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PHYSICS, M6 2017 HSC 11 MC

An AC source is connected to a transformer having a primary winding of 900 turns. Connected to the secondary winding of 450 turns is a pair of parallel plates 0.01 m apart.
 

The AC input is shown in the graph.
 

What is the maximum field strength (in `text{V m}^(-1)`) produced between the plates?

  1. `1.7`
  2. `6.8`
  3. `1.7 × 10^4`
  4. `6.8 × 10^4`
Show Answers Only

`C`

Show Worked Solution

Maximum voltage in primary coil = 340 `text{V}`  (see graph)

`(V_(p))/(V_(s))` `=(N_(p))/(N_(s))`  
`(340)/(V_(s))` `=(900)/(450)`  
`V_(s)` `=(450 xx 340)/900`  
  `=170\ text{V}`  

 
→ The maximum voltage in the secondary circuit is 170 V.

`E` `=(V)/(d)`  
  `=(170)/(0.01)`  
  `=1.7 xx10^(4)\ text{V m}^(-1)`  

 
`=>C`

PHYSICS, M6 2017 HSC 10 MC

The diagram shows a model of a generator connected to a galvanometer.
 

The loop is rotated continuously in a clockwise direction as viewed from the end nearest the galvanometer.

Which row of the table correctly identifies the type of generator and the movement of the needle of the galvanometer
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex} \ \rule[-0.5ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{1.5ex}\textit{Type of generator}\rule[-0.5ex]{0pt}{0pt}& \textit{Movement of the needle} \\
\hline
\rule{0pt}{2.5ex}\text{DC}\rule[-1ex]{0pt}{0pt}&\text{Swings between 0 and +}\\
\hline
\rule{0pt}{2.5ex}\text{AC}\rule[-1ex]{0pt}{0pt}& \text{Swings between – and 0}\\
\hline
\rule{0pt}{2.5ex}\text{DC}\rule[-1ex]{0pt}{0pt}& \text{Swings between + and –} \\
\hline
\rule{0pt}{2.5ex}\text{AC}\rule[-1ex]{0pt}{0pt}& \text{Swings between – and +} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

→ The generator has a split ring commutator so it is a DC generator.

→ Using the right hand palm rule, induced current will flow clockwise as viewed from above through the external circuit.

→ This means current will flow into the positive terminal of the galvanometer causing a deflection towards +.

\(\Rightarrow A\)

PHYSICS, M6 2018 HSC 26

Outline the similarities and differences between the effects of electric fields and gravitational fields on matter. In your answer, refer to the definitions of these fields.   (4 marks)

Show Answers Only

Differences in effects:

→ Electric fields are regions in space where a force acts on a charged particle whereas gravitational fields are regions in space where a force acts on an object with mass due to the influence of another object with mass.

→ Electric field strength is defined by `E=(F)/(q)`, or force per unit charge.

→ Gravitational field strength is defined with `g=(F)/(m)`, or force per unit mass.

→ Electric fields can produce forces of repulsion (due to like charges) while gravitational fields can only produce forces of attraction.
 

Similarities:

→ Both fields are able to produce forces of attraction.

→ Electric and gravitational fields vary in strength due to magnitudes of charge and mass respectively.

→ The strength of both fields is inversely proportional to the distance of separation from a point charge or mass.

→ Gravity is a significantly weaker fundamental force compared to electromagnetism.

Show Worked Solution

Differences in effects:

→ Electric fields are regions in space where a force acts on a charged particle whereas gravitational fields are regions in space where a force acts on an object with mass due to the influence of another object with mass.

→ Electric field strength is defined by `E=(F)/(q)`, or force per unit charge.

→ Gravitational field strength is defined with `g=(F)/(m)`, or force per unit mass.

→ Electric fields can produce forces of repulsion (due to like charges) while gravitational fields can only produce forces of attraction.
 

Similarities:

→ Both fields are able to produce forces of attraction.

