Band 5 – Page 22

Calculus, MET2 2024 VCAA 10 MC

Suppose a function $f:[0,5] \rightarrow R$ and its derivative $f^{\prime}:[0,5] \rightarrow R$ are defined and continuous on their domains. If $f^{\prime}(2)<0$ and $f^{\prime}(4)>0$, which one of these statements must be true?

$f$ is strictly decreasing on $[0,2]$.
$f$ does not have an inverse function.
$f$ is positive on $[4,5]$.
$f$ has a local minimum at $x=3$.

Show Answers Only

$B$

Show Worked Solution

$\text{As}\ \ f^{\prime}(2)<0\ \ \text{and}\ \ f^{\prime}(4)>0\ \ \text{the function must have} $

$\text{a turning point between}\ \ x=2\ \ \text{and}\ \ x=4.$

$\therefore\ f(x)\ \text{will not have an inverse function, as it is many-to-one.}$

$\Rightarrow B$

♦♦ Mean mark 38%.

Functions, MET2 2024 VCAA 6 MC

Consider the function $f(x)=\dfrac{2 x+1}{3-x}$ with domain $x \in R \backslash\{3\}$

The inverse of $f$ is

$f^{-1}(x)=\dfrac{3 x-1}{x+2}$ with domain $x \in R \backslash\{3\}$
$f^{-1}(x)=3-\dfrac{7}{x+2}$ with domain $x \in R \backslash\{-2\}$
$f^{-1}(x)=3+\dfrac{5}{x+2}$ with domain $x \in R \backslash\{-2\}$
$f^{-1}(x)=\dfrac{1-3 x}{x+2}$ with domain $x \in R \backslash\{-2\}$

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$B$

Show Worked Solution

$\text{Let}\ y=f(x)$

$\text{For inverse: swap }x\leftrightarrow y$

$x$	$=\dfrac{2y+1}{3-y}$
$3x-xy$	$=2y+1$
$3x-1$	$=2y+xy$
$y(2+x)$	$=3x-1$
$y$	$=\dfrac{3x-1}{2+x}$
	$=\dfrac{6+3x-7}{2+x}$
	$=3-\dfrac{7}{2+x}$

$\Rightarrow B$

♦ Mean mark 50%.

Functions, MET2 2024 VCAA 5 MC

Consider the functions $f:(1, \infty) \rightarrow R, \ (x)=x^2-4 x$ and $g: R \rightarrow R, \ g(x)=e^{-x}$.

The range of the composite function $g(f(x))$ is

$ (0, e^3) $
$ (0, e^3 ] $
$ (0, e^4) $
$ (0, e^4 ]$

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$D$

Show Worked Solution

$\text{dom}\left(g(f)\right)=\text{dom}(f)$

$g(f(x))=e^{-(x^2-4x)}$

$\text{From Graph }\rightarrow \text{range}\ = (0, e^{4}]$

$\Rightarrow D$

♦ Mean mark 52%.

PHYSICS, M8 2024 HSC 22

The following graph, based on the data gathered by Hubble, shows the relationship between the recessional velocity of galaxies and their distance from Earth.

Describe the significance of the graph to our understanding of the universe. (2 marks)

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How were the recessional velocities of galaxies determined? (3 marks)

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a. Recessional velocity vs distance from Earth graph:

The graph shows that the further away galaxies are from Earth, the faster these galaxies are moving away from Earth.
This relationships depicts Hubble’s law: $v=H_0 D$.
This graph provides evidence that the universe is constantly expanding as predicted by the big bang theory.

b. Determining recessional velocities:

The recessional velocities of the galaxies were determined by analysing their absorption spectras.
Light waves that are moving away from the Earth will appear to be stretched (wavelength increased) according to the doppler effect.
The absorption spectra of galaxies were compared with the spectra of the same elements on Earth, revealing that the galaxies’ spectra were redshifted.
The greater the extent of the red shift in the spectra, the greater the recessional velocity of the galaxy.

Show Worked Solution

a. Recessional velocity vs distance from Earth graph:

The graph shows that the further away galaxies are from Earth, the faster these galaxies are moving away from Earth.
This relationships depicts Hubble’s law: $v=H_0 D$.
This graph provides evidence that the universe is constantly expanding as predicted by the big bang theory.

b. Determining recessional velocities:

The recessional velocities of the galaxies were determined by analysing their absorption spectras.
Light waves that are moving away from the Earth will appear to be stretched (wavelength increased) according to the doppler effect.
The absorption spectra of galaxies were compared with the spectra of the same elements on Earth, revealing that the galaxies’ spectra were redshifted.
The greater the extent of the red shift in the spectra, the greater the recessional velocity of the galaxy.

♦ Mean mark (b) 49%.

PHYSICS, M6 2024 HSC 17 MC

The diagram shows a type of particle accelerator called a cyclotron.

Cyclotrons accelerate charged particles, following the path as shown.

An electric field acts on a charged particle as it moves through the gap between the dees. A strong magnetic field is also in place.

Once a charged particle has the required velocity, it exits the accelerator towards a target.

Which of the following is true about a charged particle in a cyclotron?

It increases speed while inside the dees.
It only accelerates while between the dees.
It undergoes acceleration inside and between the dees.
It slows down inside the dees and speeds up between the dees.

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$C$

Show Worked Solution

When the charged particle is between the dees, it will experience an acceleration due to the electric field present where $a = \dfrac{qE}{m}$.
When the charged particle is inside of the dees, the particle undergoes uniform circular motion due to the strong magnetic field in place from the electromagnets.
While the magnitude of the velocity of the charged particle does not change, the direction of the velocity does. Hence, there is a change in velocity of the particle so it is experiencing an acceleration.
A charged particle will experience a centripetal force/acceleration when moving perpendicular to a magnetic field.

$\Rightarrow C$

♦ Mean mark 45%.

PHYSICS, M7 2024 HSC 16 MC

The graph shows the relationship between the maximum kinetic energy of emitted photoelectrons and the incident photon energy for four different metal surfaces.

Light of frequency $7 \times 10^{14}\ \text{Hz}$ is incident on the metals.

From which metals are photoelectrons emitted?

$\ce{K}$, $\ce{Li}$ only
$\ce{Mg}$, $\ce{Ag}$ only
All of the metals
None of the metals

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$A$

Show Worked Solution

The photon energy for a light frequency of $7 \times 10^{14}\ \text{Hz}$ is:
$E=hf = 6.626 \times 10^{-34} \times 7 \times 10^{14} = 4.64 \times 10^{-19}\ \text{J}$
The energy of the light frequency in electron volts$=\dfrac{4.64 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.9\ \text{eV}$.
Photoelectrons will only be emitted if the energy of the incident photons is enough to overcome the work function of the metals. On the graph above, the work function of each metal is the same value as the x-intercepts of their respective graphs.
Therefore photoelectrons will be emitted from potassium and lithium as the energy of the incident photons $(2.9\ \text{eV})$ is greater than the work functions of these metals.

$\Rightarrow A$

♦ Mean mark 47%.

PHYSICS, M6 2024 HSC 15 MC

A uniform magnetic field is directed into the page. A conductor $P Q$ rotates about the end $P$ at a constant rate.

Which graph shows the emf induced between the ends of the conductor, $P$ and $Q$, as it rotates one revolution from the position shown?

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$D$

Show Worked Solution

As the conductor rotates in the magnetic field, all of the charges in the conductor will have a velocity.
This will produce a force on each positive and negative charge in the rod according to the equation $F=qvB$.
Using the right hand rule, the direction of the force applied to each positive charge in the rod during the rotation will be towards $P$. Similarly, the direction of force on each negative charge will be towards $Q$.
This separation of positive and negative charges to opposite ends of the conductor generates the emf.
As the velocity of the charges during the circular motion will remain constant, the force on the charges will be constant. Therefore, a constant emf will be produced in the conductor.

$\Rightarrow D$

♦♦ Mean mark 35%.

PHYSICS, M8 2024 HSC 14 MC

The velocity of a proton $ {\displaystyle \left({ }_1^1 \text{H}\right) } $ is twice the velocity of an alpha particle $ { \displaystyle \left({ }_2^4 \text{He}\right) } $. The proton has a de Broglie wavelength of $\lambda$.

What is the de Broglie wavelength of the alpha particle?

$\dfrac{\lambda}{8}$
$\dfrac{\lambda}{2}$
$2 \lambda$
$8 \lambda$

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$B$

Show Worked Solution

Let the mass of the proton be $m$ and the velocity of the proton be $v$.
Therefore the mass of the alpha particle will be $4m$ and the velocity will be $\dfrac{v}{2}$.
The de Broglie wavelength of the proton, $\lambda_{p} = \dfrac{h}{mv}$.
Therefore the de Broglie wavelength of the alpha particle:
$\lambda_{\alpha}=\dfrac{h}{4m \times \frac{v}{2}}=\dfrac{h}{2mv} = \dfrac{\lambda}{2}$

$\Rightarrow B$

♦ Mean mark 49%.

PHYSICS, M6 2024 HSC 10 MC

A rod carrying a current, $I$, placed in a uniform magnetic field as shown, experiences a force $F$.

How many degrees must the rod be rotated clockwise so that it experiences a force $\dfrac{F}{2}$?

30°
45°
60°
90°

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$C$

Show Worked Solution

The force experienced by a current carrying conductor in a magnetic field is given by $F=lIB\sin \theta$, where $\sin \theta$ refers to the angle between the direction of the current through the conductor and the magnetic field lines.
For the rod to experience a force of $\dfrac{F}{2}$, the value of $\sin \theta = \dfrac{1}{2}$.
$\sin \theta=\dfrac{1}{2}\ \ \Rightarrow \ \theta=\sin^{-1}\bigg(\dfrac{1}{2}\bigg)=30^{\circ}$
As the angle between the current carrying conductor and magnetic field lines originally is $90^{\circ}$, the rod must be rotated by $60^{\circ}$.

$\Rightarrow C$

♦♦ Mean mark 31%.

PHYSICS, M8 2024 HSC 7 MC

A pure sample of polonium-210 undergoes alpha emission to produce the stable isotope lead-206.

The half-life of polonium-210 is 138 days.

At the end of 276 days, what is the ratio of polonium-210 atoms to lead-206 atoms in the sample?

$1 : 4$
$1 : 3$
$1 : 2$
$1 : 1$

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$B$

Show Worked Solution

After two half-life’s of the polonium only one quarter of the sample of polonium-210 atoms will be left.
Therefore, three quarters of the polonium-210 atoms would have decayed into lead-206 atoms.
The ratio of polonium atoms to lead atoms is $\dfrac{1}{4}:\dfrac{3}{4} = 1:3$

$\Rightarrow B$

♦ Mean mark 45%.

PHYSICS, M8 2024 HSC 3 MC

Which of the following is a fundamental particle in the Standard Model of matter?

Hadron
Neutron
Photon
Proton

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$C$

Show Worked Solution

Hadrons are subatomic particles which are composed of two or more quarks. Protons and Neutrons are categorised as hadrons as they are both composed of 3 quarks.
Hadrons, protons and neutrons are not fundamental particles as they are all composed of quarks.
The photon is a fundamental particle, classified as a gauge boson and mediates the electromagnetic force.

$\Rightarrow C$

♦ Mean mark 55%.

BIOLOGY, M8 2024 HSC 35

The graph shows the results of a survey conducted to determine if children changed their method of communication after cochlear implantation.

With reference to the data, describe how cochlear implants work, and how they affect communication in children. (5 marks)

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Cochlear implants are surgical electronic devices that help restore hearing in patients with cochlear damage.
The implants are inserted directly into the cochlea and stimulate the auditory nerve by sending sound signals straight to the brain.
The graph demonstrates that the age of implantation significantly affects communication outcomes.
Early implantation (under 3 years): Dramatic decrease in sign language use, with only 10% (approximately) still signing after 5 years.
Middle age implantation (3-5 years): Moderate decrease in sign language use.
Late implantation (over 5 years): Little to no change in sign language use
The data clearly shows that earlier cochlear implantation leads to greater shifts away from sign language as the primary mode of communication.

Show Worked Solution

Cochlear implants are surgical electronic devices that help restore hearing in patients with cochlear damage.
The implants are inserted directly into the cochlea and stimulate the auditory nerve by sending sound signals straight to the brain.
The graph demonstrates that the age of implantation significantly affects communication outcomes.
Early implantation (under 3 years): Dramatic decrease in sign language use, with only 10% (approximately) still signing after 5 years.
Middle age implantation (3-5 years): Moderate decrease in sign language use.
Late implantation (over 5 years): Little to no change in sign language use
The data clearly shows that earlier cochlear implantation leads to greater shifts away from sign language as the primary mode of communication.

♦♦ Mean mark 52%.

BIOLOGY, M6 2024 HSC 34

Discuss the ethical implications and impacts on society of the use of TWO biotechnologies. (7 marks)

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Social impacts of recombinant DNA technology (plant biotechnology)

Recombinant DNA has created beneficial products like Bt corn and Bt cotton.
Bt crops require fewer pesticide applications since they produce their own insecticidal proteins.
This reduces chemical pesticide costs for farmers and decreases environmental impact from pesticide spraying.

