 40 e^{-k t}\, d t\)
  \(=-\dfrac{40}{k} \times e^{-k t}+c\)

 

\(\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{40}{k}\)

   \(x=\dfrac{40}{k}-\dfrac{40}{k} e^{-k t}=\dfrac{40}{k}\left(1-e^{-k t}\right)\ \ldots\ (1)\)
 

\(\text{Vertical Motion }\)

\(\dfrac{d \dot{y}}{dt}\) \(=-k \dot{y}-10 \quad \text{(given)}\)
\(\dfrac{d t}{d \dot{y}}\) \(=-\dfrac{1}{k} \times \dfrac{1}{\dot{y}+\frac{10}{k}}\)
\(t\) \(=-\dfrac{1}{k} \displaystyle \int \dfrac{1}{\dot{y}+\frac{10}{k}} \, d \dot{y}\)
  \(=-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}+c\)

 

\(\text{When} \ \ t=0, \, \dot{y}=-30 \ \ \Rightarrow\ \ c=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}\)

\(t\) \(=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}\)
  \(=\dfrac{1}{k} \ln \abs{\frac{-30+\frac{10}{k}}{\dot{y}+\frac{10}{k}}}\)
  \(e^{k t}\) \(=\abs{\dfrac{-30+\frac{10}{k}}{y+\frac{10}{k}}}\)
\(\dot{y}\) \(=\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10}{k}\)
\(y\) \(=\displaystyle \left(-30+\dfrac{10}{k}\right) \int e^{-kt}\, d t-\int \dfrac{10}{k}\, dt\)
  \(=-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10 t}{k}+c\)

 \(\text{When} \ \ t=0, y=0 \ \Rightarrow \  c=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)\)

  \(y\) \(=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10t}{k}\)
  \(=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right)\left(1-e^{-k t}\right)-\dfrac{10 t}{k}\ \ldots\ (2)\)

 

\(\text {Cartesian equation (using (1) above):}\)

\(x\) \(=\dfrac{40}{k}\left(1-e^{-k t}\right)\)
\(\dfrac{k x}{40}\) \(=1-e^{-k t}\)
\(e^{-k t}\) \(=1-\dfrac{k x}{40}\)
\(-k t\) \(=\ln \abs{1-\dfrac{k x}{40}}\)
\(t\) \(=-\dfrac{1}{k} \ln \abs{1-\dfrac{kx}{40}}\)

 

\(y\) \(=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right) \times \dfrac{k x}{40}+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}\)
  \(=\left(\dfrac{1-3 k}{4 k}\right) x+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}\)

Complex Numbers, EXT2 N1 2025 HSC 7 MC

The complex number \(z\) lies on the unit circle.
 

What is the range of \(\operatorname{Arg}(z-2 i)\) ?

  1. \(\dfrac{\pi}{6} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{5 \pi}{6}\)
  2. \(\dfrac{\pi}{3} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{2 \pi}{3}\)
  3. \(-\dfrac{5 \pi}{6} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{6}\)
  4. \(-\dfrac{2 \pi}{3} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{3}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{The limits of Arg}(z-2i)\ \where it intersects the unit circle are:}\)
 

\(\sin \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)

\(\text {Range Arg}(z-2 i):\)

\(-\dfrac{\pi}{2}-\dfrac{\pi}{6} \leqslant \operatorname{Arg}(z-2 i) \leqslant-\dfrac{\pi}{2}+\dfrac{\pi}{6}\)

\(-\dfrac{2 \pi}{3} \leqslant \operatorname{Arg}(2-2 i) \leqslant-\dfrac{\pi}{3}\)

\(\Rightarrow D\)

Complex Numbers, EXT2 N2 2025 HSC 6 MC

The complex numbers \(z\) and \(w\) lie on the unit circle. The modulus of  \(z+w\)  is \(\dfrac{3}{2}\).

What is the modulus of  \(z-w\) ?

  1. \(\dfrac{1}{8}\)
  2. \(\dfrac{\sqrt{7}}{2}\)
  3. \(\dfrac{3}{2}\)
  4. \(\dfrac{7}{4}\)
Show Answers Only

\(B\)

Show Worked Solution

\(|z|=|w|=1, \quad \abs{z+w}=\dfrac{3}{2} \ \text{(given)}\)

\(\text{Find}\ \ \abs{z-w}:\)

\(\abs{z+\omega}^2=(z+\omega)(\bar{z}+\bar{\omega})=z \bar{z}+z \bar{\omega}+\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z-\omega}^2=(z-\omega)(\bar{z}-\bar{\omega})=\bar{z} z-z \bar{\omega}-\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z+\omega}^2+|z-\omega|^2\) \(=2 z \bar{z}+2 \omega \bar{\omega}=2\abs{z}^2+2\abs{\omega}^2=4\)
\(\dfrac{9}{4}+\abs{z-\omega}^2\) \(=4\)
\(\abs{z-w}^2\) \(=\dfrac{7}{4}\)
\(\abs{z-\omega}\) \(=\dfrac{\sqrt{7}}{2}\)

 
\(\Rightarrow B\)

Vectors, EXT2 V1 2025 HSC 10 MC

Which of the following gives the same curve as  \(\left(\begin{array}{c}\cos (t) \\ -t \\ \sin (t)\end{array}\right)\) for  \(t \in \mathbb{R}\) ?

  1. \(\left(\begin{array}{c}\cos (2 t) \\ 2 t \\ \sin (2 t)\end{array}\right)\)
  2. \(\left(\begin{array}{c}\cos \left(t^2+\dfrac{\pi}{2}\right) \\ t^2+\dfrac{\pi}{2} \\ \sin \left(t^2+\dfrac{\pi}{2}\right)\end{array}\right)\)
  3. \(\left(\begin{array}{c}\cos \left(t^2\right) \\ -t^2 \\ \sin \left(t^2\right)\end{array}\right)\)
  4. \(\left(\begin{array}{c}\cos \left(2 t+\dfrac{\pi}{2}\right) \\ 2 t+\dfrac{\pi}{2} \\ -\sin \left(2 t+\dfrac{\pi}{2}\right)\end{array}\right)\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Find which option is a re-parametrisation of the given curve.}\)

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

\(\text{Let}\ \ u=- \left(2t + \dfrac{\pi}{2}\right)\)

\(\text{Since}\ t \in \mathbb{R}\ \ \Rightarrow \ \ u \in \mathbb{R}\)

\(\cos \left(2 t+\dfrac{\pi}{2}\right) = \cos\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \cos\,u\)

\(2 t+\dfrac{\pi}{2} = -\left( -\left(2 t+\dfrac{\pi}{2} \right) \right) = -u\)

\(-\sin \left(2 t+\dfrac{\pi}{2}\right) = \sin\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \sin\,u\)

\(\Rightarrow D\)

Complex Numbers, EXT2 N2 2025 HSC 9 MC

The points \(U, V, W\) and \(Z\) represent the complex numbers \(u, v, w\) and \(z\) respectively. It is given that  \(v+z=u+w\)  and  \(u+k i z=w+k i v\)  where  \(k \in \mathbb{R} , k>1\).