→ Electric and gravitational fields vary in strength due to magnitudes of charge and mass respectively.

→ The strength of both fields is inversely proportional to the distance of separation from a point charge or mass.

→ Gravity is a significantly weaker fundamental force compared to electromagnetism.

PHYSICS, M6 2018 HSC 24b

Three parallel wires `X`, `Y` and `Z` all carry electric currents. A force of attraction is produced between `Y` and `Z`. There is zero net force on `Y`.
 

What is the magnitude and direction of the current in `X` ?   (3 marks)

Show Answers Only

50 Amps in the same direction as `Y`.

Show Worked Solution

→  Net force on `Y` = 0 (given).

→  The force of attraction between `Y` and `X` is equal to the force of attraction between `Y` and `Z`.

`(F)/(l)` `=(mu_(0))/(2pi)(I_(1)I_(2))/(r)`  
`(mu_(0))/(2pi)(I_(Y)I_(Z))/(0.3)` `=(mu_(0))/(2pi)(I_(X)I_(Y))/(0.75)`  
`(I_(Y)I_(Z))/(0.3)` `=(I_(X)I_(Y))/(0.75)`  
`(20 xx20)/(0.3)` `=(20I_(X))/(0.75)`  
`I_(X)` `=50\ text{A}`  

 
→ The direction of current in `X` is the same as the direction of current in `Y`, as there is a force of attraction between `X` and `Y`.

ENGINEERING, PPT 2022 HSC 4 MC

The diagram shows a box being pulled along a horizontal surface at angle `theta` to the horizontal.
 

Which equation is correct for `N`?

  1. `N = W - P costheta`
  2. `N = W + P costheta`
  3. `N = W - P sintheta`
  4. `N = W + P sintheta`
Show Answers Only

`C`

Show Worked Solution
`SigmaF_V` `=0`  
`0` `= N-W + P sintheta`  
`N` `= W-P sintheta`  

 
`=>C`

Mean mark 59%.

Algebra, STD1 A3 2022 HSC 25

Sam is making cupcakes to sell at a market. It costs Sam $60 to hire a stall, and each cupcake costs $1.50 to make. Sam intends to sell each cupcake for $4.00.

The equations representing Sam's cost `($ C)` and revenue `($ R)`, are

`C=1.5 x+60`  and  `R=4 x`, where `x` is the number of cupcakes sold.

The graphs of `C` and `R` are shown below.
 


 

  1. How many cupcakes must Sam sell in order to break even?  (1 mark)

  2. If Sam sells 60 cupcakes, what profit is made?
  3. You may assume that  Profit = Revenue – Cost.  (2 marks)

Show Answers Only
  1. 24
  2. $90
Show Worked Solution

a.   Break even occurs where the two graphs intersect.

→ 24 cupcakes
 

b.   If cupcakes sold (`x`) = 60:

`C = 1.5 xx 60 + 60= $150`

`R = 4 xx 60=$240`
  
`:.\ text{Profit}\ = 240-150 = $90`

Statistics, STD1 S3 2022 HSC 23

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

The results are shown on the scatterplot below.
 

  1. The data for two new students, Alinta and Birrani, are shown in the table below. Plot their results on the scatterplot.  (2 marks)

\begin{array} {|l|c|c|}
\hline
  & \textit{Average hours of} & \textit{Average hours of} \\ & \textit{phone use per day} & \textit{sleep per day} \\
\hline
\rule{0pt}{2.5ex} \text{Alinta} \rule[-1ex]{0pt}{0pt} & 4 & 8 \\
\hline
\rule{0pt}{2.5ex} \text{Birrani} \rule[-1ex]{0pt}{0pt} & 0 & 10.5 \\
\hline
\end{array}

  1. By first fitting the line of best fit by eye on the scatterplot, estimate the average number of hours of sleep per day for a student who uses the phone for an average of 2 hours per day.  (2 marks)

Show Answers Only
  1.  
  2. 9 hours (see LOBF in diagram above)
Show Worked Solution

a.     \(\text{New data points are marks with a × on the diagram below.}\)
 

b.   \(\text{9 hours (see LOBF in diagram above)}\)

Financial Maths, STD1 F1 2022 HSC 21

A real estate agent's commission for selling houses is 2% for the first $800 000 of the sale price and 1.5% for any amount over $800 000.