Associated ethical concerns:

Farmers growing Bt corn and cotton must purchase, at a significant cost, new GM seeds each season. Traditional farmers can reuse their seeds for the following year’s crop planting.
This difference creates economic disparities in access to GM crops and the related market opportunities.

Social impacts of selective breeding/hybridisation (animal biotechnology)

Selective breeding/hybridisation has produced, for example, dairy cows capable of increased milk production.
This results in higher yields and greater food availability.
Improved profits and living standards for farmers who can access the technology.
Increased food production to support population growth.

Associated ethical issues:

High-yield dairy cows show decreased fertility.
May compromise animal welfare and quality of life.

Show Worked Solution

Social impacts of recombinant DNA technology (plant biotechnology)

Recombinant DNA has created beneficial products like Bt corn and Bt cotton.
Bt crops require fewer pesticide applications since they produce their own insecticidal proteins.
This reduces chemical pesticide costs for farmers and decreases environmental impact from pesticide spraying.

Associated ethical concerns:

Farmers growing Bt corn and cotton must purchase, at a significant cost, new GM seeds each season. Traditional farmers can reuse their seeds for the following year’s crop planting.
This difference creates economic disparities in access to GM crops and the related market opportunities.

Social impacts of selective breeding/hybridisation (animal biotechnology)

Selective breeding/hybridisation has produced, for example, dairy cows capable of increased milk production.
This results in higher yields and greater food availability.
Improved profits and living standards for farmers who can access the technology.
Increased food production to support population growth.

Associated ethical issues:

High-yield dairy cows show decreased fertility.
May compromise animal welfare and quality of life.

♦ Mean mark 56%.

BIOLOGY, M5 2024 HSC 33

Female Jack Jumper ants (Myrmecia pilosula) have a single pair of chromosomes. During meiosis, crossing over occurs. The diagram shows the crossing over and the position of three genes on the chromosomes.

Outline the significance of crossing over for the Jack Jumper ants. (2 marks)

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Draw the chromosomes of the four possible gametes after crossing over for the Jack Jumper ants occurs. Include the alleles for each gene. (2 marks)

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a. Significance of crossing over:

Genetic variation increases in the Jack Jumper ant population through recombination.
This enhanced genetic diversity improves the species’ chances of survival when faced with environmental changes, as some ants may carry beneficial adaptations.

Show Worked Solution

a. Significance of crossing over:

Genetic variation increases in the Jack Jumper ant population through recombination.
This enhanced genetic diversity improves the species’ chances of survival when faced with environmental changes, as some ants may carry beneficial adaptations.

♦ Mean mark (b) 54%.

BIOLOGY, M7 2024 HSC 32

Helicobacter pylori is a bacterium that invades the gut lining and can cause damage to the stomach as shown in the diagram.

With reference to innate and adaptive immunity, explain how the body responds after exposure to Helicobacter pylori. (7 marks)

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Damaged cells release chemicals that trigger inflammation as an initial response.
The inflammatory response causes blood vessels to dilate, increasing blood flow and allowing phagocytes (macrophages and neutrophils) to move into the infected area.
Phagocytes process H.pylori antigens and present them to helper T-cells, which launch the adaptive immune response by releasing cytokines.
This cytokine release activates both T and B cells to mount multiple specific defences.
Cytotoxic T-cells directly attack H.pylori while memory T-cells remain for secondary rapid responses.
Suppressor T-cells regulate the immune response and plasma B-cells produce H.pylori-specific antibodies. Memory B-cells persist for responding to future (secondary) infections.
Antibodies work in two ways – direct neutralisation of antigens and tagging antigens for destruction by phagocytes.

The immune response involves both innate and adaptive immunity systems working together:

Innate immunity provides rapid, immediate defence.
Adaptive immunity develops more slowly but offers long-term protection through memory cells.

Show Worked Solution

Damaged cells release chemicals that trigger inflammation as an initial response.
The inflammatory response causes blood vessels to dilate, increasing blood flow and allowing phagocytes (macrophages and neutrophils) to move into the infected area.
Phagocytes process H.pylori antigens and present them to helper T-cells, which launch the adaptive immune response by releasing cytokines.
This cytokine release activates both T and B cells to mount multiple specific defences.
Cytotoxic T-cells directly attack H.pylori while memory T-cells remain for secondary rapid responses.
Suppressor T-cells regulate the immune response and plasma B-cells produce H.pylori-specific antibodies. Memory B-cells persist for responding to future (secondary) infections.
Antibodies work in two ways – direct neutralisation of antigens and tagging antigens for destruction by phagocytes.

The immune response involves both innate and adaptive immunity systems working together:

Innate immunity provides rapid, immediate defence.
Adaptive immunity develops more slowly but offers long-term protection through memory cells.

♦♦ Mean mark 48%.

BIOLOGY, M8 2024 HSC 29

An epidemiological study was conducted to help model how many people will be affected by Type 2 diabetes globally in the future. Continuous data were collected from 1990 to 2020. From that data, the following data points were chosen to demonstrate the trend.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Year} \rule[-1ex]{0pt}{0pt} & \textit{Percentage of population affected} \\
\rule{0pt}{2.5ex} \textit{} \rule[-1ex]{0pt}{0pt} & \textit{by Type 2 diabetes (%)} \\
\hline
\rule{0pt}{2.5ex} \text{1990} \rule[-1ex]{0pt}{0pt} & \text{3.1} \\
\hline
\rule{0pt}{2.5ex} \text{2000} \rule[-1ex]{0pt}{0pt} & \text{3.7} \\
\hline
\rule{0pt}{2.5ex} \text{2010} \rule[-1ex]{0pt}{0pt} & \text{4.3} \\
\hline
\rule{0pt}{2.5ex} \text{2010} \rule[-1ex]{0pt}{0pt} & \text{5.6} \\
\hline
\end{array}

Plot the data on the grid provided and include the line of best fit. (2 marks)

A prediction of the global population numbers suggests there will be about 9 billion $(9\ 000\ 000\ 000)$ people on the planet by 2040.

Predict the number of people that will be affected by diabetes in 2040. Show working on your graph on the previous page and your calculations. (3 marks)

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b. 630 000 000 (630 million)

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♦ Mean mark (a) 56%.

b. Using the LOBF on the graph:

Population (%) with diabetes in 2040 = 7%

People with diabetes in 2040 = $\dfrac{7}{100} \times 9\ 000\ 000\ 000 = 630\ 000\ 000$

BIOLOGY, M8 2024 HSC 31

A study monitored the changes in the body temperature of a kookaburra (an Australian bird) and a human over a 24-hour period. The results of the study are shown in the graph.

At what time was the kookaburra's body temperature the lowest? (1 mark)

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Some endothermic organisms can display torpor (a significant decrease in physiological activity).

With reference to the graph, explain whether the human or the kookaburra was displaying torpor and if so, state the time this occurred. (3 marks)

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Outline an adaptation that may lead to an increase in the kookaburra's body temperature during the inactive period. (2 marks)

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a. 4 am

b. Signs of torpor:

The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity
In contrast, the kookaburra exhibited classic torpor behaviour.
Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.

c. Kookaburra adaptation:

Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.
This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier
The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.

Show Worked Solution

a. 4 am

b. Signs of torpor:

The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity
In contrast, the kookaburra exhibited classic torpor behaviour.
Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.

Mean mark (b) 56%.

c. Kookaburra adaptation:

Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.
This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier
The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.

♦ Mean mark (c) 47%.

BIOLOGY, M5 2024 HSC 30

The diagram shows a simplified version of the process of polypeptide synthesis.

Compare Process $A$ with DNA replication. (3 marks)

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Explain the importance of mRNA and tRNA in polypeptide synthesis. (5 marks)

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a. Process $A$ vs DNA replication:

Both DNA replication and transcription (Process $A$) begin with unwinding the DNA double helix.
DNA replication’s goal is to create two identical DNA molecules, with each containing one original and one new strand.
In contrast, transcription copies just one DNA strand to produce a single mRNA strand.

b. mRNA and tRNA’s role in polypeptide synthesis:

mRNA is created in the nucleus by copying a DNA template during transcription.
This mRNA molecule serves as a messenger, carrying genetic instructions (in the form of codons) from the nucleus out to ribosomes in the cytoplasm.
At the ribosome, translation kicks in – this is where the genetic code gets converted into protein.
tRNA molecules are key players here – each has an anticodon that matches up with specific codons on the mRNA strand.
The process flows like an assembly line: mRNA codons are read in sequence, tRNA molecules bring in matching amino acids, and these amino acids are linked together to form a polypeptide chain.

Show Worked Solution

a. Process $A$ vs DNA replication:

Both DNA replication and transcription (Process $A$) begin with unwinding the DNA double helix.
DNA replication’s goal is to create two identical DNA molecules, with each containing one original and one new strand.
In contrast, transcription copies just one DNA strand to produce a single mRNA strand.

b. mRNA and tRNA’s role in polypeptide synthesis:

mRNA is created in the nucleus by copying a DNA template during transcription.
This mRNA molecule serves as a messenger, carrying genetic instructions (in the form of codons) from the nucleus out to ribosomes in the cytoplasm.
At the ribosome, translation kicks in – this is where the genetic code gets converted into protein.
tRNA molecules are key players here – each has an anticodon that matches up with specific codons on the mRNA strand.
The process flows like an assembly line: mRNA codons are read in sequence, tRNA molecules bring in matching amino acids, and these amino acids are linked together to form a polypeptide chain.

♦ Mean mark (b) 60%.

BIOLOGY, M5 2024 HSC 28

Cystic fibrosis is an inherited disorder that causes damage to the lungs, digestive system and other organs in the body. A person with cystic fibrosis will have two faulty recessive alleles for the cystic fibrosis gene (CFTR) on chromosome 7.

Two healthy parents, heterozygous for cystic fibrosis, have a child that does not have cystic fibrosis. They are planning to have a second child.

Using a Punnett square, determine the probability of their second child being born with the condition. Use $R$ for the normal CFTR allele, and $r$ for the faulty CFTR allele. (3 marks)

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The defect that creates the faulty CFTR allele is often caused by the deletion of three nucleotides. The following diagram illustrates a small section of the CFTR gene and the corresponding amino acid sequence of the CFTR protein.

The following codon chart displays all the codons and the corresponding amino acids. The chart translates mRNA sequences into amino acids.

Explain how the deletion of nucleotides in the CFTR gene removes only one amino acid. Include reference to the nucleotides that code for isoleucine and phenylalanine amino acids. (4 marks)

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a. Punnett square:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \boldsymbol{R} & \boldsymbol{r} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{R} \rule[-1ex]{0pt}{0pt} & \textit{RR} & \textit{Rr} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{r}\rule[-1ex]{0pt}{0pt} & \textit{Rr} & \textit{rr} \\
\hline
\end{array}

Probability of 2nd child having cystic fibrosis = 25%.

b. Deletion of nucleotides in CFTR gene:

Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.
At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.
In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.
Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.

Show Worked Solution

a. Punnett square:

Probability of 2nd child having cystic fibrosis = 25%.

b. Deletion of nucleotides in CFTR gene:

Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.
At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.
In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.
Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.

♦ Mean mark (b) 56%.

BIOLOGY, M7 2024 HSC 27

Milk pasteurisation (heating to approximately 70°C) was gradually introduced in America from the early 1900s. The graph shows the number of disease outbreaks in relation to raw (unpasteurised) and pasteurised milk in America from 1900-1975.

Explain the trends observed in the graph. In your response, refer to the role of Pasteur's work in pasteurisation. (5 marks)

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Louis Pasteur’s research was pivotal in debunking the theory of spontaneous generation and establishing our understanding of microorganisms.
His work revealed that microbes present in milk could be responsible for disease outbreaks.
He demonstrated that exposing substances to high temperatures effectively kills microorganisms, which is why heating milk to 70°C eliminates many harmful bacteria.
This scientific foundation – the presence of microbes in milk and their vulnerability to heat – explains the effectiveness of milk pasteurisation in preventing disease outbreaks.
The historical data presented in the graph supports this, showing significantly fewer disease outbreaks linked to pasteurised milk compared to raw milk.
A notable decline in raw milk-related outbreaks occurred after 1945, though this may also be attributed to decreased raw milk consumption during that period.
While pasteurised milk has generally proven safer, some disease outbreaks have still occurred with pasteurised products. These cases typically result from issues in the pasteurisation process itself or problems during subsequent storage and transportation of the milk.

Show Worked Solution

Louis Pasteur’s research was pivotal in debunking the theory of spontaneous generation and establishing our understanding of microorganisms.
His work revealed that microbes present in milk could be responsible for disease outbreaks.
He demonstrated that exposing substances to high temperatures effectively kills microorganisms, which is why heating milk to 70°C eliminates many harmful bacteria.
This scientific foundation – the presence of microbes in milk and their vulnerability to heat – explains the effectiveness of milk pasteurisation in preventing disease outbreaks.
The historical data presented in the graph supports this, showing significantly fewer disease outbreaks linked to pasteurised milk compared to raw milk.
A notable decline in raw milk-related outbreaks occurred after 1945, though this may also be attributed to decreased raw milk consumption during that period.
While pasteurised milk has generally proven safer, some disease outbreaks have still occurred with pasteurised products. These cases typically result from issues in the pasteurisation process itself or problems during subsequent storage and transportation of the milk.