Which quadrilateral best describes \(UVWZ\) ?

  1. Parallelogram
  2. Rectangle
  3. Rhombus
  4. Square
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Quadrilateral}\ UVWZ\ \ \Rightarrow\ \ \text{Diagonals are \(UW\) and \(VZ\)} \).

\(\text{Given}\ \ v+z=u+w\ \ \Rightarrow\ \ \dfrac{v+z}{2}=\dfrac{u+w}{2}\)

\(\text{Mid-points of diagonals are equal (diagonals bisect).}\)

\(u+kiz\) \(=w+kiv\)  
\(u-w\) \(=ki(v-z)\)  

\(\therefore UW\ \text{and}\ VZ\ \text{are perpendicular.}\)

\(\Rightarrow C\)

Proof, EXT2 P1 2025 HSC 16a

Consider the equation

\(z^n \cos\left[n \theta\right]+z^{n-1} \cos \left[(n-1) \theta\right]+z^{n-2} \cos \left[(n-2) \theta\right]+\cdots+z\, \cos\left[\theta\right]=1\)

where  \(z \in \mathbb{C} , \theta \in \mathbb{R} \), and \(n\) is a positive integer.

Using a proof by contradiction and the triangle inequality, or otherwise, prove that all the solutions to the equation lie outside the circle  \(\abs{z}=\dfrac{1}{2}\)  on the complex plane.   (4 marks)

Show Answers Only

\(\text{Proof by contradiction}\)

\(\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}\)

\(z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1\)
 

\(\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):\)

   \(\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|\)

\(\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|\)
 

\(1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|\)

\(1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)\)

\(1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) \)

\(1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\  a=2^{-n}, r=2}\)

\(1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}\)

\(1 \leqslant 1-2^{-n}\)

\(2^{-n} \leqslant 0 \ \ \text {(which is not true)}\)

\(\therefore \ \text{By contradiction, the original statement is correct}\)

Show Worked Solution

\(\text{Proof by contradiction}\)

\(\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}\)

\(z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1\)
 

\(\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):\)

   \(\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|\)

\(\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|\)
 

\(1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|\)

\(1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)\)

\(1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) \)

\(1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\  a=2^{-n}, r=2}\)

\(1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}\)

\(1 \leqslant 1-2^{-n}\)

\(2^{-n} \leqslant 0 \ \ \text {(which is not true)}\)

\(\therefore \ \text{By contradiction, the original statement is correct}\)

Complex Numbers, EXT2 N2 2025 HSC 15c

  1. Show that
  2.     \(\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.\)   (3 marks)

  3. Use De Moivre's theorem to show that the sixth roots of \(-1\) are given by 
  4.    \(\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)\)  for  \(k=0,1,2,3,4,5\).   (2 marks)

  5. Hence, or otherwise, show the solutions to  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)  are 
  6. \(z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)\), and \(i \cot \left(\dfrac{11 \pi}{12}\right)\).   (2 marks)

Show Answers Only

i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

 

Show Worked Solution

i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

Mechanics, EXT2 M1 2025 HSC 15b

A particle moves in simple harmonic motion about the origin with amplitude \(A\), and it completes two cycles per second. When it is \(\dfrac{1}{4}\) metres from the origin, its speed is half its maximum speed.

Find the maximum positive acceleration of the particle during its motion.   (4 marks)

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\(a_{\text{max}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Show Worked Solution

\(v^2=-n^2\left(x^2-A^2\right)\)

\(\operatorname{Period}\ (T)=\dfrac{1}{2} \ \Rightarrow \ \dfrac{2 \pi}{n}=\dfrac{1}{2} \ \Rightarrow \ n=4 \pi\)

\(\text{Max velocity occurs at}\ \  x=0:\)

   \(v_{\text{max}}^2=-(4 \pi)^2\left(0-A^2\right)=16 \pi^2 A^2\)

   \(v_{\text{max}}=\sqrt{16 \pi^2 A^2}=4 \pi A\)
 

\(\text{At} \ \ x=\dfrac{1}{4}, \ v=\dfrac{1}{2} \times 4 \pi A=2 \pi A\)

\((2 \pi A)^2\) \(=-(4 \pi)^2\left(\dfrac{1}{16}-A^2\right)\)
\(\dfrac{A^2}{4}\) \(=A^2-\dfrac{1}{16}\)
\(\dfrac{3 A^2}{4}\) \(=\dfrac{1}{16}\)
\(A^2\) \(=\dfrac{1}{12}\)
\(A\) \(=\dfrac{1}{\sqrt{12}}\)

 

\(\text{Max positive acceleration occurs at} \ \ x=-\dfrac{1}{\sqrt{12}}:\)

\(a_{\text{max}}=-n^2 x=-(4 \pi)^2 \times-\dfrac{1}{\sqrt{12}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Vectors, EXT2 V1 2025 HSC 15a

The adjacent sides of a parallelogram are represented by the vectors  \(\underset{\sim}{a}=4 \underset{\sim}{i}+3 \underset{\sim}{j}-\underset{\sim}{k}\)  and  \(\underset{\sim}{b}=2 \underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k}\).

Show that the area of the parallelogram is \(6 \sqrt{10}\) square units.   (4 marks)

Show Answers Only

\(A=\dfrac{1}{2}\abs{a}\abs{b} \sin \theta\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \  \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)
 

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=\dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Show Worked Solution

\(A=\dfrac{1}{2}\abs{a}\abs{b} \sin \theta\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \  \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)
 

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=\dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Complex Numbers, EXT2 N2 2025 HSC 14c

Let \(w\) be a complex number such that  \(1+w+w^2+\cdots+w^6=0\).

  1. Show that \(w\) is a 7th root of unity.   (1 mark)

The complex number  \(\alpha=w+w^2+w^4\)  is a root of the equation  \(x^2+b x+c=0\), where \(b\) and \(c\) are real and \(\alpha\) is not real.