Calculate the commission earned in selling a house for $1 500 000.  (2 marks)

Show Answers Only

$26 500

Show Worked Solution
`text{Commission}` `=800\ 000 xx 2text{%} + (1\ 500\ 000-800\ 000) xx 1.5text{%}`  
  `=800\ 000 xx 0.02 + 700\ 000 xx 0.015`  
  `=16\ 000 + 10\ 500`  
  `=$26\ 500`  

Networks, STD1 N1 2022 HSC 20

The table below shows the distances, in kilometres, between a number of towns.
 

  1. Using the vertices given, draw a weighted network diagram to represent the information shown in the table.  (2 marks)
     

     
  2. A tourist wishes to visit each town.
  3. Draw the minimum spanning tree which will allow for this AND determine its length.  (3 marks)
     

Show Answers Only
  1.  
     
     
  2.   
     
  3. `1015\ text{km}`
Show Worked Solution

a. 

b.   `text{Using Prim’s algorithm (starting at}\ Y):`

`text{1st edge:}\ YC`

`text{2nd edge:}\ CB`

`text{3rd edge:}\ SB`

`text{4th edge:}\ YM`

`text{Length of minimum spanning tree}`

`=275 + 150+60+530`

`=1015\ text{km}`

Statistics, STD1 S3 2022 HSC 16

A concert organiser is interested in the relationship between the distance from the stage, in metres, and the loudness of the sound measured in decibels.

The data the concert organiser collected is shown on the graph.
 

  1. Is the relationship between distance and loudness linear or non-linear?  (1 mark)

  2. Based on this dataset, at approximately what distance from the stage would the sound be at 90 decibels?  (1 mark)

Show Answers Only
  1. `text{Non-linear}`
  2. `text{6 metres}`
Show Worked Solution

a.   The graph is not a straight line, therefore, non-linear.

b.   Distance =  6 metres.

→ Intersection of line of best fit and horizontal line at decibels (y-axis) = 90


♦ Mean mark part (b) 46%.

Measurement, STD1 M4 2022 HSC 15

Tom is 25 years old, and likes to keep fit by exercising.

  1. Use this formula to find his maximum heart rate (bpm).
  2.       Maximum heart rate = 220 – age in years
  3. Tom's maximum heart rate is .................... bpm.  (1 mark)
  4. Tom will get the most benefit from this exercise if his heart rate is between 50% and 85% of his maximum heart rate.
  5. Between what two heart rates should Tom be aiming for to get the most benefit from his exercise?  (2 marks)

Show Answers Only
  1. `195\ text{bpm}`
  2. `text{98 – 166 bpm}`
Show Worked Solution
a.    `text{Max heart rate}` `=220-25`
    `=195\ text{bpm}`

 

b.   `text{50% max heart rate}\ = 0.5 xx 195 = 97.5\ text{bpm}`

`text{85% max heart rate}\ = 0.85 xx 195 = 165.75\ text{bpm}`

`:.\ text{Tom should aim for between 98 and 166 bpm in exercise.}`

ENGINEERING, AE 2020 HSC 27a

Explain why engineering production drawings prepared for Australian use should adhere to AS 1100 drawing standards.   (2 marks)

Show Answers Only

→ AS1100 standards are important as they provide an unambiguous, standard method of communicating engineering drawings.

→ This standard can be applied nationally and understood internationally due to the strict rules which must be adhered to.