BIOLOGY, M6 2024 HSC 19 MC

The diagram represents some experimental steps used in the production of large amounts of human growth hormone.

What makes this technology successful?

DNA ligase can cut human growth hormone genes from human cells.
Human growth hormone can cause E. coli to grow and mature rapidly.
Human plasmids containing the gene of interest can be inserted into bacteria.
Restriction enzymes can produce sticky ends on both bacterial and human DNA.

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$D$

Show Worked Solution

The technology is possible because restriction enzymes create compatible sticky ends on both the bacterial plasmid and human DNA containing the growth hormone gene.
This allows them to be joined together via complementary base pairing before DNA ligase permanently connects them, which is essential for successful gene transfer.

$\Rightarrow D$

♦ Mean mark 53%.

BIOLOGY, M6 2024 HSC 18 MC

The following diagram models a population of glowworms in an isolated cave. The letter $B$ and $b$ represent the alleles for a gene in an individual.

Which row of the table correctly identifies the process and reason for the change in the gene pool?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Process} \rule[-1ex]{0pt}{0pt}& \textit{Reason} \\
\hline
\rule{0pt}{2.5ex}\text{Gene flow} & \text{Change in frequency} \\
& \text{of the $b$ allele due to a mutation} \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Genetic drift} & \text{Change in frequency} \\
& \text{of the $b$ allele due to random events} \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Gene flow} & \text{Introduction of more $b$ alleles due } \\
& \text{to new members moving into the population} \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Genetic drift} & \text{Introduction of more $b$ alleles due } \\
& \text{to this allele being more advantageous} \rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

The diagram shows genetic drift, where the $b$ allele becomes fixed in the population over time through random chance events in a small, isolated population (as indicated by the cave setting).
The context of an isolated cave means that the changes could not have been the result of natural selection or gene flow from outside populations.

$\Rightarrow B$

♦♦ Mean mark 33%.

BIOLOGY, M8 2024 HSC 17 MC

Over 12 months, the prevalence of a non-infectious disease will increase in a population if

the total population increases.
disease recovery time decreases.
the incidence rate of the disease decreases.
the survival time of individuals with the disease increases.

Show Answers Only

$D$

Show Worked Solution

Prevalence = number of existing cases in a population at a given time.
Factors affecting prevalence include new cases (incidence), duration of disease, death/recovery rates and population changes.
If people with the disease live longer, they remain in the “prevalent cases” count for a longer time, increasing overall prevalence.

$\Rightarrow D$

♦ Mean mark 51%.

BIOLOGY, M8 2024 HSC 16 MC

In a person with a particular visual disorder, light from a distant object focuses in front of the retina.

How can this disorder be corrected?

Laser surgery to reshape the retina
Use of a diverging lens in front of the eye
Use of a converging lens in front of the eye
Laser surgery to make the cornea more curved

Show Answers Only

$B$

Show Worked Solution

A diverging (concave) lens will spread light rays slightly before they enter the eye, moving the focal point back to the retina.

$\Rightarrow B$

♦ Mean mark 53%.

BIOLOGY, M5 2024 HSC 14 MC

The graph shows the quantity of DNA in one mammalian cell over time.

Which of the statements about the cell's activity is correct?

The cell is dividing during section $O$.
The cell is inactive for the first two hours.
The cell takes one hour to replicate its DNA.
The cell becomes diploid only during section $N$.

Show Answers Only

$A$

Show Worked Solution

Section $O$ shows the DNA quantity decreasing sharply, which occurs during cell division (mitosis).
At this stage, the parent cell splits into two daughter cells, each receiving half of the DNA content.

$\Rightarrow A$

♦ Mean mark 50%.

CHEMISTRY, M5 2024 HSC 39

Water and octan-1-ol do not mix. When an aqueous solution of bromoacetic acid $\left(\ce{BrCH2COOH}\right)$ is shaken with octan-1-ol, an equilibrium system is established between bromoacetic acid dissolved in the octan-1-ol and in the water.

$\ce{BrCH2COOH(aq) \rightleftharpoons BrCH2COOH}\textit{(octan-l-ol)}$

The equilibrium constant expression for this system is

$K_{e q}=\dfrac{\left[\ce{BrCH2COOH}\textit{(octan-l-ol)}\right]}{\left[\ce{BrCH2COOH}\textit{(aq)}\right]}$.

An aqueous solution of bromoacetic acid with an initial concentration of 0.1000 mol L $^{-1}$ is shaken with an equal volume of octan-1-ol. Bromoacetic acid does not dissociate in octan-1-ol but does dissociate in water, with $K_a=1.29 \times 10^{-3}$. When the system has reached equilibrium, the $\left[\ce{H+}\right]$ is $9.18 \times 10^{-3} \text{ mol L}^{-1}$.

Calculate the equilibrium concentration of aqueous bromoacetic acid and hence, or otherwise, calculate the $K_{eq}$ for the octan-1-ol and water system. (4 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

$0.390$

Show Worked Solution

The ionisation of bromoacetic acid in water is:
$\ce{BrCH2COOH(aq) \rightleftharpoons BrCH2COO^-(aq) + H^+(aq)}$
At equilibrium $\ce{[BrCH2COO^-(aq)] = [H^+(aq)]} = 9.18 \times 10^{-3}\ \text{mol L}^{-1}$ as the are formed in a $1:1$.

$K_{a}$	$=\dfrac{\ce{[H^+][BrCH2COO^-]}}{\ce{[BrCH2COOH]_{eq}}}$
$\ce{[BrCH2COOH]_{eq}}$	$=\dfrac{\ce{[H^+][BrCH2COO^-]}}{K_a}$
	$=\dfrac{(9.18 \times 10^{-3})^2}{1.29 \times 10^{-3}}$
	$=0.06533\ \text{mol L}^{-1}$

$\ce{[BrCH2COOH]_{\text{total}}}=\ce{[BrCH2COOH(aq)]_{eq} + [BrCH2COO^-(aq)]}$

$\ce{+ [BrCH2COOH(octan-1-ol)]_{eq}}$

$\ce{[BrCH2COOH(octan-1-ol)]_{eq}}$	$=0.1000-0.06533-9.18 \times 10^{-3}$
	$=0.02549\ \text{mol L}^{-1}$

Since the volume of the aqueous solution of bromoacetic acid and octane is the same, the concentration values between the water and octane solutions can be added/subtracted in one equation and mole calculations are not required.
$K_{eq} = \dfrac{\ce{[BrCH2COOH(octan-1-ol)]_{eq}}}{\ce{[BrCH2COOH(aq)]_{eq}}}= \dfrac{0.02549}{0.06533}=0.390\ \text{(3 sig. fig.)}$

♦♦ Mean mark 27%.
COMMENT: Students who identified the acid conc in the organic solvent often succeeded in this question.

CHEMISTRY, M8 2024 HSC 38

Compounds $\text{A}$ and $\text{B}$ are isomers with formula $\ce{C3H7X}$, where $\ce{X}$ is a halogen. The mass spectrum for compound $\text{A}$ is shown.

Compounds $\text{A}$ and $\text{B}$ undergo substitution reactions in the presence of hydroxide ions, producing alcohols $\text{C}$ and $\text{D}$. Compound $\text{D}$ can be oxidised to a ketone; compound $\text{C}$ can also be oxidised, but does not produce a ketone.

Compound $\text{E}$ can be produced by refluxing 3-methylbutanoic acid, with one of the alcohols $\text{C}$ or $\text{D}$, in the presence of a catalyst.

The ${ }^1 \text{H NMR}$ spectrum for compound $\text{E}$ contains the following features.

Draw the structure of compounds $\text{A}$, $\text{B}$ and $\text{E}$. Explain your answer with reference to the information provided. (7 marks)

--- 20 WORK AREA LINES (style=lined) ---

Show Answers Only

The mass spectrum has two peaks to the far right of the spectrum of similar height at 122 m/z and 124 m/z. This is due to the halogen having two isotopes with the same relative abundance.
Isotope $\ce{X}$ must be bromine, whose isotopes are $\ce{Br-79}$ and $\ce{Br-81}$. The two molecular ion peaks both correspond correctly to the molar mass of the parent molecule, $\ce{C3H7Br}$, depending on which isomer is present in the compound.
$MM\ce{(C3H7Br-79)} = 3 \times 12 + 1 \times 7 + 79 = 122$
$MM\ce{(C3H7Br-81)} = 3 \times 12 + 1 \times 7 + 81 = 124$
The two isomers of $\ce{C3H7Br}$ are 1-bromopropane and 2-bromopropane and when they undergo a substitution reaction with $\ce{OH^{-}}$, they will produce propan-1-ol and propan-2-ol respectively.
Only secondary alcohols will oxidise to produce a ketone. Hence compound $\text{D}$ must be propan-2-ol and compound $\text{C}$ must be propan-1-ol which can be oxidised to an aldehyde and the aldehyde can be oxidised to a carboxylic acid.
Therefore, compound $\text{B}$ is 2-bromopropane and compound $\text{A}$ is 1-bromopropane as during the substitution reaction the bromine atom is substituted with the hydroxide ion.
When 3-methylbutanoic acid is reacted with alcohol $\text{C}$ (propan-1-ol) the following reaction takes place:

When 3-methylbutanoic acid is reacted with alcohol $\text{D}$ (propan-2-ol) the following reaction takes place:

There are 6 unique hydrogen environments present in compound $\text{E}$. Therefore compound $\text{E}$ must be the ester produced in the reaction between 3-methylbutanoic acid and propan-1-ol (it has 6 hydrogen environments vs propan-2-ol ester which has 5).
Compound $\text{E}$ is propyl 3-methylbutanoate.
This can be confirmed by the integration (ratio of hydrogens in each environment) and peak splitting columns (number of splits = number of adjacent hydrogens + 1)
The shift at 0.96 is due to environment 1 which has six hydrogen atoms and has one neighbouring hydrogen atom (produces a doublet).
The shift at 2.1 is due to environment 2 which has one hydrogen 1 atom and has 8 neighbouring hydrogen atoms a (produces a mutliplet of 9)(.
The shift at 4.0 is due to environment 4. This $\ce{CH2}$ group is bonded to an oxygen atom corresponding to a large chemical shift between 3.2–5.0. It also has 2 neighbouring hydrogens and so produces a triplet.
Other answers could have included a further explanation regarding the integration and peak splitting of all the hydrogen environments and their relative chemical shifts.

Show Worked Solution

The mass spectrum has two peaks to the far right of the spectrum of similar height at 122 m/z and 124 m/z. This is due to the halogen having two isotopes with the same relative abundance.
Isotope $\ce{X}$ must be bromine, whose isotopes are $\ce{Br-79}$ and $\ce{Br-81}$. The two molecular ion peaks both correspond correctly to the molar mass of the parent molecule, $\ce{C3H7Br}$, depending on which isomer is present in the compound.
$MM\ce{(C3H7Br-79)} = 3 \times 12 + 1 \times 7 + 79 = 122$
$MM\ce{(C3H7Br-81)} = 3 \times 12 + 1 \times 7 + 81 = 124$
The two isomers of $\ce{C3H7Br}$ are 1-bromopropane and 2-bromopropane and when they undergo a substitution reaction with $\ce{OH^{-}}$, they will produce propan-1-ol and propan-2-ol respectively.
Only secondary alcohols will oxidise to produce a ketone. Hence compound $\text{D}$ must be propan-2-ol and compound $\text{C}$ must be propan-1-ol which can be oxidised to an aldehyde and the aldehyde can be oxidised to a carboxylic acid.
Therefore, compound $\text{B}$ is 2-bromopropane and compound $\text{A}$ is 1-bromopropane as during the substitution reaction the bromine atom is substituted with the hydroxide ion.
When 3-methylbutanoic acid is reacted with alcohol $\text{C}$ (propan-1-ol) the following reaction takes place:

When 3-methylbutanoic acid is reacted with alcohol $\text{D}$ (propan-2-ol) the following reaction takes place:

There are 6 unique hydrogen environments present in compound $\text{E}$. Therefore compound $\text{E}$ must be the ester produced in the reaction between 3-methylbutanoic acid and propan-1-ol (it has 6 hydrogen environments vs propan-2-ol ester which has 5).
Compound $\text{E}$ is propyl 3-methylbutanoate.
This can be confirmed by the integration (ratio of hydrogens in each environment) and peak splitting columns (number of splits = number of adjacent hydrogens + 1)
The shift at 0.96 is due to environment 1 which has six hydrogen atoms and has one neighbouring hydrogen atom (produces a doublet).
The shift at 2.1 is due to environment 2 which has one hydrogen 1 atom and has 8 neighbouring hydrogen atoms a (produces a mutliplet of 9)(.
The shift at 4.0 is due to environment 4. This $\ce{CH2}$ group is bonded to an oxygen atom corresponding to a large chemical shift between 3.2–5.0. It also has 2 neighbouring hydrogens and so produces a triplet.
Other answers could have included a further explanation regarding the integration and peak splitting of all the hydrogen environments and their relative chemical shifts.

♦ Mean mark 52%.

Networks, GEN2 2024 VCAA 15

An upgrade to the supermarket requires the completion of 11 activities, $A$ to $K$.

The directed network below shows these activities and their completion time, in weeks.

The minimum completion time for the project is 29 weeks.

Write down the critical path. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Which activity can be delayed for the longest time without affecting the minimum completion time of the project? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

Use the following information to answer parts c-e.