  1. Find the other root of  \(x^2+b x+c=0\)  in terms of positive powers of \(w\).  (2 marks)

  2. Find the numerical value of \(c\).  (1 mark)

Show Answers Only

i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \text{(given)}\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7=1\) \(=1\)

ii.   \(w^6+w^5+w^3\)

iii.  \(2\)

Show Worked Solution

i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \ \text{(given,}\ w\neq 1)\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7\) \(=1\)

 
ii.    \(\text {Find the other root of:} \ \ x^2+b x+c=0\)

\(\text{Since \(b, c\) are real (given),}\)

\(\text{Using conjugate root theory, other root}\ =\bar{\alpha}\)

\(\bar{\alpha}\) \(=\overline{w+w^2+w^4}\)
  \(=\overline{w}+\overline{w^2}+\overline{w^4}\)
  \(=\dfrac{1}{w}+\dfrac{1}{w^2}+\dfrac{1}{w^4} \quad\left( \bar{w}=\dfrac{1}{w} \ \text{since} \ \ \abs{w}=1\right)\)
  \(=\dfrac{w^7}{w}+\dfrac{w^7}{w^2}+\dfrac{w^7}{w^4}\)
  \(=w^6+w^5+w^3\)

 

iii.    \(\text{Product of roots}=\dfrac{c}{a}=c\)

\(c\) \(=\left(w+w^2+w^4\right)\left(w^6+w^5+w^3\right)\)
  \(=w^7+w^6+w^4+w^8+w^7+w^5+w^{10}+w^9+w^7\)
  \(=1+w^6+w^4+\left(w^7 \cdot w\right)+1+w^5+\left(w^7 \cdot w^3\right)+\left(w^7 \cdot w^2\right)+1\)
  \(=2+\underbrace{1+w+w^2+w^3+w^4+w^5+w^6}_{=0}\)
  \(=2\)

Calculus, EXT2 C1 2025 HSC 14a

Let  \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta\)  for integers  \(n \geq 0\).

  1. Show that  \(I_n=\dfrac{1}{2 n-1}-I_{n-1}\) for \(n>0\),  given that  \(\dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\).   (3 marks)

  2. Hence, or otherwise, calculate \(I_2\).   (1 mark)

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i.    \(\text{See Worked Solutions}\)

ii.   \(\dfrac{\pi}{4}-\dfrac{2}{3}\)

Show Worked Solution

i.    \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta \ \ \Rightarrow \ \ \)\(I_{n-1}=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)

\(\text{Show} \ \ I_n=\dfrac{1}{2n-1}-I_{n-1} \ \ \text{for} \ \ n>0\)

\(\text{Given} \ \ \dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\ \ldots\ (1)\)

\(I_n\) \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \cot ^2 \theta \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta\left(\operatorname{cosec}^2 \theta-1\right) \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \operatorname{cosec}^2 \theta \, d \theta-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)
  \(=-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \dfrac{d}{d \theta}(\cot \theta) d \theta-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} \theta}{2 n-1}\right]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} (\frac{\pi}{2})}{2 n-1}-\dfrac{\cot ^{2 n-1} (\frac{\pi}{4})}{2 n-1}\right]-I_{n-1}\)
  \(=-\left[0-\dfrac{1}{2 n-1}\right]-I_{n-1}\)
  \(=\dfrac{1}{2n-1}-I_{n-1}\)

 

ii.     \(I_2\) \(=\dfrac{1}{3}-I_1\)
    \(=\dfrac{1}{3}-\left[ \dfrac{1}{2-1}-\displaystyle \int_{\tfrac{\pi}{2}}^{\tfrac{\pi}{2}} 1 \, d\theta \right]\)
    \(=\dfrac{1}{3}-\left[1-\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)\right]\)
    \(=\dfrac{\pi}{4}-\dfrac{2}{3}\)

Calculus, EXT2 C1 2025 HSC 13d

Evaluate  \(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\), by first using the substitution  \(u=\dfrac{\pi}{2}-x\).   (4 marks)

Show Answers Only

 

Show Worked Solution

\(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\)

\(\text{Let} \ \ u=\dfrac{\pi}{2}-x \ \ \Rightarrow\ \ \dfrac{du}{dx}=-1 \ \ \Rightarrow\ \ du=-dx\)

\(\text{Limits:} \ \ u=\dfrac{\pi}{2}\ \ \Rightarrow\ \ x=0, \ \ u=0\ \ \Rightarrow\ \ x=\dfrac{\pi}{2}\)

\(I\) \(=-\displaystyle \large{\int_{\small{\dfrac{\pi}{2}}}^0}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\sin \left(\dfrac{\pi}{2}-x\right)+\cos \left(\dfrac{\pi}{2}-x\right)}\,dx\)
  \(=\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\cos x+\sin x}\, d x\)

 

\(\text{Add \(I\) (swap variable from \(x\) to \(u\)) to original integral:}\)

\(2I=\dfrac{\pi}{2} \displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{1}{1+\sin u+\cos u}\, d u\)

\(\text{Substitute} \ \ t=\tan \left(\frac{u}{2}\right), \ \sin u=\dfrac{2t}{1+t^2}, \ \cos u=\dfrac{1-t^2}{1+t^2}\)

\(d t=\dfrac{1}{2} \sec ^2\left(\frac{u}{2}\right)\, du \ \ \Rightarrow\ \ du=\dfrac{2}{1+\tan ^2\left(\frac{u}{2}\right)}\, dt=\dfrac{2}{1+t^2}\, d t\)

\(\text{Limits:} \ \ n=\dfrac{\pi}{2}\  \Rightarrow \ t=1, \ n=0 \ \Rightarrow \ t=0\)

\(2I\) \(=\dfrac{\pi}{2} \displaystyle \int_0^1 \dfrac{1}{1+\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}} \times \frac{2}{1+t^2}\,d t\)
\(I\) \(=\displaystyle\frac{\pi}{4} \int_0^1 \frac{2}{1+t^2+2 t+1-t^2}\, d t\)
  \(=\displaystyle \frac{\pi}{4} \int_0^1 \frac{1}{1+t}\, d t\)
  \(=\dfrac{\pi}{4}\Bigl[\ln (1+t)\Bigr]_0^1\)
  \(=\dfrac{\pi}{4}(\ln 2-\ln 1)\)
  \(=\dfrac{\pi \, \ln 2}{4}\)

BIOLOGY, M8 2025 HSC 32

A population lives across three regions, \(A,\ B\) and \(C\).
  

People in community \(B\) developed an environmental disease. An epidemiological study was carried out to determine the risk of developing the disease due to age at exposure. The results of this study are shown in the graph.
  

Design an epidemiological study that could be used to produce the results shown in the graph. Justify the features of your design.   (7 marks)

Show Answers Only

Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Participants: Recruit individuals from community B across three age groups: 10-year-olds, 20-year-olds and 30-year-olds at the time of exposure to the environmental factor. This is justified because the graph displays separate curves for exposure at these three ages.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

Longitudinal Follow-up: Monitor all participants annually for disease development over 60 years. This is justified because the graph tracks disease risk across this timeframe and shows when risk peaks and declines.

Data Collection: Document whether each participant develops the disease and calculate the percentage of each age cohort affected at yearly intervals. This is justified because the y-axis shows risk as a percentage.

Control Variables: Ensure all participants experience similar levels of environmental exposure in community B. This is justified because the study isolates age at exposure as the only variable affecting disease risk.

Statistical Analysis: Calculate risk percentages for each time point after exposure for each age group. This is justified because it produces the three distinct curves showing risk declining differently based on initial exposure age.