→ It is far easier to communicate ideas with visuals rather than words, and using these standards makes this communication clearer and easier to understand.

Show Worked Solution

→ AS1100 standards are important as they provide an unambiguous, standard method of communicating engineering drawings.

→ This standard can be applied nationally and understood internationally due to the strict rules which must be adhered to.

→ It is far easier to communicate ideas with visuals rather than words, and using these standards makes this communication clearer and easier to understand.

ENGINEERING, PPT 2020 HSC 26d

The diagram shows an electrical circuit that is used to control the speed of a DC electric traction motor.
 

Explain, with reference to the diagram, how the speed of the motor can be controlled. (3 marks)

Show Answers Only

→ The motor’s speed and torque can be increased or decreased by varying the input current.

→ Since current and resistance are inversely proportional, the current can be increased by decreasing resistance and vice versa.

→ The above diagram shows 4 resistors that can be used or avoided by flicking a switch for each.

→ The more resistors that are switched out of the circuit, the more current flows to the motor and the speed of the motor increases.

Show Worked Solution

→ The motor’s speed and torque can be increased or decreased by varying the input current.

→ Since current and resistance are inversely proportional, the current can be increased by decreasing resistance and vice versa.

→ The above diagram shows 4 resistors that can be used or avoided by flicking a switch for each.

→ The more resistors that are switched out of the circuit, the more current flows to the motor and the speed of the motor increases.

BIOLOGY, M7 2019 HSC 32

Use the following data to answer parts (a) and (b).

Dengue fever and malaria are examples of infectious diseases transmitted between humans by mosquitoes. Dengue fever is caused by a virus transmitted by mosquitoes of the genus Aedes. Malaria is caused by a single-celled organism transmitted by mosquitoes of the genus Anopheles.

The following data provide information about the global incidence of these two diseases over time.
 


 

  1. Based on the data provided, identify trends in the global disease burden for both malaria and dengue fever.   (3 marks)

  2. Analyse factors that could have contributed to the change in global distribution of both dengue fever and malaria over the last 100 years. Support your answer with reference to the data provided.   (7 marks)

Show Answers Only

a.   Trends in global disease burden:

→ The distribution on Dengue Fever has significantly increased since 1950, with continents such as Australia, Europe, South America and Africa now having a significant number of reported cases.

→ This spread is seen both north and south of the equator.

→ Malaria has a declining number of countries with reported cases of the disease.

→ However, population growth means the number of individuals at risk is increasing, having almost doubled between 1965 and 2010.

→ We can conclude however, that there is an increasing number of people at risk of malaria, but representing a smaller portion of the global population.
 

b.  Analysis of factors that could have contributed to the change in global distribution of both dengue fever and malaria

Both diseases are transmitted by a mosquito vector. The prevalence of mosquitoes in each area would therefore have a large influence on the number of infected individuals.

→ The increase in international travel allows mosquitoes to spread across the world.

→ The increasing population of the world along with rapid urbanisation gives rise to more urban mosquito habitats.

These factors can be associated with the distribution seen on the Dengue Fever map, however the countries with reported cases of Malaria has shrunk.

→ The mosquitoes responsible for Dengue Fever and Malaria respectively, are from a different genus of mosquito. It is likely that more is known about Anopheles than Aedes.

→ The use of quarantine and pesticides on the Anopheles is most likely being used to contain the spread of malaria.

Medical advances such as vaccines also have the potential to be an effective measure of containing diseases.

→ From both data sets, it would be accurate to conclude, however, that medicinal products had better results when maintaining malaria compared to dengue fever.

→ Antimalarial tablets or a malaria vaccine may be types of malaria medicinal products that may be used to maintain the disease.

→ The rapid evolution of the dengue virus, ineffective antivirals or potentially that there is no known vaccine may be some of the reasons why medicinals may not be effective for dengue fever.

Show Worked Solution

a.   Trends in global disease burden:

→ The distribution on Dengue Fever has significantly increased since 1950, with continents such as Australia, Europe, South America and Africa now having a significant number of reported cases.