A change is made to the order of activities.

The table below shows the activities and their new latest starting times in weeks.

\begin{array}{|c|c|}
\hline
\textbf{Activity} & \textbf{Latest Starting}\\
&\textbf{time} \text{(weeks)}\\
\hline A & 0 \\
\hline B & 2 \\
\hline C & 10 \\
\hline D & 9 \\
\hline E & 13 \\
\hline F & 14 \\
\hline G & 18 \\
\hline H & 17 \\
\hline I & 19 \\
\hline J & 25 \\
\hline K & 22 \\
\hline
\end{array}

A dummy activity is now required in the network.

On the directed network below, draw a directed edge to represent the dummy activity. Include a label. (1 mark)
--- 1 WORK AREA LINES (style=lined) ---

What is the new minimum completion time of the project? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
The owners of the supermarket want the project completed earlier.
They will pay to reduce the time of some of the activities.
A reduction in completion time of an activity will incur an additional cost of $10 000 per week.
Activities can be reduced by a maximum of two weeks.
The minimum number of weeks an activity can be reduced to is seven weeks.
What is the minimum amount the owners of the supermarket will have to pay to reduce the completion time of the project as much as possible? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $A, C, H, J$

b. $\text{Activity E}$

d. $\text{30 weeks}$

e. $$50\,000$

Show Worked Solution

a. $\text{Critical path: }A, C, H, J$

b. $\text{Activity with the largest float time can be delayed the longest.}$

$\text{Consider Activity E:}$

$\text{EST = 11, LST}= 18-4=14\rightarrow\ \text{Float time = 3 weeks}$

$\therefore\ \text{Activity E can be delayed the longest.}$

♦♦ Mean mark (b) 34%.

♦♦♦ Mean mark (c) 10%.

d. $\text{New minimum completion time is 30 weeks.}$

♦♦♦ Mean mark (d) 27%.

e. $\text{Activities that can be reduced:}$

$-A\ \text{can be reduced by 2 weeks}$

$-B, D\ \text{can each be reduced by 1 week each}$

$-H\ \text{can be reduced by 1 week}$

$\text{Total reduction = 5 weeks}$

$\Rightarrow \ \text{Minimum payment}=$50\,000$

♦♦♦ Mean mark (e) 7%.

Networks, GEN2 2024 VCAA 14

A manufacturer $(M)$ makes deliveries to the supermarket $(S)$ via a number of storage warehouses, $L, N, O, P, Q$ and $R$. These eight locations are represented as vertices in the network below.

The numbers on the edges represent the maximum number of deliveries that can be made between these locations each day.

When considering the possible flow of deliveries through this network, many different cuts can be made.
Determine the capacity of Cut 1, shown above. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Determine the maximum number of deliveries that can be made each day from the manufacturer to the supermarket. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
The manufacturer wants to increase the number of deliveries to the supermarket.
This can be achieved by increasing the number of deliveries between one pair of locations.
Complete the following sentence by writing the locations on the lines provided:
To maximise this increase, the number of deliveries should be increased between
locations ____ and ____. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $46$

b. $37$

c. $\text{R and S}$

Show Worked Solution

a. $13+18+6+9=46$

$\text{(Reverse flow}\ Q → O\ \text{is not counted.)}$

$\text{Max deliveries (min cut)}\ =13+5+11+8=37$

♦ Mean mark (b) 29%.

c. $\text{The number of deliveries should be increased between}$

$\text{locations R and S.}$

♦ Mean mark (c) 22%.

CHEMISTRY, M6 2024 HSC 36

14.7 g of solid sodium hydrogen carbonate ($MM$ = 84.008 g mol$^{-1}$) was reacted with 120 mL of 1.50 mol L$^{-1}$ hydrochloric acid solution (density 1.02 g mL$^{-1}$) in a calorimeter. The temperature of the solution before and after reaction was recorded.

$
\begin{array}{|c|c|}
\hline \begin{array}{c}
\textit {Initial temperature of } \\
\textit {hydrochloric acid solution } \\
\left({ }^{\circ} C\right)
\end{array} & \begin{array}{c}
\textit {Final temperature of } \\
\textit {reaction solution } \\
\left({ }^{\circ} C\right)
\end{array} \\
\hline 21.5 & 11.5 \\
\hline
\end{array}
$

Assume that all $\ce{CO2}$ produced is lost from the reaction solution and that the specific heat capacity of the reaction solution is 3.80 J K$^{-1}$ g$^{-1}$.

What is the enthalpy of reaction, in kJ mol$^{-1}$? (5 marks)

--- 15 WORK AREA LINES (style=lined) ---

Show Answers Only

$+28.1\ \text{kJ mol}^{-1}$

Show Worked Solution

$\ce{NaHCO3(s) + HCl(aq) -> NaCl(aq) + H2O(l) + CO2(g)}$

$n(\ce{NaHCO3}) = \dfrac{14.7}{84.008} = 0.0175\ \text{mol}$

$n(\ce{HCl}) = 1.5 \times 0.120 = 0.180\ \text{mol}$

$\ce{HCl}$ and $\ce{NaHCO3}$ react in a $1:1$ ratio $\Rightarrow$ $\ce{NaHCO3}$ is the limiting reagent.

$n(\ce{CO2}) = 0.175\ \text{mol}$

Find mass of $\ce{CO2}$ lost to the surroundings

$n \times MM = 0.175 \times (12.01 +2(16.00)) = 7.70\ \text{g}$

Mass of the final solution $= 14.7 + (120 \times 1.02)-7.7 = 129.4\ \text{g}$

$\Delta H$	$=\dfrac{-q}{n}$
	$=\dfrac{-mc\Delta t}{n}$
	$=\dfrac{-129.4 \times 3.8 \times (11.5-21.5)}{0.175}$
	$=+2810\ \text{J mol}^{-1}$
	$=+28.1\ \text{kJ mol}^{-1}$

♦ Mean mark 55%.

CHEMISTRY, M7 2024 HSC 35

Unknown samples of three carboxylic acids, labelled $\text{X , Y}$ and $\text{Z}$, are analysed to determine their identities.

Both $\text{Y}$ and $\text{Z}$ react rapidly with bromine in the absence of UV light, but $\text{X}$ does not. A 0.100 g sample of $\text{Y}$ reacts with the same amount of bromine as a 0.200 g sample of $\text{Z}$.
Separate 0.100 g samples of $\text{X , Y}$ and $\text{Z}$ are titrated with 0.0617 mol L$^{-1}$ sodium hydroxide solution. The titre volumes are shown.

$
\begin{array}{|l|c|c|c|}
\hline \textit{Acid } & X & Y & Z \\
\hline \begin{array}{l}
\text {Volume of } \ce{NaOH \text{(mL)}} \\
\end{array} & 21.88 & 22.49 & 22.49 \\
\hline
\end{array}
$

Both $\text{Y}$ and $\text{Z}$ can undergo hydration reactions in the presence of a suitable catalyst. Two products are possible for the hydration of $\text{Y}$, but only one product is possible with $\text{Z}$.

Identify which structures 1, 2 and 3 in the table are acids $\text{X , Y}$ and $\text{Z}$. Justify your answer with reference to the information provided. (7 marks)

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Show Answers Only

Both sample $\text{Y}$ and $\text{Z}$ undergo an addition reaction with bromine and a hydration reaction. Therefore these samples must contain a $\ce{C=C}$ bond.
As sample $\text{X}$, undergoes neither of these reactions, it must have no $\ce{C=C}$ bond, thus sample $\text{X}$ is structure $2$.
Both structure $1$ and structure $3$ contain 1 $\ce{C=C}$ each $\Rightarrow$ they will react in a $1:1$ with $\ce{Br2}$. Hence the same number of moles of the carboxylic acid samples will react with the bromine.
Since the mass of sample $\text{Z}$ that reacts with the bromine is double the mass of sample $\text{Y}$, the molar mass of sample $\text{Z}$ must be double the molar mass of sample $\text{Y}$ following the formula $m = n \times MM$.
Therefore, sample $\text{Y}$ is structure 1 and sample $\text{Z}$ is structure 3.

Other information provided that could support identification includes:

The two products formed for the hydration of $\text{Y}$ is due to the asymmetry of structure 1 and the single product formed in the hydration of $\text{Z}$ is due to the symmetrical nature of structure 3.

The titration values are consistent with the proposed samples and their corresponding structures.

$\ce{n(NaOH)}$ reacted with $\text{X} = 0.0617 \times 0.02188 = 1.35 \times 10^{-3}\ \text{mol}$
$\text{n}_{\text{X}}= \dfrac{0.100}{74.078}= 0.00135\ \text{mol}$. Therefore $\text{X}$ reacts in a $1:1$ molar ratio as it is a monoprotic acid.
$\ce{n(NaOH)}$ reacted with $\text{Y}$ and $\text{Z} = 0.0617 \times 0.02249 = 1.39 \times 10^{-3}\ \text{mol}$
$\text{n}_{\text{Y}}= \dfrac{0.100}{72.062}= 0.00139\ \text{mol}$. Therefore $\text{Y}$ reacts in a $1:1$ molar ratio as it is a monoprotic acid.
$\text{n}_{\text{Z}}= \dfrac{0.100}{144.124}= 0.0006938\ \text{mol}$. Therefore $\text{Z}$ reacts in a $2:1$ molar ratio with $\ce{NaOH}$ as it is a diprotic acid.
The equal volumes of $\text{Y}$ and $\text{Z}$ used in the titration can be attributed to $\text{Z}$ having twice the molar mass of $\text{Y}$ and being a diprotic acid.

Show Worked Solution

Both sample $\text{Y}$ and $\text{Z}$ undergo an addition reaction with bromine and a hydration reaction. Therefore these samples must contain a $\ce{C=C}$ bond.
As sample $\text{X}$, undergoes neither of these reactions, it must have no $\ce{C=C}$ bond, thus sample $\text{X}$ is structure $2$.
Both structure $1$ and structure $3$ contain 1 $\ce{C=C}$ each $\Rightarrow$ they will react in a $1:1$ with $\ce{Br2}$. Hence the same number of moles of the carboxylic acid samples will react with the bromine.
Since the mass of sample $\text{Z}$ that reacts with the bromine is double the mass of sample $\text{Y}$, the molar mass of sample $\text{Z}$ must be double the molar mass of sample $\text{Y}$ following the formula $m = n \times MM$.
Therefore, sample $\text{Y}$ is structure 1 and sample $\text{Z}$ is structure 3.

Other information provided that could support identification includes:

The two products formed for the hydration of $\text{Y}$ is due to the asymmetry of structure 1 and the single product formed in the hydration of $\text{Z}$ is due to the symmetrical nature of structure 3.

The titration values are consistent with the proposed samples and their corresponding structures.

$\ce{n(NaOH)}$ reacted with $\text{X} = 0.0617 \times 0.02188 = 1.35 \times 10^{-3}\ \text{mol}$
$\text{n}_{\text{X}}= \dfrac{0.100}{74.078}= 0.00135\ \text{mol}$. Therefore $\text{X}$ reacts in a $1:1$ molar ratio as it is a monoprotic acid.
$\ce{n(NaOH)}$ reacted with $\text{Y}$ and $\text{Z} = 0.0617 \times 0.02249 = 1.39 \times 10^{-3}\ \text{mol}$
$\text{n}_{\text{Y}}= \dfrac{0.100}{72.062}= 0.00139\ \text{mol}$. Therefore $\text{Y}$ reacts in a $1:1$ molar ratio as it is a monoprotic acid.
$\text{n}_{\text{Z}}= \dfrac{0.100}{144.124}= 0.0006938\ \text{mol}$. Therefore $\text{Z}$ reacts in a $2:1$ molar ratio with $\ce{NaOH}$ as it is a diprotic acid.
The equal volumes of $\text{Y}$ and $\text{Z}$ used in the titration can be attributed to $\text{Z}$ having twice the molar mass of $\text{Y}$ and being a diprotic acid.

♦ Mean mark 55%.

CHEMISTRY, M8 2024 HSC 33

Acetone can be reduced, as shown.

Identify the shape around the central carbon atom in each molecule. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Explain how ${ }^{13} \text{C NMR}$ spectroscopy could be used to monitor the progress of this reaction. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

a. Acetone:

Double bond and 2 single bonds coming off the central carbon atom $\Rightarrow$ trigonal planar.

Product:

Contains single bonds coming off the central carbon atom (\Rightarrow\) tetrahedral. (Note: the hydrogen bonded to the central carbon atom in the product molecule is not shown due to the skeletal structure)

b. ${ }^{13} \text{C NMR}$ Spectroscopy:

${ }^{13} \text{C NMR}$ will differentiate between molecules with different carbon environments. This produces different signals on the ${ }^{13} \text{C NMR}$ spectrum.
The acetone would produce two signals on the ${ }^{13} \text{C NMR}$ spectrum. The first signal would be due to the $\ce{CH3}$ groups either side of the central carbon between 20-50 ppm. The second signal would be from the carbonyl group between 190-220 ppm.
The product of the reduction would also produce two signals on the ${ }^{13} \text{C NMR}$ spectrum. The carbon with the hydroxyl group attached to it would produce a signal between 50-90 ppm and the $\ce{CH3}$ groups either side would produce a signal between 5-40 ppm.
The reaction can be monitored by observing the disappearance of the carbonyl signal (190-220 ppm) and appearance of the hydroxyl signal (50-90 ppm) as acetone is reduced to the product.