Show Worked Solution

Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Participants: Recruit individuals from community B across three age groups: 10-year-olds, 20-year-olds and 30-year-olds at the time of exposure to the environmental factor. This is justified because the graph displays separate curves for exposure at these three ages.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

Longitudinal Follow-up: Monitor all participants annually for disease development over 60 years. This is justified because the graph tracks disease risk across this timeframe and shows when risk peaks and declines.

Data Collection: Document whether each participant develops the disease and calculate the percentage of each age cohort affected at yearly intervals. This is justified because the y-axis shows risk as a percentage.

Control Variables: Ensure all participants experience similar levels of environmental exposure in community B. This is justified because the study isolates age at exposure as the only variable affecting disease risk.

Statistical Analysis: Calculate risk percentages for each time point after exposure for each age group. This is justified because it produces the three distinct curves showing risk declining differently based on initial exposure age.

BIOLOGY, M6 2025 HSC 30

PAI-1 protein is encoded by the SERPINE 1 gene in humans. Anopheles mosquitoes have been genetically modified to express PAI-1, which blocks the entry of the malarial Plasmodium into the mosquito gut. This disrupts the Plasmodium life cycle, resulting in reduced transmission of malaria. 

  1. Describe a process that could be used to produce mosquitoes which express PAI-1.   (4 marks)

  2. 'Genetic technologies are beneficial for society.'
  3. Evaluate this statement.   (7 marks)

Show Answers Only

a.    Mosquito Production

  • The SERPINE 1 gene is isolated from human DNA using restriction enzymes that cut at specific recognition sites.
  • The same restriction enzymes are used to cut mosquito DNA, creating complementary sticky ends.
  • The human PAI-1 gene is inserted into the mosquito DNA using DNA ligase to form recombinant DNA.
  • The recombinant DNA is introduced into mosquito eggs or embryos using a vector or microinjection technique.
  • Modified mosquitoes are screened to confirm PAI-1 gene expression and successful integration into the genome.
  • Transgenic mosquitoes produce PAI-1 protein that blocks Plasmodium entry into the gut, disrupting malaria transmission.

b.    Evaluation Statement

Genetic technologies are highly beneficial for society when evaluated against health improvements and food security criteria. Despite some ethical and environmental concerns requiring careful management, the overall benefits are substantial.

Health Benefits
Evidence supporting includes:

  • Genetically modified mosquitoes expressing PAI-1 significantly reduce malaria transmission, potentially saving millions of lives annually.
  • Recombinant DNA technology produces insulin and vaccines, improving accessibility to life-saving treatments for diabetes and infectious diseases.
  • Gene therapy offers potential cures for inherited genetic disorders, dramatically improving quality of life for affected individuals.

The health criterion strongly meets beneficial status because these technologies address major global health challenges.

Food Security and Agricultural Benefits
Evidence supporting includes:

  • Genetically modified crops like Bt cotton and Golden Rice increase crop yields and nutritional content, addressing food scarcity.
  • Drought-resistant GM crops enable farming in challenging environments, supporting population growth and farmer livelihoods.

However, concerns exist about reduced genetic diversity and corporate control over seeds, creating inequalities in access.

Final Evaluation

Weighing these factors shows genetic technologies are substantially beneficial for society. The health improvements and food security gains outweigh the manageable ethical concerns. While challenges like biodiversity impacts and equitable access require ongoing attention, the overall societal benefit remains considerable through life-saving medical applications and enhanced food production.

Show Worked Solution

a.    Mosquito Production

  • The SERPINE 1 gene is isolated from human DNA using restriction enzymes that cut at specific recognition sites.
  • The same restriction enzymes are used to cut mosquito DNA, creating complementary sticky ends.
  • The human PAI-1 gene is inserted into the mosquito DNA using DNA ligase to form recombinant DNA.
  • The recombinant DNA is introduced into mosquito eggs or embryos using a vector or microinjection technique.
  • Modified mosquitoes are screened to confirm PAI-1 gene expression and successful integration into the genome.
  • Transgenic mosquitoes produce PAI-1 protein that blocks Plasmodium entry into the gut, disrupting malaria transmission.

b.    Evaluation Statement

Genetic technologies are highly beneficial for society when evaluated against health improvements and food security criteria. Despite some ethical and environmental concerns requiring careful management, the overall benefits are substantial.

Health Benefits
Evidence supporting includes:

  • Genetically modified mosquitoes expressing PAI-1 significantly reduce malaria transmission, potentially saving millions of lives annually.
  • Recombinant DNA technology produces insulin and vaccines, improving accessibility to life-saving treatments for diabetes and infectious diseases.
  • Gene therapy offers potential cures for inherited genetic disorders, dramatically improving quality of life for affected individuals.

The health criterion strongly meets beneficial status because these technologies address major global health challenges.

Food Security and Agricultural Benefits
Evidence supporting includes:

  • Genetically modified crops like Bt cotton and Golden Rice increase crop yields and nutritional content, addressing food scarcity.
  • Drought-resistant GM crops enable farming in challenging environments, supporting population growth and farmer livelihoods.

However, concerns exist about reduced genetic diversity and corporate control over seeds, creating inequalities in access.

Final Evaluation

Weighing these factors shows genetic technologies are substantially beneficial for society. The health improvements and food security gains outweigh the manageable ethical concerns. While challenges like biodiversity impacts and equitable access require ongoing attention, the overall societal benefit remains considerable through life-saving medical applications and enhanced food production.

BIOLOGY, M5 2025 HSC 27

The table shows the number of eggs produced by females in TWO different groups of animals.

\(\textit{Group A}\) \(\textit{Group B}\)
\(\quad\textit{Animal}\quad\) \(\quad\textit{Number of Eggs}\quad\) \(\quad\textit{Animal}\quad\) \(\quad\textit{Number of Eggs}\quad\)
  \(\text{Monotremes}\quad\)  \(1-3\)   \(\text{Crabs}\)   \(1000-2000\)
  \(\text{Snakes}\) \(1-100\)   \(\text{Sea Urchins}\quad\)    \(100\, 000\ \text{to 2 million}\) 
  \(\text{Birds}\) \(1-17\)   \(\text{Squid}\)   \(2000-3000\)

  
Compare the types of fertilisation that occur in group A and group B animals with reference to the data provided.   (4 marks) 

Show Answers Only

Group A Animals – Internal Fertilisation

  • Group A animals (monotremes, snakes, birds) use internal fertilisation where sperm meets egg inside the female’s body.
  • The data shows Group A produces few eggs (1-100), as the protected internal environment increases fertilisation success and offspring survival.
  • These animals typically provide greater parental care, further improving survival rates.