→ This spread is seen both north and south of the equator.

→ Malaria has a declining number of countries with reported cases of the disease.

→ However, population growth means the number of individuals at risk is increasing, having almost doubled between 1965 and 2010.

→ We can conclude however, that there is an increasing number of people at risk of malaria, but representing a smaller portion of the global population.

b.  Analysis of factors that could have contributed to the change in global distribution of both dengue fever and malaria

Both diseases are transmitted by a mosquito vector. The prevalence of mosquitoes in each area would therefore have a large influence on the number of infected individuals.

→ The increase in international travel allows mosquitoes to spread across the world.

→ The increasing population of the world along with rapid urbanisation gives rise to more urban mosquito habitats.

These factors can be associated with the distribution seen on the Dengue Fever map, however the countries with reported cases of Malaria has shrunk.

→ The mosquitoes responsible for Dengue Fever and Malaria respectively, are from a different genus of mosquito. It is likely that more is known about Anopheles than Aedes.

→ The use of quarantine and pesticides on the Anopheles is most likely being used to contain the spread of malaria.

Medical advances such as vaccines also have the potential to be an effective measure of containing diseases.

→ From both data sets, it would be accurate to conclude, however, that medicinal products had better results when maintaining malaria compared to dengue fever.

→ Antimalarial tablets or a malaria vaccine may be types of malaria medicinal products that may be used to maintain the disease.

→ The rapid evolution of the dengue virus, ineffective antivirals or potentially that there is no known vaccine may be some of the reasons why medicinals may not be effective for dengue fever.


♦ Mean mark (b) 52%.

BIOLOGY, M7 2019 HSC 31

  1. Outline ONE adaptation of a specific pathogen that facilitates its entry into a host.   (2 marks)

  2. Explain how the mode of transmission of pathogens influences the spread of diseases.   (3 marks)

Show Answers Only

a.   → Helicobacter pylori is a bacteria that causes stomach ulcers.

→ It has flagellum which allows it to move in the stomach and penetrate the stomach wall.
 

Other answers could include:

→ Salmonella and its ability to adapt to host blood temperature.
 

b.   → Diseases will be able to spread faster and easier with certain modes of transmission.

→ Airborne disease such as influenza are able to spread faster as the virus can be passed through droplets of air by infected individuals sneezing/coughing.

→ Diseases that can only spread via direct contact will have lower infection rates as there is a less effective mode of transmission.
 

Other answers could include the effectiveness of modes such as 

→ Vectors and their presence in an area influencing infection rate.

→ Foodborne/waterborne diseases.

→ Zoonotic diseases.

Show Worked Solution

a.   → Helicobacter pylori is a bacteria that causes stomach ulcers.

→ It has flagellum which allows it to move in the stomach and penetrate the stomach wall.
 

Other answers could include:

→ Salmonella and its ability to adapt to host blood temperature.
 


♦♦ Mean mark (a) 36%.

b.   → Diseases will be able to spread faster and easier with certain modes of transmission.

→ Airborne disease such as influenza are able to spread faster as the virus can be passed through droplets of air by infected individuals sneezing/coughing.

→ Diseases that can only spread via direct contact will have lower infection rates as there is a less effective mode of transmission.
 

Other answers could include the effectiveness of modes such as 

→ Vectors and their presence in an area influencing infection rate.

→ Foodborne/waterborne diseases.

→ Zoonotic diseases.

Measurement, STD1 M4 2022 HSC 14

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

   
 

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

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

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

  4. Between which times in the day was Jamie travelling at the fastest speed? Justify your answer, without calculations.  (2 marks)

Show Answers Only

a.   `140\ text{km}`
 

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

`:. 10\ text{am}`
 

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

  

d.   `text{Fastest speed → graph is the steepest}`

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

Show Worked Solution

a.   `140\ text{km}`
 

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

`:. 10\ text{am}`
 

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

  

d.   `text{Fastest speed → graph is the steepest}`

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


♦♦ Mean mark part (b) 31%.
♦♦ Mean mark part (c) 35%.
♦♦ Mean mark part (d) 32%.