Show Worked Solution

a. Acetone:

Double bond and 2 single bonds coming off the central carbon atom $\Rightarrow$ trigonal planar.

Product:

Contains single bonds coming off the central carbon atom (\Rightarrow\) tetrahedral. (Note: the hydrogen bonded to the central carbon atom in the product molecule is not shown due to the skeletal structure)

♦ Mean mark (a) 44%.

b. ${ }^{13} \text{C NMR}$ Spectroscopy:

${ }^{13} \text{C NMR}$ will differentiate between molecules with different carbon environments. This produces different signals on the ${ }^{13} \text{C NMR}$ spectrum.
The acetone would produce two signals on the ${ }^{13} \text{C NMR}$ spectrum. The first signal would be due to the $\ce{CH3}$ groups either side of the central carbon between 20-50 ppm. The second signal would be from the carbonyl group between 190-220 ppm.
The product of the reduction would also produce two signals on the ${ }^{13} \text{C NMR}$ spectrum. The carbon with the hydroxyl group attached to it would produce a signal between 50-90 ppm and the $\ce{CH3}$ groups either side would produce a signal between 5-40 ppm.
The reaction can be monitored by observing the disappearance of the carbonyl signal (190-220 ppm) and appearance of the hydroxyl signal (50-90 ppm) as acetone is reduced to the product.

CHEMISTRY, M5 2024 HSC 32

Calculate the concentration of cadmium ions in a saturated solution of cadmium($\text{II}$) phosphate, $\ce{Cd3\left(PO4\right)2}, \ K_{sp}=2.53 \times 10^{-33}$. (4 marks)

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$3.56 \times 10^{-7}\ \text{mol L}^{-1}$

Show Worked Solution

$\ce{Cd3(PO4)2(s) \rightleftharpoons 3Cd^{2+}(aq) + 2PO4^{3-}(aq)}$

$\Rightarrow K_{sp} = \ce{[Cd^{2+}]^3[PO4^{3-}]^2} = 2.53 \times 10^{-33}$

$\text{Let the molar solubility of }\ce{Cd3(PO4)2} \text{ be } x\ \text{mol L}^{-1}$
$\ce{[Cd^{2+}]} = 3x,\ \ \ce{[PO4^{3-}]} = 2x$

$\ce{[Cd^{2+}]^3[PO4^{3-}]^2}$	$=2.53 \times 10^{-33}$
$(3x)^3 \times (2x)^2$	$=2.53 \times 10^{-33}$
$108x^5$	$=2.53 \times 10^{-33}$
$x$	$=\sqrt[5]{\dfrac{2.53 \times 10^{-33}}{108}}$
	$=1.11856 \times 10^{-7}$

$\ce{[Cd^{2+}]} = 3 \times 1.1186 \times 10^{-7} = 3.56 \times 10^{-7}\ \text{mol L}^{-1}\ \ \text{(3 sig.fig)}$

♦ Mean mark 55%.

CHEMISTRY, M5 2024 HSC 30

An equilibrium mixture of hydrogen, carbon dioxide, water and carbon monoxide is in a closed, 1 L container at a fixed temperature as shown:

$\ce{H2(g) +CO_2(g) \rightleftharpoons H2O(g) +CO(g)} \quad \quad K_{eq}=1.600$

The initial concentrations are $\left[\ce{H2}\right]=1.000 \text{ mol L}^{-1}, \left[ \ce{CO2}\right]=0.500\ \text{mol L}^{-1},\ \left[\ce{H2O}\right]=0.400 \text{ mol L}^{-1}$ and $[ \ce{CO} ]=2.000 \text{ mol L} ^{-1}$.

An unknown amount of $\ce{CO(g)}$ was added to the same container, and the temperature was kept constant. After the new equilibrium had been established, the concentration of $\ce{H2O(g)}$ was found to be 0.200 mol L$^{-1}$.

Using this information, calculate the unknown amount (in mol) of $\ce{CO(g)}$ that was added to the container. (4 marks)

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$4.92\ \text{mol}$

Show Worked Solution

$\ce{n_{intial}(CO(g))} = 0.400 \text{ and } \ce{n_{final}(CO(g))} = 0.200$.
$\text{Change in the number of moles in } \ce{CO(g)} = 0.400-0.200 = 0.200\ \text{mol in } 1\ \text{L}$

\begin{array} {|c|c|c|c|c|}
\hline & \ce{H2(g)} & \ce{CO2(g)} & \ce{H2O(g)} & \ce{CO2(g)} \\
\hline \text{Initial} & 1 & 0.5 & 0.4 & 2 + x \\
\hline \text{Change} & +0.2 & +0.2 & -0.2 & -0.2 \\
\hline \text{Equilibrium} & 1.2 & 0.7 & 0.2 & 1.8 + x \\
\hline \end{array}

Since all substances are present in a 1 L container, the concentrations of each substance is equal to the number of moles of that substance present at equilibrium

$K_{eq}$	$=\dfrac{\ce{[H2O(g)][CO(g)]}}{\ce{[H2(g)][CO2(g)]}}$
$1.600$	$=\dfrac{0.2 \times (1.8+x)}{1.2 \times 0.7}$
$1.8 + x$	$=1.6 \times \dfrac{1.2 \times 0.7}{0.2}$
$x$	$=6.72-1.8$
	$=4.92\ \text{mol}$

4.92 mol of $\ce{CO(g)}$ were added to the container.

♦ Mean mark 55%.

CHEMISTRY, M5 2024 HSC 26

The equilibrium equation for the reaction of iodine with hydrogen cyanide in aqueous solution is given.

$\ce{I_2(aq) + HCN(aq)\rightleftharpoons ICN(aq) + I^{-}(aq) + H^{+}(aq)}$

At $t=0$ min, $\ce{I2}$ was added to a mixture of $\ce{HCN, I^{-}}$ and $\ce{H^{+}}$, bringing $\left[ \ce{I2}\right]$ to 2.0 × 10$^{-5}$ mol L$^{-1}$. After 3 minutes, the system was at equilibrium, and an analysis of the mixture found that half of the $\ce{I2}$ had reacted.

On the axes provided, sketch a graph to show how $\left[\ce{I_2}\right]$ changes in the solution between $t=0$ min and $t=6$ min. (2 marks)

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Using collision theory, explain the rate of reaction between $t=0$ min and $t=6$ min. Refer to the $\left[ \ce{I2}\right]$ in your answer. (3 marks)

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b. Initially, the high $\ce{[I2]}$ results in a large number of collisions between reactants.

This results in an initially high reaction rate for the forward reaction.
Between $t=0$ and $t=3$, the concentration of $\ce{I2}$ decreases. Less collisions between reactant particles occur leading to a decrease in the rate of the forward reaction.
Between $t=0$ and $t=3$, as the concentration of reactants decrease, the concentration of the products increase. This leads to an increase in the number of successful collisions between product particles and a subsequent increase in the rate of the reverse reaction.
Between $t=3$ and $t=6$, the concentration of $\ce{I2}$ remains constant. This is due to the system reaching a state of dynamic equilibrium so the frequency of successful collisions between the reactants is equal to the frequency of successful collisions between the products. i.e. the rates of the forward and reverse reactions are equal.

Show Worked Solution

b. Initially, the high $\ce{[I2]}$ results in a large number of collisions between reactants.

This results in an initially high reaction rate for the forward reaction.
Between $t=0$ and $t=3$, the concentration of $\ce{I2}$ decreases. Less collisions between reactant particles occur leading to a decrease in the rate of the forward reaction.
Between $t=0$ and $t=3$, as the concentration of reactants decrease, the concentration of the products increase. This leads to an increase in the number of successful collisions between product particles and a subsequent increase in the rate of the reverse reaction.
Between $t=3$ and $t=6$, the concentration of $\ce{I2}$ remains constant. This is due to the system reaching a state of dynamic equilibrium so the frequency of successful collisions between the reactants is equal to the frequency of successful collisions between the products. i.e. the rates of the forward and reverse reactions are equal.

Mean mark (b) 56%.

CHEMISTRY, M6 2024 HSC 20 MC

The concentration of ascorbic acid $\left(M M=176.124\ \text{g mol}^{-1}\right)$ in solution $\text{A}$ was determined by titration.

- A 25.00 mL sample of solution $\text{A}$ was titrated with potassium hydroxide solution.
- 50.00 mg of ascorbic acid was added to a second 25.00 mL sample of solution $\text{A}$, which was titrated in the same way.

Titration volumes for both titrations are given.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Solution} \rule[-1ex]{0pt}{0pt} & \textit{Titre}\text{ (mL)} \\
\hline
\rule{0pt}{2.5ex} \text{25.00 mL solution A} \rule[-1ex]{0pt}{0pt} & 17.50 \\
\hline
\rule{0pt}{2.5ex} \text{25.00 mL solution A+}& 33.10 \\
\text{50.00 mg of ascorbic acid}& \text{} \\
\hline
\end{array}

What is the concentration of ascorbic acid in solution $\text{A}$?

$5.352 \times 10^{-3} \text{ mol L}^{-1}$
$6.004 \times 10^{-3} \text{ mol L}^{-1}$
$1.012 \times 10^{-2} \text{ mol L}^{-1}$
$1.274 \times 10^{-2} \text{ mol L}^{-1}$

Show Answers Only

$D$

Show Worked Solution

$\ce{n(ascorbic acid)} = \dfrac{0.05}{176.124} = 0.00028389\ \text{mol}$
Volume of $\ce{KOH}$ to neutralise the two samples = 15.6 mL.
15.6 mL was required to neutralise the extra 0.00028389 mol of ascorbic acid.
Concentration of potassium hydroxide $=\dfrac{\text{n}}{\text{V}}= \dfrac{0.00028389}{0.0156} = 0.018198\ \text{mol L}^{-1}$
$\ce{n(KOH)}$ used in the first titration $\ce{= c \times V} = 0.018198 \times 0.0175 = 0.0003185\ \text{mol} = \ce{n(ascorbic acid)}$
$\ce{[ascrobic acid]} = \dfrac{0.0003185}{0.0250} = 0.01274 = 1.274 \times 10^{-2}\ \text{mol L}^{-1}$

$\Rightarrow D$

♦ Mean mark 43%.

CHEMISTRY, M8 2024 HSC 19 MC

Which of the following compounds produces TWO doublets in the ${ }^1 \text{H NMR}$ spectrum?

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$A$

Show Worked Solution

For two doublets to occur, the substance must have two chemically different hydrogen environments that both have one neighbouring hydrogen on an adjacent carbon atom.

This is displayed by option $\text{A}$ where the hydrogen atoms in environments 1 and 3 both have only one neighbouring hydrogen atom each.

$\Rightarrow A$

♦ Mean mark 48%.

CHEMISTRY, M5 2024 HSC 18 MC

A reaction mixture, not at equilibrium, is composed of both $\ce{N_2O_4(g)}$ and $\ce{NO_2(g)}$ in a closed container. The reaction quotient for the system, $Q$, is given.

$Q=\dfrac{\left[\ce{NO_2}\right]^2}{\left[\ce{N_2O_4}\right]}$

The rate of the forward reaction is initially greater than the rate of the reverse reaction.

Which diagram shows how $Q$ changes over time for this mixture?

Show Answers Only

$B$

Show Worked Solution

When calculating the reaction quotient of a chemical reaction, the formula is $\dfrac{products}{reactants}$.
The equation for the reaction taking place is:
$\ce{N2O4(g) \rightleftharpoons 2NO2(g)}$
As the rate of the forward reaction is greater than the rate of the reverse reaction, $\ce{[NO2]^2}$ will increase and $\ce{[N2O4]}$ will decrease.
Hence the value for the reaction quotient, $\dfrac{\left[\ce{NO_2}\right]^2}{\left[\ce{N_2O_4}\right]}$, will increase.
As it states both $\ce{N2O4(g)}$ and $\ce{NO2(g)}$ are present in the intial system, the value for $Q$ will not be zero.

$\Rightarrow B$

♦ Mean mark 43%.

CHEMISTRY, M6 2024 HSC 17 MC

20 mL of a 0.1 mol L$^{-1}$ solution of an acid is titrated against a 0.1 mol L$^{-1}$ solution of sodium hydroxide. A graph of pH against the volume of sodium hydroxide for this experiment is shown.

Which of the following acids was used in the titration?

\begin{align*}
\begin{array}{l}
\ & \\
\textbf{A.}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|c|c|c|}
\hline
\quad \textit{Acid}\quad & \quad pK_{a1}\quad & \quad pK_{a2}\quad \\
\hline
1& 4.76 & – \\
\hline
2 & \text{Strong} & – \\
\hline
3 & 1.91 & 6.30 \\
\hline
4 & 4.11 & 9.61 \\
\hline
\end{array}
\end{align*}

Show Answers Only

$C$

Show Worked Solution

There are two equivalence points on the graph, so the acid is diprotic and will have two $pK_a$ values.
The reaction at the first equivalence point is:
$\ce{H2A -> HA^- + H^+}$
The reaction at the second equivalence point is:
$\ce{HA^- -> A^{2-} + H^+}$
The $pK_a$ values for each equivalence point will be equal to the pH values at the half-equivalence points.
The first half-equivalence point occurs at 10 mL, where the corresponding pH is 1.91 $\Rightarrow pK_{a1}$ = 1.91
The second half-equivalence point will occur halfway between the first two equivalence points, at 30 mL
The corresponding pH is 6.30 $\Rightarrow pK_{a2}$ = 6.30

$\Rightarrow C$

♦ Mean mark 41%.
COMMENT: Student’s did well understanding the relationship between the $pK_a$ values and half-equivalence points.