Group B Animals – External Fertilisation

  • Group B animals (crabs, sea urchins, squid) use external fertilisation where gametes are released into water.
  • The data shows Group B produces vastly more eggs (1,000-2 million) to compensate for low survival rates.
  • External fertilisation exposes gametes to predation, dilution and environmental factors, reducing fertilisation success.
  • These organisms provide minimal parental care, relying on high egg numbers for species continuity.

Comparison

The dramatic difference in egg numbers (1-100 vs 1,000-2,000,000) directly reflects the fertilisation strategy.

Internal fertilisation prioritises quality through protection and parental care, while external fertilisation relies on quantity to overcome environmental challenges.

Show Worked Solution

Group A Animals – Internal Fertilisation

  • Group A animals (monotremes, snakes, birds) use internal fertilisation where sperm meets egg inside the female’s body.
  • The data shows Group A produces few eggs (1-100), as the protected internal environment increases fertilisation success and offspring survival.
  • These animals typically provide greater parental care, further improving survival rates.

Group B Animals – External Fertilisation

  • Group B animals (crabs, sea urchins, squid) use external fertilisation where gametes are released into water.
  • The data shows Group B produces vastly more eggs (1,000-2 million) to compensate for low survival rates.
  • External fertilisation exposes gametes to predation, dilution and environmental factors, reducing fertilisation success.
  • These organisms provide minimal parental care, relying on high egg numbers for species continuity.

Comparison

The dramatic difference in egg numbers (1-100 vs 1,000-2,000,000) directly reflects the fertilisation strategy.

Internal fertilisation prioritises quality through protection and parental care, while external fertilisation relies on quantity to overcome environmental challenges.

BIOLOGY, M8 2025 HSC 25

The graph shows the changes in UV level in a single day.
  

The Cancer Council suggests that sun protection is needed whenever the UV level is 3 or above. The information provided on a sunscreen product suggests that sunscreen should be used between 10 am and 4 pm.

Using the graph, evaluate the information provided on the sunscreen product with regard to the Cancer Council suggestion.   (3 marks)

Show Answers Only

Evaluation Judgment:

The sunscreen product recommendation is partially effective but requires improvement for optimal sun protection.

Supporting Evidence:

  • The graph shows UV levels exceed 3 from approximately 8 am to 6 pm.
  • The sunscreen recommendation (10 am to 4 pm) misses two critical hours of UV exposure.
  • Between 8-10 am and 4-6 pm, UV levels remain above 3, requiring protection.

Conclusion:

The product advice inadequately protects users during morning and late afternoon exposure periods when UV damage still occurs.

Show Worked Solution

Evaluation Judgment:

The sunscreen product recommendation is partially effective but requires improvement for optimal sun protection.

Supporting Evidence:

  • The graph shows UV levels exceed 3 from approximately 8 am to 6 pm.
  • The sunscreen recommendation (10 am to 4 pm) misses two critical hours of UV exposure.
  • Between 8-10 am and 4-6 pm, UV levels remain above 3, requiring protection.

Conclusion:

The product advice inadequately protects users during morning and late afternoon exposure periods when UV damage still occurs.

BIOLOGY, M6 2025 HSC 23

Compare the processes of artificial insemination and artificial pollination.   (3 marks)

Show Answers Only

Similarities:

  • Both processes involve the transfer of gametes from one organism to another to produce offspring with desired characteristics.

Differences:

  • Artificial insemination occurs only in animals, where sperm is collected and inserted into the female reproductive tract.
  • Artificial pollination occurs only in (flowering) plants, where pollen is manually transferred between flowers.
  • Artificial insemination is used to improve livestock genetics and breeding programs, while artificial pollination is used to enhance crop production and plant breeding.

Show Worked Solution

Similarities:

  • Both processes involve the transfer of gametes from one organism to another to produce offspring with desired characteristics.

Differences:

  • Artificial insemination occurs only in animals, where sperm is collected and inserted into the female reproductive tract.
  • Artificial pollination occurs only in (flowering) plants, where pollen is manually transferred between flowers.
  • Artificial insemination is used to improve livestock genetics and breeding programs, while artificial pollination is used to enhance crop production and plant breeding.

Calculus, EXT2 C1 2025 HSC 13a

It is given that  \(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\, dx\).

Show that  \(\displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=k A\), where \(k\) and \(m\) are constants.   (3 marks)

Show Answers Only

\(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\,d x\)

\(\text{Let} \ \ u=m-x\)

\(\dfrac{d u}{d x}=-1 \ \ \Rightarrow \ \ du=-d x\)
 

\(\text{When} \ \ x=4, \ u=m-4\)

\(\text{When} \ \ x=2, \ u=m-2\)

\(A\) \(=-\displaystyle\int_{m-2}^{m-4} \frac{e^{m-u}}{m-u-1}\, du\)
  \(=-e^m \displaystyle \int_{m-2}^{m-4} \frac{e^{-u}}{m-u-1}\, du\)
  \(=-e^m \times \left[\displaystyle \int_{m-4}^{m-2} \frac{e^{-u}}{u-m+1}\, du\right]\)

 

\(\Rightarrow \displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=kA \ \ (\text{where}\ \ k=-e^{-m})\)

Show Worked Solution

\(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\,d x\)

\(\text{Let} \ \ u=m-x\)

\(\dfrac{d u}{d x}=-1 \ \ \Rightarrow \ \ du=-d x\)
 

\(\text{When} \ \ x=4, \ u=m-4\)

\(\text{When} \ \ x=2, \ u=m-2\)

\(A\) \(=-\displaystyle\int_{m-2}^{m-4} \frac{e^{m-u}}{m-u-1}\, du\)
  \(=-e^m \displaystyle \int_{m-2}^{m-4} \frac{e^{-u}}{m-u-1}\, du\)
  \(=-e^m \times \left[\displaystyle \int_{m-4}^{m-2} \frac{e^{-u}}{u-m+1}\, du\right]\)

 

\(\Rightarrow \displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=kA \ \ (\text{where}\ \ k=-e^{-m})\)

Algebra, STD2 A4 2025 HSC 39

After a dose of a medication, the amount of the medication remaining in a person can be modelled by the equation  \(y=k a^x\),  where \(x\) is the number of hours after taking the dose, and \(y\) is the amount remaining in milligrams (mg).

The graph shows the amount of the medication remaining in a person after \(x\) hours. Two points are also shown on the graph.
 