Measurement, STD1 M5 2022 HSC 5 MC

Two similar figures are shown.
 

What is the value of `x` ?

  1. 6
  2. 8
  3. 18
  4. 27
Show Answers Only

`C`

Show Worked Solution

`text{Scale factor}\ =3/2 =1.5`

`:.\ x = 1.5 xx 12 = 18`
  

`text{Alternate solution}`

`text{Using sides of similar figures in the same ratio:}`

`x/12` `=3/2`  
`x` `=12 xx (3/2)`  
`x` `=18`  

   
`=> C`

Probability, STD1 S2 2022 HSC 3 MC

A jar contains 12 red, 10 black and 13 white lollies.

Alex picks out a red lolly and eats it. He then randomly picks a second lolly.

What is the probability that the second lolly is also red?

  1. `(11)/(34)`
  2. `(11)/(35)`
  3. `(12)/(34)`
  4. `(12)/(35)`
Show Answers Only

`A`

Show Worked Solution
`P(E)` `=text{favourable outcomes}/text{total outcomes}`  
  `=(12-1)/((12-1)+10+13)`  
  `=11/34`  

 
`=>A`

BIOLOGY, M7 2019 HSC 16 MC

The diagram shows the concentration of an antibody to a particular pathogen.
 

Two students are exposed to the pathogen. Student `X` had previously been vaccinated against this pathogen. Student `Y` had never been exposed to it.

Which row of the table shows the most likely levels of antibody in the blood of each student a week after exposure to the pathogen on this occasion?

\begin{align*}
\begin{array}{l}
\rule[-1ex]{0pt}{0pt} \ \textit{Concentration of antibody in the blood}\ \text{(mg L\(^{-1}\))} \\
\end{array} \\
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array} 
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \quad \quad \textit{Student X}\quad\quad \ \ \rule[-1ex]{0pt}{0pt}& \ \ \quad \quad \textit{Student Y} \quad\quad \ \ \\
\hline
\rule{0pt}{2.5ex}250\rule[-1ex]{0pt}{0pt}&1500\\
\hline
\rule{0pt}{2.5ex}1500\rule[-1ex]{0pt}{0pt}& 1000\\
\hline
\rule{0pt}{2.5ex}1000\rule[-1ex]{0pt}{0pt}& 1500 \\
\hline
\rule{0pt}{2.5ex}1500\rule[-1ex]{0pt}{0pt}& 250 \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution

→ Student X has had previous exposure to the pathogen through vaccination and will respond with a high level of antibodies (similar to a secondary exposure).

→ Student Y has no previous exposure to the pathogen and will produce antibodies similar to that following a primary exposure (see graph).

\(\Rightarrow D\)

BIOLOGY, M6 2019 HSC 24

Explain the loss of biodiversity that may result from TWO biotechnologies used in agriculture.   (5 marks)

Show Answers Only

→ Biodiversity can refer to both the diversity of species within an ecosystem, but also the variety of genomes and allele frequencies present within a species.

→ Biotechnologies used in agriculture may reduce biodiversity on both scales.

→ Artificial insemination involves the process of manually injecting sperm of favoured genetic traits into females of favoured genetic traits to have a greater chance of producing offspring of a desired type.

→ The repeated use of this process will result in all offspring becoming genetically similar, reducing the diversity within the species.

→ Artificial insemination has an equivalent process in plants which can result in the loss of genetic diversity in a similar way.

→ Transgenic organisms are those whose genomes have been modified by having a gene from another species inserted into it.

→ The mass use of this process will produce genetically advanced animals or crops which out-compete wild and native types, reducing the biodiversity of its species.