CHEMISTRY, M5 2024 HSC 15 MC

The thermal decomposition of lithium peroxide $\ce{(Li_2O_2)}$ is given by the equation shown.

$\ce{2Li_2O_2(s)\rightleftharpoons 2 Li_2 O(s) + O_2(g)}$

Mixtures of $\ce{Li_2O_2}$, $\ce{Li_2O}$ and $\ce{O_2}$ were allowed to reach equilibrium in two identical, closed containers, $\text{P}$ and $\text{Q}$, at the same temperature. The amount of $\ce{Li_2O_2(s)}$ in container $\text{P}$ is double that in container $\text{Q}$ . The amount of $\ce{Li_2O(s)}$ is the same in each container.

What is the ratio of $\left[ \ce{O_2(g)}\right]$ in container $\text{P}$ to $\left[\ce{O_2(g)}\right]$ in container $\text{Q}$?

$1: 1$
$2: 1$
$3: 2$
$5: 4$

Show Answers Only

$A$

Show Worked Solution

When calculating the $K_{eq}$ of a system, substances in solid states are all given a value of $1$.
The equilibrium constant of the above reaction is $K_{eq} = \ce{[O_2(g)]}$.
As both mixtures reached equilibrium, the $K_{eq}$ values for each mixture is the same, hence the ratio of $\ce{[O2(g)]}$ in each container is $1:1$.

$\Rightarrow A$

♦♦ Mean mark 39%.

CHEMISTRY, M5 2024 HSC 11 MC

Which is the correct expression for calculating the solubility (in mol L$^{-1}$) of lead$\text{(II}$) iodide in a 0.1 mol L$^{-1}$ solution of $\ce{NaI}$ at 25°C?

$\dfrac{9.8 \times 10^{-9}}{2 \times 0.1}$
$\dfrac{9.8 \times 10^{-9}}{(2 \times 0.1)^2}$
$\dfrac{9.8 \times 10^{-9}}{0.1}$
$\dfrac{9.8 \times 10^{-9}}{(0.1)^2}$

Show Answers Only

$D$

Show Worked Solution

$\ce{PbI2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)}$

Let the solubility of $\ce{PbI2} = x\ \text{mol L}^{-1}$

\begin{array} {|c|c|c|c|}
\hline \text{Concentration (mol/L)} & \ce{PbI2} & \ce{Pb^{2+}} & \ce{2I^-} \\
\hline \text{Initial} & – & 0 & 0.1 \\
\hline \text{Change} & – & +x & +2x \\
\hline \text{Equilibrium} & – & x & 0.1 +2x \\
\hline \end{array}

As $x$ is very small, assume $0.1 +2x \approx 0.1$

$K_{sp} \ce{=[Pb^{2+}][I^-]^2} =9.8 \times 10^{-9}$

$K_{sp} = 0.1^2 \times x =9.8 \times 10^{-9}$

$\therefore x=\dfrac{9.8 \times 10^{-9}}{0.1^2}$

$\Rightarrow D$

♦ Mean mark 55%.

Matrices, GEN2 2024 VCAA 12

When the construction company established the construction site at the beginning of 2023, it employed 390 staff to work on the site.

The staff comprised 330 construction workers $(C)$, 50 foremen $(F)$ and 10 managers $(M)$.

At the beginning of each year, staff can choose to stay in the same job, move to a different job on the site, or leave the site $(L)$ and not return.

The transition diagram below shows the proportion of staff who are expected to change their job at the site each year.

This situation can be modelled by the recurrence relation

$S_{n+1}=T S_n$, where

$T$ is the transitional matrix, $S_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L \end{aligned}$ and $n$ is the number of years after 2023.

Calculate the predicted percentage decrease in the number of foremen $(F)$ on the site from 2023 to 2025. (1 mark)

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Determine the total number of staff on the site in the long term. (1 mark)

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To encourage more construction workers $(C)$ to stay, the construction company has given workers an incentive to move into the job of foreman $(F)$.

Matrix $R$ below shows the ways in which staff are expected to change their jobs from year to year with this new incentive in place.

\begin{aligned}
& \quad \quad \ \ \textit{this year} \\
& \quad C \quad \ \ F \quad \ \ M \quad L\\
R = & \begin{bmatrix}
0.4 & 0.2 & 0 & 0 \\
0.4 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.2 & 0.4 & 0.3 & 1
\end{bmatrix}\begin{array}{l}
C\\
F\\
M\\
L
\end{array} \quad \textit{next year}
\end{aligned}

The site always requires at least 330 construction workers.

To ensure that this happens, the company hires an additional 190 construction workers $(C)$ at the beginning of 2024 and each year thereafter.

The matrix $V_{n+1}$ will then be given by

$V_{n+1}=R V_n+Z$, where

$V_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \quad\quad\quad Z=\left[\begin{array}{c}190 \\ 0 \\ 0 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \ \ $ and $n$ is the number of years after 2023.

How many more staff are there on the site in 2024 than there were in 2023 ? (1 mark)

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Based on this new model, the company has realised that in the long term there will be more than 200 foremen on site.
In which year will the number of foremen first be above 200? (1 mark)

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a. $14\%$

b. $\text{Zero}$

c. $101$

d. $2027$

Show Worked Solution

a. $\text{Using CAS:}$

$\text{Transition matrix}(T):\ \begin{aligned}
& \quad \quad \quad \ \ \ \ \textit{from} \\
& \quad \ \ \ \ C \ \ \ \ \ F \ \ \ \ \ \ M \ \ \ \ \ \ L \ \ \\
\ \ \textit{to}\ \ \ & \begin{array}{l}
C \\
F \\
M \\
L
\end{array}\begin{bmatrix}
0.3 & 0.2 & 0 & 0 \\
0.2 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.5 & 0.4 & 0.3 & 1
\end{bmatrix}
\end{aligned}$

♦♦♦ Mean mark (a) 25%.

$S_{2023}=\begin{bmatrix}
330 \\
50 \\
10 \\
0 \\
\end{bmatrix}\ \ ,\ \ S_{2024}=T\times S_{2023}=\begin{bmatrix}
109 \\
80 \\
13 \\
188 \\
\end{bmatrix}$

$ S_{2025}=T\times S_{2024}=\begin{bmatrix}
48.7 \\
43 \\
19.9 \\
278.4 \\
\end{bmatrix}$

$\text{% decrease in foremen}\ (F)=\dfrac{50-43}{50}\times 100\%=14\%$

b. $\text{Test for 10 years:}$

$\text{Using CAS:}$

$S_{2024}=T^{10}\times S_{2023}=\begin{bmatrix}
0.490748 \\
0.734232 \\
0.488858 \\
388.286 \\
\end{bmatrix}$

$\text{In the long term, there will be zero employees on site.}$

♦♦♦ Mean mark (b) 24%.

c. $\text{Using CAS:}$

$V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}$

$\text{Difference in staff from}\ 2023-2024$

$=(332+146+13)-390$

$=101\ \text{more staff.}$

$\text{NOTE: 89 not included in calculation as these are the staff}$

$\text{who have left the company during the year.}$

♦♦♦ Mean mark (c) 12%.

d. $\text{Using CAS:}$

$V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}\ \ ,\ \ V_{2025}=R\times V_{2024}+Z=\begin{bmatrix}
352 \\
167.2 \\
33.1 \\
217.7 \\
\end{bmatrix}\ \\$

♦♦♦ Mean mark (d) 26%.

$V_{2026}=R\times V_{2025}+Z=\begin{bmatrix}
364.24\\
187.48 \\
43.37 \\
364.91 \\
\end{bmatrix}\ \ ,\ \ V_{2027}=R\times V_{2026}+Z=\begin{bmatrix}
373.192 \\
200.54 \\
50.507 \\
525.761 \\
\end{bmatrix}\ \\$

$\therefore\ \text{In 2027 the number of foremen will be over 200.}$

Matrices, GEN2 2024 VCAA 11

A population of a native animal species lives near the construction site.

To ensure that the species is protected, information about the initial female population was collected at the beginning of 2023. The birth rates and the survival rates of the females in this population were also recorded.

This species has a life span of 4 years and the information collected has been categorised into four age groups: 0-1 year, 1-2 years, 2-3 years, and 3-4 years.

This information is displayed in the initial population matrix, $R_0$, and the Leslie matrix, $L$, below.

$R_0=\left[\begin{array}{c}70 \\ 80 \\ 90 \\ 40\end{array}\right] \quad \quad L=\left[\begin{array}{cccc}0.4 & 0.75 & 0.4 & 0 \\ 0.4 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.5 & 0\end{array}\right]$

Using the information above
i. complete the following transition diagram. (1 mark)
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ii. complete the following table, showing the initial female population, and the predicted female population after one year, for each of the age groups. (1 mark)

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It is predicted that if this species is not protected, the female population of each of the four age groups will rapidly decrease within the next 10 years.
After how many years is it predicted that the total female population of this species will first be half the initial female population? (1 mark)

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a.i.

a.ii.

	$\textbf{Age Group}$
	$\ 0-1\ \text{year}\ $	$\ 1-2\ \text{years}\ $	$\ 2-3\ \text{years}\ $	$\ 3-4\ \text{years}\ $
$\ \text{Initial population}\ $	$70$	$80$	$90$	$40$
$\ \text{Population after}\ $ $\ \text{one year}$	$124$	$28$	$56$	$45$

b. $\text{5 years}$

Show Worked Solution

a.i.

♦♦♦ Mean mark (a) 24%.

a.ii. $\text{Population after 1 yr calculations}$

$0-1\ \text{year}\ =0.4\times 70+0.75\times 80+0.4\times 90=124$

$1-2\ \text{years}\ =0.4\times 70=28$

$2-3\ \text{years}\ =0.7\times 80=56$

$3-4\ \text{years}\ =0.5\times 90=45$

	$\textbf{Age Group}$
	$\ 0-1\ \text{year}\ $	$\ 1-2\ \text{years}\ $	$\ 2-3\ \text{years}\ $	$\ 3-4\ \text{years}\ $
$\ \text{Initial population}\ $	$70$	$80$	$90$	$40$
$\ \text{Population after}\ $ $\ \text{one year}$	$124$	$28$	$56$	$45$

b. $\text{Using CAS:}$

$R_1=L\times R_0=\begin{bmatrix}
124 \\
28 \\
56 \\
45 \end{bmatrix}\ \ \text{Total = 253}\ ,\ \ R_2=L\times R_1=\begin{bmatrix}
93 \\
49.6 \\
19.6 \\
28 \end{bmatrix}\ \ \text{Total = 190.2}$

$R_3=L\times R_2=\begin{bmatrix}
82.24 \\
37.2 \\
34.72 \\
9.8 \end{bmatrix}\ \ \text{Total = 163.96}\ ,\ \ R_4=L\times R_3=\begin{bmatrix}
74.684 \\
32.896 \\
26.04 \\
17.36 \end{bmatrix}\ \ \text{Total =150.98}$

$R_5=L\times R_4=\begin{bmatrix}
64.9616 \\
29.8736 \\
23.0272 \\
13.02 \end{bmatrix}\ \ \text{Total = 130.8824}$

$\therefore\ \text{Total female population less than 140 after 5 years}$

♦ Mean mark (b) 39%.

Matrices, GEN2 2024 VCAA 10

To access the southern end of the construction site, Vince must enter a security code consisting of five numbers.

The security code is represented by the row matrix $W$.

The element in row $i$ and column $j$ of $W$ is $w_{i j}$.

The elements of $W$ are determined by the rule $(i-j)^2+2 j$.

Complete the following matrix showing the five numbers in the security code. (1 mark)

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To access the northern end of the construction site, Vince enters a different security code, consisting of eight numbers.

This security code is represented by the row matrix $X$.

The element in row $i$ and column $j$ of $X$ is $x_{i j}$.

The elements of $X$ are also determined by the rule $(i-j)^2+2 j$.

What is the last number in this security code to access the northern end of the construction site? (1 mark)

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a. $W=\ [2\quad 5\quad 10\quad 17\quad 26\ ]$

b. $x_{18}=65$

Show Worked Solution

a.	$W$	$=\ [(0)^2+2\ \ \ (-1)^2+4\ \ \ (-2)^2+36\ \ \ (-3)^2+8\ \ \ (-4)^2+10\ ]$
		$=\ [\ 2\quad 5\quad 10\quad 17\quad 26\ ]$

b. $x_{18}=(1-8)^2+2\times 8=65$

Mean mark (a) 54%.
♦ Mean mark (b) 50%

Financial Maths, GEN2 2024 VCAA 8

Emi takes out a reducing balance loan of $500 000.

The interest rate is 5.3% per annum, compounding monthly.

Emi makes regular monthly repayments of $3071.63 for the duration of the loan, with only the final repayment amount being slightly different from all the other repayments.