Using the information provided, find the amount of medication that remains in a person when  \(x=4\).   (3 marks)

Show Answers Only

\(5.4 \ \text{mg}\)

Show Worked Solution

\(\text{Given}\ (0,15) \ \text{lies on graph:}\)

   \(15=k \times a^{0} \ \ \Rightarrow \ \ k=15\)

\(\text{Find \(a\) given \((2,9)\) lies an graph:}\)

\(9\) \(=15 \times a^2\)
   \(a^2\) \(=\dfrac{9}{15}\)
\(a\) \(=\sqrt{\dfrac{9}{15}}\)

   

\(\text{Find \(y\) when  \(\ x=4\):}\)

\(y=15 \times\left(\sqrt{\dfrac{9}{15}}\right)^4=5.4 \ \text{mg}\)

Measurement, STD2 M7 2025 HSC 38

A car’s fuel efficiency is 30 miles per US gallon.

\begin{array}{|ll|}
\hline 
\rule{0pt}{2.5ex} 1 \ \text{US gallon}=3.8 \  \text{litres } \rule[-1ex]{0pt}{0pt}& \text{(correct to } 2 \text { significant figures)} \\
\rule{0pt}{2.5ex} 1 \ \text{mile}=1.6 \ \text{km} \rule[-1ex]{0pt}{0pt}& \text{(correct to } 2 \text { significant figures)} \\
\hline
\end{array}

Calculate the car’s fuel efficiency in litres per 100 km, correct to 1 decimal place.   (3 marks)

Show Answers Only

\(\text{Fuel efficiency}=7.9 \ \text{litres / 100 km}\)

Show Worked Solution

\(30 \ \text{miles}=30 \times 1.6=48 \ \text{km}\)

\(\text{Convert 48 km per 3.8 litres into km/litre:}\)

\(\dfrac{48}{3.8}=12.631 \ldots \ \text{km/litre}\)

\(\text{Fuel used in 100 km}=\dfrac{100}{12.631\ldots}=7.917\ldots=7.9 \ \text{litres (1 d.p.)}\)
 

\(\therefore \ \text{Fuel efficiency}=7.9 \ \text{litres per 100 km}\)

Measurement, STD2 M1 2025 HSC 32

Solid spheres are placed inside a square-based pyramid as shown.
 

The base of the pyramid has side lengths of 14 cm . The height of the pyramid is \(h\) cm. The radius of each sphere is 1.5 cm.

The amount of empty space remaining inside the pyramid after 30 spheres have been placed inside the pyramid is 634 cm³.

What is the height, \(h\), of the pyramid? Give your answer correct to the nearest centimetre.   (3 marks)

Show Answers Only

\(h=16 \ \text{cm}\)

Show Worked Solution

\(\text{Volume (pyramid)}=\dfrac{1}{3} \times A h=\dfrac{1}{3} \times 14 \times 14 \times h=\dfrac{196}{3} h\)

\(\text{Volume (spheres)}=30 \times \dfrac{4}{3} \pi\left(\dfrac{3}{2}\right)^3=135 \pi\)

\(\text{Empty space }=634 \ \text {(given)}\)
 

\(\text{Equating the volumes:}\)

\(\dfrac{196}{3} h\) \(=135 \pi+634\)
\(h\) \(=(135 \pi+634) \times \dfrac{3}{196}\)
  \(=16.19 \ldots\)
  \(=16 \ \text{cm (nearest cm)}\)

Financial Maths, STD2 F1 2025 HSC 31

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

At the end of the 2024-2025 financial year, Alex's tax payable was $47 420, excluding the Medicare levy.

What was Alex's taxable income?   (3 marks)

Show Answers Only

\(\text{Taxable income}\ =\$ 178\, 600\)

Show Worked Solution

\(\text{Tax payable}=\$ 47\,420\)

\(\text{Let}\ \ I =\text{Alex’s taxable income.}\)

\(47\, 420\) \(=31\,288+0.37(I-135\,000)\)
\(16\,132\) \(=0.37(I-135\,000)\)
\(16\,132\) \(=0.37 I-49\,950\)
\(0.37 I\) \(=66\, 082\)
\(I\) \(=\dfrac{66\,082}{0.37}=\$ 178\, 600\)

BIOLOGY, M6 2025 HSC 16 MC

The diagram shows a process that was used to make multiple clones in sheep.

The identical offspring are clones of

  1. each other.
  2. only sheep B.
  3. the surrogates.
  4. both sheep A and sheep B.
Show Answers Only

\(A\)

Show Worked Solution
  • A is correct: Embryo splitting produces offspring that are genetic clones of each other.

Other Options:

  • B is incorrect: Clones share genetics from both parents, not just sheep B.
  • C is incorrect: Surrogates provide environment only, no genetic contribution.
  • D is incorrect: Each clone is not identical to the parents; they’re identical to each other.

BIOLOGY, M8 2025 HSC 13 MC

The prevalence of a non-infectious disease has remained constant for over 10 years. A new treatment for this disease prolongs the life of people with the disease, but does not cure them.

Which row in the table shows the effect of the treatment on both the incidence and prevalence of this disease?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\ 
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Change in incidence}\rule[-1ex]{0pt}{0pt}& \textit{Change in prevalence} \\
\hline
\rule{0pt}{2.5ex}\text{None}\rule[-1ex]{0pt}{0pt}&\text{Increases}\\
\hline
\rule{0pt}{2.5ex}\text{None}\rule[-1ex]{0pt}{0pt}& \text{Decreases}\\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{None} \\
\hline
\rule{0pt}{2.5ex}\text{Decreases}\rule[-1ex]{0pt}{0pt}& \text{None} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • A is correct: Treatment doesn’t affect new cases but increases prevalence by prolonging life.

Other Options:

  • B is incorrect: Prolonging life increases prevalence, not decreases it.
  • C is incorrect: Treatment doesn’t cause more new cases to occur.
  • D is incorrect: Treatment doesn’t prevent new cases from occurring.

BIOLOGY, M6 2025 HSC 12 MC

The following represents a karyotype for an individual with Down syndrome.

 

What conclusions can be drawn about the individual from this karyotype?

  1. A male with a deletion mutation
  2. A female with a deletion mutation
  3. A male with a chromosomal mutation
  4. A female with a chromosomal mutation
Show Answers Only

\(D\)

Show Worked Solution
  • D is correct: XX sex chromosomes indicate female; trisomy 21 is a chromosomal mutation.

Other Options:

  • A is incorrect: Down syndrome involves an extra chromosome, not a deletion.
  • B is incorrect: Down syndrome is caused by an extra chromosome 21, not deletion.
  • C is incorrect: The presence of two X chromosomes indicates a female individual.

BIOLOGY, M7 2025 HSC 11 MC

In 2018, the Victorian government reported the target of vaccinating more than 95% of children below the age of five had been achieved.

What is the benefit of achieving a 95% vaccination rate for an infectious disease?

  1. Only 5% of individuals will catch the disease.
  2. Only 5% of individuals will be protected by the vaccine.
  3. This will protect the remaining 5% who are not vaccinated.
  4. The vaccinated individuals can transfer immunity to the remaining 5%.
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: High vaccination rate provides herd immunity protecting unvaccinated individuals.

Other Options:

  • A is incorrect: Herd immunity means fewer than 5% will catch disease.
  • B is incorrect: 95% are protected by vaccine, not 5%.
  • D is incorrect: Vaccinated individuals cannot transfer immunity to others through vaccination.