→ The use of transgenic organisms such as Bt cotton has also resulted in the reduction of its main pest, Bollworm. While a pest to the crop and the farmers, Bt cotton has resulted in the reduction of Bollworm in areas where it is prominent, reducing the diversity of the ecosystem it resides in.

 
Other answers could include

→ Selective breeding.

→ Cloning (both gene and whole organism).

Show Worked Solution

→ Biodiversity can refer to both the diversity of species within an ecosystem, but also the variety of genomes and allele frequencies present within a species.

→ Biotechnologies used in agriculture may reduce biodiversity on both scales.

→ Artificial insemination involves the process of manually injecting sperm of favoured genetic traits into females of favoured genetic traits to have a greater chance of producing offspring of a desired type.

→ The repeated use of this process will result in all offspring becoming genetically similar, reducing the diversity within the species.

→ Artificial insemination has an equivalent process in plants which can result in the loss of genetic diversity in a similar way.

→ Transgenic organisms are those whose genomes have been modified by having a gene from another species inserted into it.

→ The mass use of this process will produce genetically advanced animals or crops which out-compete wild and native types, reducing the biodiversity of its species.

→ The use of transgenic organisms such as Bt cotton has also resulted in the reduction of its main pest, Bollworm. While a pest to the crop and the farmers, Bt cotton has resulted in the reduction of Bollworm in areas where it is prominent, reducing the diversity of the ecosystem it resides in.

 
Other answers could include

→ Selective breeding.

→ Cloning (both gene and whole organism).


Mean mark 54%.

BIOLOGY, M8 2019 HSC 23

Explain how educational programs can be effective in reducing the incidence of non-infectious diseases. Support your answer with examples.   (5 marks)

Show Answers Only

→ Educational programs show individuals the risks of exposure to certain factors, encouraging them to change their behaviour to prevent their effects.

→ The Grim Reaper campaign shows a series of ‘scare’ advertisements, which encourage check ups and safe sex to reduce the incidence of HIV/AIDS.

→ Slip Slop Slap is an advertisement campaign, run by the Cancer Council to help reduce the incidence of melanoma by encouraging sun safe behaviours such as applying sunscreen and wearing a hat and shirt outdoors.

Show Worked Solution

→ Educational programs show individuals the risks of exposure to certain factors, encouraging them to change their behaviour to prevent their effects.

→ The Grim Reaper campaign shows a series of ‘scare’ advertisements, which encourage check ups and safe sex to reduce the incidence of HIV/AIDS.

→ Slip Slop Slap is an advertisement campaign, run by the Cancer Council to help reduce the incidence of melanoma by encouraging sun safe behaviours such as applying sunscreen and wearing a hat and shirt outdoors.

ENGINEERING, CS 2020 HSC 26a

Describe methods used to protect civil structures against corrosion. Support your answer with examples.   (3 marks)

Show Answers Only

→ The use of sacrificial anodes. For example a block of zinc can be connected to a civil structure so any corrosion occurs at this anode.

→ Surface coating such as organic based paints or salt films can protect the surface of the metal structure.
 

Other examples could include:

→ Galvanic protection.

→ Ceramic coating or oxide films.

Show Worked Solution

→ The use of sacrificial anodes. For example a block of zinc can be connected to a civil structure so any corrosion occurs at this anode.

→ Surface coating such as organic based paints or salt films can protect the surface of the metal structure.
 

Other examples could include:

→ Galvanic protection.

→ Ceramic coating or oxide films.

ENGINEERING, TE 2020 HSC 25a

Describe potential legal and ethical implications of a professional engineer failing to undertake ongoing training.   (3 marks)

Show Answers Only

→ Engineers have both ethical and legal obligations to keep up to date on safety, design and engineering best practice.

→ If an engineer fails to undertake required ongoing training they could breach these serious legal and professional obligations.