Determine the total cost of Emi's loan, rounding your answer to the nearest cent, and state the number of payments required to fully repay the loan. (2 marks)

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$\text{Total cost}\ =$884\,633.62\ ,\ 288\ \text{payments}$

Show Worked Solution

$\text{Using CAS: Find}\ N\ \text{when}\ FV=0$

♦♦ Mean mark 29%.

$\text{Using CAS: Find amount still owing if 288 payments are made.}$

$\text{Final payment}\ =$3071.63+$4.18 =$3075.81$

$\text{Total cost of loan}\ =3071.63\times 287 + 3075.81=$884\,633.62$

$\text{Total number of payments}\ =288$

Financial Maths, GEN2 2024 VCAA 7

Emi decides to invest a $300 000 inheritance into an annuity.

Let $E_n$ be the balance of Emi's annuity after $n$ months.

A recurrence relation that can model the value of this balance from month to month is

$E_0=300\,000, \quad E_{n+1}=1.003 E_n-2159.41$

Showing recursive calculations, determine the balance of the annuity after two months. Round your answer to the nearest cent. (1 mark)

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For how many years will Emi receive the regular payment? (1 mark)

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Calculate the annual compound interest rate for this annuity. (1 mark)

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If Emi wanted the annuity to act as a perpetuity, what monthly payment, in dollars, would she receive? (1 mark)

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Show Answers Only

a. $$297\,477.40$

b. $15\ \text{years}$

c. $3.6\%$

d. $$900$

Show Worked Solution

a. $E_0=300\,000$

$E_1=1.003\times 300\,000-2159.41=$298\,740.59$

$E_2=1.003\times 298\,740.59-2159.41=$297\,477.4018\approx $297\,477.40$

♦ Mean mark (a) 49%.

b. $\text{Using CAS:}$

$\text{Number of years}\ =\dfrac{180}{12}=15\ \text{years}$

♦ Mean mark (b) 42%.

c. $\text{Annual interest rate}\ =(1.003-1)\times 12\times 100\% = 3.6\%$

d. $\text{Method 1: Using CAS}$

$\text{Monthly payment}\ =$900$

$\text{Method 2:}$

$\text{Amount must be equal to the amount of monthly interest earned.}$

$\therefore\ \text{Monthly payment}\ =300\,000\times 0.003=$900$

♦ Mean mark (d) 47%.

Vectors, EXT2 V1 2024 HSC 15a

Consider the three vectors $\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}$ and $\underset{\sim}{c}=\overrightarrow{O C}$, where $O$ is the origin and the points $A, B$ and $C$ are all different from each other and the origin.

The point $M$ is the point such that $\dfrac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})=\overrightarrow{O M}$.

Show that $M$ lies on the line passing through $A$ and $B$. (1 mark)

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The point $G$ is the point such that $\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})=\overrightarrow{O G}$.
Show that $G$ lies on the line passing through $M$ and $C$, and lies between $M$ and $C$. (2 marks)

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The complex numbers $x, w$ and $z$ are all different and all have modulus 1.
Using part (ii), or otherwise, show that $\dfrac{1}{3}(x+w+z)$ is never a cube root of $x w z$. (2 marks)

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Show Answers Only

i. $\text{Equation of line through \(A$ and $B$}\)

$\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}$

	$\overrightarrow{O M}$	$=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}$
		$=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}$
		$=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})$
		$=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}$

$\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1$.

ii. $\text{Equation of line through \(M$ and $C$}\)

$\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}$

	$\overrightarrow{O G}$	$=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})$
		$=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}$
		$=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)$
		$=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}$

$\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2$

$\ \ \overrightarrow{O G} \neq \overrightarrow{O C} \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}$

$\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M$.

iii. $\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}$

$\abs{w}=\abs{x}=\abs{z}=1$

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

$G \equiv \dfrac{1}{3}(x+w+z)$

$G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}$

$\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1$

$\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1$

$\Rightarrow \ \text{All cube roots have modulus = 1.}$

$\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of \(xwz$.}\)

Show Worked Solution

i. $\text{Equation of line through \(A$ and $B$}\)

$\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}$

	$\overrightarrow{O M}$	$=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}$
		$=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}$
		$=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})$
		$=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}$

$\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1$.

ii. $\text{Equation of line through \(M$ and $C$}\)

$\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}$

	$\overrightarrow{O G}$	$=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})$
		$=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}$
		$=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)$
		$=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}$

$\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2$

$\ \ \overrightarrow{O G} \neq \overrightarrow{O C} \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}$

$\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M$.

♦ Mean mark (ii) 43%.

iii. $\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}$

♦♦♦ Mean mark (iii) 10%.

$\abs{w}=\abs{x}=\abs{z}=1$

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

$G \equiv \dfrac{1}{3}(x+w+z)$

$G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}$

$\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1$

$\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1$

$\Rightarrow \ \text{All cube roots have modulus = 1.}$

$\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of \(xwz$.}\)

Financial Maths, GEN2 2024 VCAA 6

Emi invested profits of $10 000 into a savings account that earns interest compounding fortnightly, for one year.

The effective interest rate, rounded to two decimal places, is 5.07%.

Assume that there are exactly 26 fortnights in a year.

What is the nominal percentage rate of interest for the account?
Round your answer to two decimal places. (1 mark)

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Explain why the nominal interest rate appears lower than the effective interest rate. (1 mark)

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Show Answers Only

a. $4.95\%\ \text{(2 d.p.)}$

b. $\text{Nominal and effective interest rates are equal if there is only one}$

$\text{compounding period per year. If there are more compounding}$

$\text{periods (as in this example) the effective rate will be higher than}$

$\text{the nominal rate.}$

Show Worked Solution

a. $\text{Using CAS: FINANCE}\ \rightarrow \ \text{Interest Conversion}\ \rightarrow \ \text{Nominal Interest Rate}$

$\text{nom}(5.07, 26) = 4.95037\%\approx 4.95\%\ \text{(2 d.p.)}$

$\text{Using formula:}$

Mean mark (a) 51%.

$r_e$	$=\left[\left(1+\dfrac{r}{100n}\right)^n-1\right]\times 100\%$
$\dfrac{5.07}{100}$	$=\left[\left(1+\dfrac{r}{2600}\right)^{26}-1\right]$
$\left(1+\dfrac{r}{2600}\right)^{26}$	$=1.0507$
$\dfrac{r}{2600}$	$=(1.0507)^\frac{1}{26}-1$
$r$	$=\left((1.0507)^\frac{1}{26}-1\right)\times 2600$
	$=4.95036\dots\approx 4.95\%\ \text{(2 d.p.)}$

b. $\text{Nominal and effective interest rates are equal if there is only one}$

$\text{compounding period per year. If there are more compounding}$

$\text{periods (as in this example) the effective rate will be higher than}$

$\text{the nominal rate.}$

♦♦♦ Mean mark (b) 26%.

Financial Maths, GEN2 2024 VCAA 5

Emi operates a mobile dog-grooming business.

The value of her grooming equipment will depreciate.

Based on average usage, a rule for the value, in dollars, of the equipment, $V_n$, after $n$ weeks is

$V_n=15000-60 n$

Assume that there are exactly 52 weeks in a year.

By what amount, in dollars, does the value of the grooming equipment depreciate each week? (1 mark)

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Emi plans to replace the grooming equipment after four years.
What will be its value, in dollars, at this time? (1 mark)

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$V_n$ is the value of the grooming equipment, in dollars, after $n$ weeks.
Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$ that can model this value from one week to the next. (1 mark)

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The value of the grooming equipment decreases from one year to the next by the same percentage of the original $15 000 value.
What is this annual flat rate percentage? (1 mark)

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a. $$60$

b. $$2520$

c. $V_0=15\,000 , \ \ V_{n+1}=V_n-60$

d. $20.8\%$

Show Worked Solution

a. $$60$

b. $n=4\times 52=208$

$V_{208}$	$=15\,000-60\times208$
	$=$2520$

c. $V_0=15\,000 , \ \ V_{n+1}=V_n-60$

d. $\text{Flat rate}\ =\dfrac{60}{15\,000}\times 52\times 100\%=20.8\%$

♦ Mean mark (d) 42%.

Data Analysis, GEN2 2024 VCAA 4

The time series plot below shows the gold medal-winning height for the women's high jump, $\textit{Wgold}$, in metres, for each Olympic year, $year$, from 1952 to 1988.

A five-median smoothing process will be used to smooth the time series plot above.

The first two points have been placed on the graph with crosses (X) and joined by a dashed line (---).

Complete the five-median smoothing by marking smoothed values with crosses (X) joined by a dashed line (---) on the time series plot above. (1 mark)

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Identify two qualitative features that best describe the time series plot above. (1 mark)

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Show Answers Only

b. $\text{Random fluctuations, increasing trend.}$

Show Worked Solution

a. $\text{Medians are:}$

$\textbf{1960}: 1.67, 1.76, \colorbox{lightblue}{1.82}, 1.85, 1.90$

$\textbf{1964}: 1.76, 1.82, \colorbox{lightblue}{1.85}, 1.90, 1.92$

$\textbf{1968}: 1.82, 1.85, \colorbox{lightblue}{1.90}, 1.92, 1.93$

$\textbf{1972}: 1.82, 1.90, \colorbox{lightblue}{1.92}, 1.93, 1.97$

$\textbf{1976}: 1.82, 1.92, \colorbox{lightblue}{1.93}, 1.97, 2.02$

$\textbf{1980}: 1.92, 1.93, \colorbox{lightblue}{1.97}, 2.01, 2.02$

b. $\text{Qualitative features:}$

$\text{- Random fluctuations}$

$\text{- Increasing trend.}$

♦♦♦ Mean mark (b) 24%.

Data Analysis, GEN2 2024 VCAA 3

The Olympic gold medal-winning height for the women's high jump, $\textit{Wgold}$, is often lower than the best height achieved in other international women's high jump competitions in that same year.

The table below lists the Olympic year, $\textit{year}$, the gold medal-winning height, $\textit{Wgold}$, in metres, and the best height achieved in all international women's high jump competitions in that same year, $\textit{Wbest}$, in metres, for each Olympic year from 1972 to 2020.

A scatterplot of $\textit{Wbest}$ versus $\textit{Wgold}$ for this data is also provided.

When a least squares line is fitted to the scatterplot, the equation is found to be:

$Wbest =0.300+0.860 \times Wgold$

The correlation coefficient is 0.9318

Name the response variable in this equation. (1 mark)

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Draw the least squares line on the scatterplot above. (1 mark)

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Determine the value of the coefficient of determination as a percentage. (1 mark)
Round your answer to one decimal place.
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Describe the association between $\textit{Wbest}$ and $\textit{Wgold}$ in terms of strength and direction. (1 mark)

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\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text { strength } \rule[-1ex]{0pt}{0pt} & \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
\hline
\rule{0pt}{2.5ex}\text { direction } \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

Referring to the equation of the least squares line, interpret the value of the slope in terms of the variables $\textit{Wbest}$ and $\textit{Wgold}$. (1 mark)

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In 1984, the $\textit{Wbest}$ value was 2.07 m for a $\textit{Wgold}$ value of 2.02 m .
Show that when this least squares line is fitted to the scatterplot, the residual value for this point is 0.0328. (2 marks)

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The residual plot obtained when the least squares line was fitted to the data is shown below. The residual value from part f is missing from the residual plot.

1. Complete the residual plot by adding the residual value from part f, drawn as a cross ( X ), to the residual plot above. (1 mark)
  
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2. In part b, a least squares line was fitted to the scatterplot. Does the residual plot from part g justify this? Briefly explain your answer. (1 mark)
  
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In 1964, the gold medal-winning height, $\textit{Wgold}$, was 1.90m . When the least squares line is used to predict $\textit{Wbest}$, it is found to be 1.934 m .
Explain why this prediction is not likely to be reliable. (1 mark)
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Show Answers Only

a. $Wbest$

c. $86.8\%$

d. $\text{Strong, positive}$

e. $Wbest\ \text{will increase, on average, by 0.86 metres for every metre of increase in}\ Wgold.$

f.	$Wbest$	$=0.300 +0.86\times 2.02$
		$=2.0372$

$\therefore\ \text{Residual}\ =2.07-2.0372=0.0328$

g.i.

g.ii. $\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}$

h. $\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}$

$\text{and therefore cannot be relied upon.}$

Show Worked Solution

a. $Wbest$

b. $\text{Using points:}\ (1.90, 1.934)\ \text{and}\ (2.00, 2.02)$

Mean mark (b) 51%.

c. $r=0.9318\ \ \Rightarrow\ \ r^2=0.9318^2=0.8682\dots$

$\therefore\ \text{Coefficient of determination} \approx 86.8\%$

d. $\text{Strong, positive}$

e. $Wbest\ \text{will increase, on average, by 0.86 metres for every metre of increase in}\ Wgold.$

f.	$Wbest$	$=0.300 +0.86\times 2.02$
		$=2.0372$

$\therefore\ \text{Residual}\ =2.07-2.0372=0.0328$

♦ Mean mark (f) 48%.

g.i.

g.ii. $\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}$

♦ Mean mark (g)(i) 47%.
♦ Mean mark (g)(ii) 40%.

h. $\text{This prediction is outside the data range (1972–2020 → extrapolation)}$

$\text{and therefore cannot be relied upon.}$

♦ Mean mark (h) 50%.