BIOLOGY, M7 2025 HSC 10 MC

The graph shows the number of recorded deaths, due to measles, before and after the measles vaccine was included in the National Immunisation Program (NIP).
  

Which of the following is a trend shown in the graph?

  1. Most people have been immunised against measles since 1975.
  2. Over the last 100 years the number of deaths has consistently fallen.
  3. The number of measles-induced deaths has fallen to near zero since measles vaccine was added to the NIP.
  4. The National Immunisation Program was the primary reason for the great reduction in measle-induced deaths.
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: Graph clearly shows deaths fell to near zero after 1975.

Other Options:

  • A is incorrect: Graph shows deaths, not immunisation rates; cannot infer this.
  • B is incorrect: Deaths fluctuated considerably before 1975, not consistent decline.
  • D is incorrect: Deaths were already declining before NIP; cannot claim it’s primary reason.

BIOLOGY, M5 2025 HSC 8 MC

When a red camellia flower is crossed with a white camellia flower, all the offspring are covered in both red and white petals.

What is the reason for this occurrence?

  1. One gene is controlling multiple characteristics.
  2. Environmental factors affect the phenotype of camellia flowers.
  3. Alleles for both red and white colour in camellia flowers are recessive.
  4. Petal colour in camellia flowers is controlled by a co-dominance pattern of inheritance.
Show Answers Only

\(D\)

Show Worked Solution
  • D is correct: Both alleles are expressed equally producing red and white petals.

Other Options:

  • A is incorrect: One gene controls one characteristic; this is pleiotropy not co-dominance.
  • B is incorrect: Environmental factors don’t explain both parental colours appearing in offspring.
  • C is incorrect: Both alleles are expressed, indicating co-dominance not recessive inheritance.

BIOLOGY, M8 2025 HSC 7 MC

An animal's body temperature and the air temperature of the animal's environment were measured every 4 hours, and the following data were recorded.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ \ \ \  \ \ \  {Time} \ \ \ \ \ \  \ \ \  \rule[-1ex]{0pt}{0pt} & {Body \ temperature} \rule[-1ex]{0pt}{0pt} & {Air  \  temperature} \\
{} & \text{(°C)} & \text{(°C)} \\
\hline
\rule{0pt}{2.5ex} \text{4 am} \rule[-1ex]{0pt}{0pt} & \text{41.3} \rule[-1ex]{0pt}{0pt} & \text{19.2}\\
\hline
\rule{0pt}{2.5ex} \text{8 am} \rule[-1ex]{0pt}{0pt} & \text{41.1} \rule[-1ex]{0pt}{0pt} & \text{18.8}\\
\hline
\rule{0pt}{2.5ex} \text{12 pm} \rule[-1ex]{0pt}{0pt} & \text{40.8} \rule[-1ex]{0pt}{0pt} & \text{21.5}\\
\hline
\rule{0pt}{2.5ex} \text{4 pm} \rule[-1ex]{0pt}{0pt} & \text{41.4} \rule[-1ex]{0pt}{0pt} & \text{26.4}\\
\hline
\rule{0pt}{2.5ex} \text{8 pm} \rule[-1ex]{0pt}{0pt} & \text{41.2} \rule[-1ex]{0pt}{0pt} & \text{27.5}\\
\hline
\rule{0pt}{2.5ex} \text{12 am} \rule[-1ex]{0pt}{0pt} & \text{41.5} \rule[-1ex]{0pt}{0pt} & \text{23.0}\\
\hline
\end{array}

Based on this data, which row of the table indicates what type of animal it is and why?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\ & \\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\ & \\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\ & \\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\ & \\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex} {Type \ of \ animal}\rule[-1ex]{0pt}{0pt}& {Reason } \\
\hline
\rule{0pt}{2.5ex}\text{Ectotherm}\rule[-1ex]{0pt}{0pt}&\text{Body temperature is around 41°C and varies} \\ & \text{with the air temperature. }\\
\hline
\rule{0pt}{2.5ex}\text{Ectotherm}\rule[-1ex]{0pt}{0pt}& \text{Body temperature is around 41°C and is} \\ & \text{always above the air temperature. }\\
\hline
\rule{0pt}{2.5ex}\text{Endotherm }\rule[-1ex]{0pt}{0pt}& \text{Body temperature is relatively constant} \\ & \text{despite changes in air temperature. } \\
\hline
\rule{0pt}{2.5ex}\text{Endotherm}\rule[-1ex]{0pt}{0pt}& \text{Body temperature is relatively constant, and} \\ & \text{air temperature is relatively constant. } \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: Body temperature remains constant around 41°C despite air temperature fluctuations.

Other Options:

  • A is incorrect: Body temperature doesn’t vary with air temperature; remains constant.
  • B is incorrect: Ectotherms cannot maintain constant body temperature above air temperature.
  • D is incorrect: Air temperature varies significantly from 18.8°C to 27.5°C.

BIOLOGY, M5 2025 HSC 4 MC

A storm randomly kills 80% of the frog population on an island. The allele frequencies in the frog population after the storm are notably different to those of the population before the storm.

What is the process that has led to the change in allele frequencies?

  1. Gene flow
  2. Genetic drift
  3. Natural selection
  4. Survival of the fittest
Show Answers Only

\(B\)

Show Worked Solution
  • B is correct: Random events changing allele frequencies in populations is genetic drift.

Other Options:

  • A is incorrect: Gene flow involves movement of alleles between populations through migration.
  • C is incorrect: Natural selection involves differential survival based on advantageous traits.
  • D is incorrect: Survival of fittest is non-random selection, not random mortality.

Financial Maths, STD1 F2 2025 HSC 25

Bobbie plans to invest $25 000 for 10 years and is offered two investment options.

Option \(A\):  Earns interest at a rate of 5% per annum, compounded monthly.

Option \(B\):  Earns simple interest at a rate of 8% per annum.

Which investment option will provide Bobbie with the best return value at the end of 10 years? Justify your answer with calculations.   (4 marks)

Show Answers Only

\(\text{Option A:}\)

\(PV=$25\ 000,\ \ \text{monthly interest rate: }r=\dfrac{0.05}{12},\ \ n=12\times 10=120\)

\(FV=25\ 000\Bigg(1+\dfrac{0.05}{12}\Bigg)^{120}\approx $41\ 175\)

  
\(\text{Option B:}\)

\(\text{Interest}=25\ 000\times 0.08\times 10=$20\ 000\)

\(\text{Total}=25\ 000+20\ 000=$45\ 000\)

\(\therefore\ \text{Option B gives the best return.}\)

Show Worked Solution

\(\text{Option A:}\)

\(PV=$25\ 000,\ \ \text{monthly interest rate: }r=\dfrac{0.05}{12},\ \ n=12\times 10=120\)

\(FV=25\ 000\Bigg(1+\dfrac{0.05}{12}\Bigg)^{120}\approx $41\ 175\)

  
\(\text{Option B:}\)

\(\text{Interest}=25\ 000\times 0.08\times 10=$20\ 000\)

\(\text{Total}=25\ 000+20\ 000=$45\ 000\)

\(\therefore\ \text{Option B gives the best return.}\)

Financial Maths, STD1 F3 2025 HSC 24

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

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

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

Calculate Kat’s monthly repayment.   (5 marks)

Show Answers Only

\($835.31\)

Show Worked Solution

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

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

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

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

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

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

Algebra, STD1 A3 2025 HSC 23

It costs $465 to register a passenger car and $350 to register a motorcycle.

Let    \(P\) \(\ =\ \text{the number of passenger cars, and}\)
  \(B\) \(\ =\ \text{the number of motorcycles}\)

 
Write TWO linear equations that represent the relationship below.

  • There are 11 times as many passenger cars as motorcycles.
  • The total registration fees for passenger cars and motorcycles is $494 million.   (2 marks)
Show Answers Only

\(\text{Equation 1: }\ P=11B\)

\(\text{Equation 2: }\ 465P+350B=494\ 000\ 000\)

Show Worked Solution

\(\text{Equation 1: }\ P=11B\)

\(\text{Equation 2: }\ 465P+350B=494\ 000\ 000\)

Measurement, STD1 M3 2025 HSC 22

An isosceles triangle is drawn inside a circle as shown. The base of the triangle is 4.8 cm long, the length of other sides is 4 cm and the height is \(h\) cm.
 

  1. Calculate the height, \(h\), of triangle \(ABC\).   (2 marks)
  1. The area of triangle \(ABC\) is 7.68 cm².
  2. The radius of the circle is 2.5 cm.
  3. Express the area of triangle \(ABC\) as a percentage of the area of the circle, correct to 1 decimal place.   (2 marks)
Show Answers Only

a.    \(h=3.2\ \text{cm}\)

b.    \(39.1\%\)

Show Worked Solution

a.   \(\text{Since}\ \Delta ABC\ \text{is isosceles:}\)

\(BM\ \text{bisects}\ AC\ \ \Rightarrow\ \ AM=MC=2.4\)
 

\(\text{Using Pythagoras:}\)

\(h^2=4^2-2.4^2=16-5.76=10.24\)

\(h = \sqrt{10.24} = 3.2\ \text{cm}\)
 

b.  \(\text{Area of circle}=\pi\times 2.5^2\)

\(\text{Percentage}\) \(=\dfrac{\text{Area of triangle}}{\text{Area of circle}}\times 100\%\)
  \(=\dfrac{7.68}{\pi\times 2.5^2}\times 100\%\)
  \(=39.11…\%\)
  \(\approx 39.1\%\ \text{(1 decimal place)}\)

Measurement, STD1 M3 2025 HSC 20

A map of a park containing a duck pond is shown.

A fence is built passing through the points \(A\), \(B\) and \(C\) around the duck pond.
 

  1. Using the scale provided on the map, calculate the length of the fence \(AB\).   (2 marks)
  1. The length of \(AB\) is equal to the length of \(BC\).
  2. Use Pythagoras’ theorem to calculate the length of \(AC\) in metres. Give your answer correct to 3 significant figures.   (3 marks)
  3. What is the true bearing of point \(A\) from point \(C\) ?   (2 marks)
Show Answers Only

a.    \(75\ \text{metres}\)

b.    \(106\ \text{metres}\)

c.    \(225^\circ\)

Show Worked Solution

a.   \(\text{Scale: 1 grid width = 5 metres}\)

\(AB = 15 \times 5 = 75\ \text{metres}\)
 

b.   \(\text{Using Pythagoras:}\)

\(AC^2=AB^2+BC^2\)

\(AC^2=75^2+75^2=11250\)

  \(\therefore\ AC\) \(=\sqrt{11250}\)
    \(=106.066…\)
    \(\approx 106\ \text{m (3 sig fig)}\)

  
c.    \(\text{Since }AB=BC:\)

\(\angle BAC=\angle BCA=45^\circ\)

\(\text{Bearing of \(A\) from \(C\)}\ =180+45=225^\circ\)

Financial Maths, STD1 F1 2025 HSC 19

At the end of the 2024−2025 financial year, Alex’s taxable income was $148 600.

  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}\textit{    Taxable income}\rule[-1ex]{0pt}{0pt} & \textit{    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 Alex’s tax payable.   (3 marks)

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

Show Answers Only

a.    \($36\ 320\)

b.    \($2972\)

Show Worked Solution
a.     \(\text{Tax payable}\) \(=31\ 288+0.37\times(148\ 600-135\ 000)\)
    \(=31\ 288+0.37\times 13\ 600\)
    \(=31288+5032\)
    \(=$36\ 320\)

 

b.    \(\text{Medicare}\) \(=0.02\times 148\ 600\)
    \(=$2972\)

Measurement, STD1 M4 2025 HSC 18

Ramon took 1.25 hours to travel the length of a freeway. The length of the freeway is 100 km.

  1. Ramon’s total travel time was made up of:
    • 20 minutes to travel to the start of the freeway
    • the time taken to travel the length of the freeway
    • 35 minutes after exiting the freeway to get to his destination.
  1. What was the total time Ramon spent travelling? Give your answer in hours and minutes.   (2 marks)
  1. What was Ramon’s average speed on the freeway?   (1 mark)
Show Answers Only

a.    \(2\ \text{hours}\ 10\ \text{minutes}\)

b.    \(80\ \text{km/h}\)

Show Worked Solution
a.     \(\text{Total time}\) \(=20+75+35\)
    \(=130\ \text{minutes}\)
    \(=2\text{ h }10\text{ min}\)

 

b.     \(\text{Average speed}\) \(=\dfrac{\text{distance}}{\text{time}}\)
    \(=\dfrac{100}{1.25}\)
    \(=80\ \text{km/h}\)

Algebra, STD1 A2 2025 HSC 16

The mass \((M )\) of a box with a square base, in grams, is directly proportional to the area of its base, in cm².
 

A box with a square base of side length 5 cm has a mass of 500 g.

What is the mass of a similar box with a square base of side length 3 cm?    (3 marks)

Show Answers Only

\(M=180\ \text{grams}\)

Show Worked Solution

\(\text{Area of square base }= s^2\)

\(M \propto s^2\ \ \Rightarrow \ \ M=k\times s^2\)

\(\text{Find \(k\) given \(\ s=5\ \) when \(\ M=500\):}\)

\(500\) \(=k\times 5^2\)
\(25k\) \(=500\)
\(k\) \(=\dfrac{500}{25}=20\)

 
\(\text{Find \(M\) when  \(s=3\):}\)

\(M=20\times 3^2=180\ \text{grams}\)