BIOLOGY, M7 2024 HSC 13 MC

Which of the following identifies plant responses to pathogens?

Increased phagocytosis and programmed cell death
Increased number of stomata and programmed cell death
Production of antihistamines and increased thickness of cell walls
Production of antimicrobial substances and increased thickness of cell walls

Show Answers Only

$D$

Show Worked Solution

Plants respond to pathogens by producing antimicrobial substances and strengthening their cell walls for enhanced physical defence.
The other options either describe animal immune responses (phagocytosis, antihistamines) or incorrect plant responses (increased stomata).

$\Rightarrow D$

♦ Mean mark 47%.

BIOLOGY, M5 2024 HSC 6 MC

The diagram shows the karyotypes of a body cell for a male and a female fruit fly.

How many chromosomes will the egg of a female fruit fly have?

Show Answers Only

$B$

Show Worked Solution

In the diagram, the female fruit fly has 8 chromosomes in body cells (2 sex chromosomes + 6 autosomes across pairs I, II, and III)
During gamete formation, the female fruit fly’s cells undergo meiosis which halves the total chromosome number.
Therefore her eggs will contain half that number, resulting in 4 chromosomes total.

$\Rightarrow B$

♦ Mean mark 50%.

BIOLOGY, M7 2024 HSC 8 MC

Trypanosomes (Trypanosoma brucei) are protozoans that cause African sleeping sickness in humans. The diagram shows the way that the disease is transmitted to humans.

Which row of the table identifies the pathogen, vector and method of disease transmission to humans?

\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|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Pathogen} & \ \ \textit{Vector} & \ \ \textit{Method of disease} \\
\textit{} \rule[-1ex]{0pt}{0pt} & \textit{} & \ \ \textit{transmission} \\
\hline
\rule{0pt}{2.5ex}\text{Trypanosomes} \rule[-1ex]{0pt}{0pt}& \text{Tsetse fly}& \text{Direct} \\
\hline
\rule{0pt}{2.5ex}\text{Tsetse fly} \rule[-1ex]{0pt}{0pt}& \text{Cow} & \text{Direct} \\
\hline
\rule{0pt}{2.5ex}\text{Trypanosomes} \rule[-1ex]{0pt}{0pt}& \text{Tsetse fly} & \text{Indirect}\\
\hline
\rule{0pt}{2.5ex}\text{Tsetse fly} \rule[-1ex]{0pt}{0pt}& \text{Cow} & \text{Indirect}\\
\hline
\end{array}
\end{align*}

Show Answers Only

$C$

Show Worked Solution

African sleeping sickness is caused by trypanosome parasites (pathogen).
The pathogens are transmitted indirectly to humans through the bite of tsetse flies (vector).

$\Rightarrow C$

♦ Mean mark 48%.

Networks, GEN1 2024 VCAA 40 MC

A project has 15 activities, $A-O$, that need to be completed.

The directed network that represents this project is shown below.

The activities are not labelled.

The activity table that could represent this project is

Show Answers Only

$B$

Show Worked Solution

$\text{A focus are would be the two activities with 4 immediate predecessors.}$

$\text{Using trial and error:}$

♦♦ Mean mark 32%.

$\Rightarrow B$

Matrices, GEN1 2024 VCAA 37 MC

The network below represents paths through a park from the carpark to a lookout.

The vertices represent various attractions, and the numbers on the edges represent the distances between them in metres.

The shortest path from the carpark to the lookout is 34 m .

This can be achieved when

$x=8$ and $y=8$
$x=9$ and $y=7$
$x=10$ and $y=6$
$x=11$ and $y=5$

Show Answers Only

$D$

Show Worked Solution

$\text{By trial and error}$

$\text{Consider Option D:}\ \ \ x=11, y=5$

$\text{Shortest path}\ = 10+11+8+5=34\ \checkmark$

$\Rightarrow D$

♦ Mean mark 40%

Networks, GEN1 2024 VCAA 35 MC

Consider the following graph.

The number of faces is

Show Answers Only

$B$

Show Worked Solution

$\text{Method 1: Make the diagram planar}$

♦ Mean mark 52%.

$\text{Method 2: Euler’s Rule}$

$\text{Vertices}\ =7,\ \text{Edges}\ =11$

$v+f$	$=e+2$
$7+f$	$=11+2$
$f$	$=6$

$\Rightarrow B$

Matrices, GEN1 2024 VCAA 31 MC

The matrix below shows the results of a round-robin chess tournament between five players: $H, I, J, K$ and $L$. In each game, there is a winner and a loser.

Two games still need to be played.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad \quad \quad \ \textit{loser}\\
&\quad \quad\quad \quad \quad \ \ \ H \quad \quad \ \ \ I \quad \quad \quad J \quad \quad \ \ \ K \quad \quad \quad L \\
& \textit{winner} \quad \begin{array}{ccccc}
H\\
\\
I\\
\\
J\\
\\
K\\
\\
L
\end{array}
\begin {bmatrix}
0 \quad & \quad 1 \quad & \quad 0 \quad & \quad 1 \quad & \quad 0 \\
\\
0 \quad& 0 & \ldots & 1 & \quad \ldots \\
\\
1 \quad& \ldots & 0 & 1 & \quad 0 \\
\\
0 \quad& 0 & 0 & 0 & \quad 1 \\
\\
1 \quad& \ldots & 1 & 0 & \quad 0
\end{bmatrix}\\
&
\end{aligned}

A '1' in the matrix shows that the player named in that row defeated the player named in that column. For example, the 1 in row 4 shows that player $K$ defeated player $L$.

A '...' in the matrix shows that the player named in that row has not yet competed against the player in that column.

At the end of the tournament, players will be ranked by calculating the sum of their one-step and two-step dominances.

The player with the highest sum will be ranked first. The player with the second-highest sum will be ranked second, and so on.

Which one of the following is not a potential outcome after the final two games have been played?

Player $I$ will be ranked first.
Player $I$ will be ranked fifth.
Player $J$ will be ranked first.
Player $J$ will be ranked fifth.

Show Answers Only

$D$

Show Worked Solution

$\text{Games to be played:}\ \ I\ \text{vs}\ J, \ \ I\ \text{vs}\ L$

$\text{Let the results matrix be}\ D$

♦ Mean mark 53%.

$\text{Case 1: Player}\ I\ \text{wins both games}$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 2 \\
\ 2 & 0 & 2 & 2 & 2 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 1 & 1 & 2 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=6\\
\ \ I=8\\
\ \ J=5\\
\ \ K=3\\
\ \ L=6
\end{array}$

$\text{Case 2: Player}\ I\ \text{loses to}\ J, \text{defeats}\ L$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 2 \\
\ 1 & 0 & 1 & 1 & 2 \\
\ 1 & 2 & 0 & 3 & 2 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 2 & 1 & 2 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=8\\
\ \ K=3\\
\ \ L=7
\end{array}$

$\text{Case 3: Player}\ I\ \text{defeats}\ J, \text{loses to}\ L$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 1 \\
\ 1 & 0 & 1 & 2 & 1 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 2 & 2 & 3 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=5\\
\ \ K=4\\
\ \ L=9
\end{array}$

$\text{Case 4: Player}\ I\ \text{loses both games}$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 1 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 2 & 0 & 3 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 3 & 1 & 3 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=4\\
\ \ I=2\\
\ \ J=7\\
\ \ K=4\\
\ \ L=9
\end{array}$

$\text{In all cases Player }I\ \text{or}\ K\ \text{will be ranked 5th.}$

$\therefore\ \text{Option D is not a potential outcome.}$

$\Rightarrow D$

Matrices, GEN1 2024 VCAA 29 MC

A tennis team consists of five players: Quinn, Rosie, Siobhan, Trinh and Ursula.

When the team competes, players compete in the order of first, then second, then third, then fourth.

The fifth player has a bye (does not compete).

On week 1 of the competition, the players competed in the following order.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad\textbf{First} \quad\rule[-1ex]{0pt}{0pt}& \quad \textbf{Second} \quad& \quad\textbf{Third} \quad& \quad\textbf{Fourth}\quad & \quad\textbf{Bye} \quad\\
\hline
\rule{0pt}{2.5ex} \text{Quinn} \rule[-1ex]{0pt}{0pt}& \text {Rosie} & \text {Siobhan} & \text { Trinh } & \text {Ursula} \\
\hline
\end{array}

This information can be represented by matrix $G_1$, shown below.

$G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}$

Let $G_n$ be the order of play in week $n$.

The playing order changes each week and can be determined by the rule $G_{n+1}=G_n \times P$

$\text{where}\quad \\P=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \end{bmatrix}$

Which player has a bye in week 4 ?

Quinn
Rosie
Siobhan
Trinh

Show Answers Only

$C$

Show Worked Solution

$G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}$

$G_2=\begin{bmatrix} S & Q & T & U & R \end{bmatrix}$

$G_3=\begin{bmatrix} T & S & U & R & Q \end{bmatrix}$

$G_4=\begin{bmatrix} U & T & R & Q & S \end{bmatrix}$

$\text{5th player has a bye.}$

$\therefore\ \text{Siobhan does not compete in week 4}$

$\Rightarrow C$

♦ Mean mark 49%.

Matrices, GEN1 2024 VCAA 27 MC

Consider the following matrix, where $h \neq 0$.

\begin{bmatrix}
4 & g \\
8 & h
\end{bmatrix}

The inverse of this matrix does not exist when $g$ is equal to

$-2 h$
$\dfrac{h}{2}$
$h$
$2 h$

Show Answers Only

$B$

Show Worked Solution

$\text{Inverse will not exist if determinant = 0}$

$\text{Find}\ g\ \text{when}\ \ 4h-8g=0:$

$g=\dfrac{h}{2}$

$\Rightarrow B$

♦ Mean mark 50%.

\(x\)	\(=\dfrac{2y+1}{3-y}\)
\(3x-xy\)	\(=2y+1\)
\(3x-1\)	\(=2y+xy\)
\(y(2+x)\)	\(=3x-1\)
\(y\)	\(=\dfrac{3x-1}{2+x}\)
	\(=\dfrac{6+3x-7}{2+x}\)
	\(=3-\dfrac{7}{2+x}\)

\(K_{a}\)	\(=\dfrac{\ce{[H^+][BrCH2COO^-]}}{\ce{[BrCH2COOH]_{eq}}}\)
\(\ce{[BrCH2COOH]_{eq}}\)	\(=\dfrac{\ce{[H^+][BrCH2COO^-]}}{K_a}\)
	\(=\dfrac{(9.18 \times 10^{-3})^2}{1.29 \times 10^{-3}}\)
	\(=0.06533\ \text{mol L}^{-1}\)

\(\Delta H\)	\(=\dfrac{-q}{n}\)
	\(=\dfrac{-mc\Delta t}{n}\)
	\(=\dfrac{-129.4 \times 3.8 \times (11.5-21.5)}{0.175}\)
	\(=+2810\ \text{J mol}^{-1}\)
	\(=+28.1\ \text{kJ mol}^{-1}\)

\(\ce{[Cd^{2+}]^3[PO4^{3-}]^2}\)	\(=2.53 \times 10^{-33}\)
\((3x)^3 \times (2x)^2\)	\(=2.53 \times 10^{-33}\)
\(108x^5\)	\(=2.53 \times 10^{-33}\)
\(x\)	\(=\sqrt[5]{\dfrac{2.53 \times 10^{-33}}{108}}\)
	\(=1.11856 \times 10^{-7}\)

\(K_{eq}\)	\(=\dfrac{\ce{[H2O(g)][CO(g)]}}{\ce{[H2(g)][CO2(g)]}}\)
\(1.600\)	\(=\dfrac{0.2 \times (1.8+x)}{1.2 \times 0.7}\)
\(1.8 + x\)	\(=1.6 \times \dfrac{1.2 \times 0.7}{0.2}\)
\(x\)	\(=6.72-1.8\)
	\(=4.92\ \text{mol}\)

\(\ce{[BrCH2COOH(octan-1-ol)]_{eq}}\)	\(=0.1000-0.06533-9.18 \times 10^{-3}\)
	\(=0.02549\ \text{mol L}^{-1}\)

	\(\textbf{Age Group}\)
	\(\ 0-1\ \text{year}\ \)	\(\ 1-2\ \text{years}\ \)	\(\ 2-3\ \text{years}\ \)	\(\ 3-4\ \text{years}\ \)
\(\ \text{Initial population}\ \)	\(70\)	\(80\)	\(90\)	\(40\)
\(\ \text{Population after}\ \) \(\ \text{one year}\)	\(124\)	\(28\)	\(56\)	\(45\)

	\(\textbf{Age Group}\)
	\(\ 0-1\ \text{year}\ \)	\(\ 1-2\ \text{years}\ \)	\(\ 2-3\ \text{years}\ \)	\(\ 3-4\ \text{years}\ \)
\(\ \text{Initial population}\ \)	\(70\)	\(80\)	\(90\)	\(40\)
\(\ \text{Population after}\ \) \(\ \text{one year}\)	\(124\)	\(28\)	\(56\)	\(45\)

	\(\overrightarrow{O M}\)	\(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
		\(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
		\(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
		\(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)