→ This exposes the engineer to potential litigation and could possibly expose both them and others to injury or death, which is an ethical issue that must be considered.

Show Worked Solution

→ Engineers have both ethical and legal obligations to keep up to date on safety, design and engineering best practice.

→ If an engineer fails to undertake required ongoing training they could breach these serious legal and professional obligations.

→ This exposes the engineer to potential litigation and could possibly expose both them and others to injury or death, which is an ethical issue that must be considered.

ENGINEERING, CS 2020 HSC 24d

An exploded view of a yoke and turnbuckle is shown.
 

Complete the assembled front view in the space below. Apply AS 1100 drawing standards. Do NOT show hidden detail. Do NOT add dimensions.  (3 marks)
 

Show Answers Only

Notes:

→ Thread must be properly drawn to the correct length

→ Nut dimensions must be correct according to AS1100

→ Question states no hidden detail or dimensions are to be shown

→ Centrelines must still be shown

→ Ensure the drawing is completed from the correct viewing direction

Show Worked Solution

Notes:

→ Thread must be properly drawn to the correct length

→ Nut dimensions must be correct according to AS1100

→ Question states no hidden detail or dimensions are to be shown

→ Centrelines must still be shown

→ Ensure the drawing is completed from the correct viewing direction

ENGINEERING, PPT 2020 HSC 24b

The photograph shows two stainless steel yokes which attach a turnbuckle to a baseplate fixed to a concrete pad.
 

   

Provide a step-by-step description of an appropriate method to manufacture the yoke.  (3 marks)

Show Answers Only

→ The metal for the yoke is stamped out of a piece of steel plate.

→ A press is then used to punch out the slot and the holes.

→ The plate is then formed around a mandrel or mould to develop the final shape.

→ The piece can then be further machined if required.

Show Worked Solution

→ The metal for the yoke is stamped out of a piece of steel plate.

→ A press is then used to punch out the slot and the holes.

→ The plate is then formed around a mandrel or mould to develop the final shape.

→ The piece can then be further machined if required.

Mean mark 59%.

ENGINEERING, CS 2020 HSC 24a

A photograph of a cast concrete pad is shown.
 

Describe why concrete would have been selected for this situation.  (3 marks)

Show Answers Only

→ Concrete was selected for this situation due to its high compressive strength and formability.

→ Irregular shapes such as the pad shown can be formed relatively easily onsite, or preformed offsite and transported.

→ This enables greater flexibility in the shape, design and construction of civil structures.

→ Concrete can also be reinforced if necessary, to increase tensile strength.

Show Worked Solution

→ Concrete was selected for this situation due to its high compressive strength and formability.

→ Irregular shapes such as the pad shown can be formed relatively easily onsite, or preformed offsite and transported.

→ This enables greater flexibility in the shape, design and construction of civil structures.

→ Concrete can also be reinforced if necessary, to increase tensile strength.

Calculus, EXT2 C1 2022 HSC 15c

Using the substitution  `x=tan^(2)theta`, evaluate

          `int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx`  (4 marks)

Show Answers Only

`pi/2-1`

Show Worked Solution

`x=tan^(2)theta`

`dx/(d theta)=2sec^2theta\ tantheta\ \ =>\ \ dx=2sec^2theta\ tantheta\ d theta`

`text{When}\ x=0, \ theta=0`

`text{When}\ x=1, \ theta=pi/4`


Mean mark 55%.
`sin^(-1)sqrt((x)/(1+x))` `=sin^(-1)sqrt((tan^(2)theta)/(1+tan^(2)theta))`  
  `=sin^(-1)sqrt((tan^(2)theta)/(text{sec}^(2)theta))`  
  `=sin^(-1)sqrt((sin^(2)theta))`  
  `=sin^(-1)(sintheta)`  
  `=theta`  

 

`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx=int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`
 

`text{Integrating by parts:}`

`u` `=theta` `u′` `=1`
`v′` `=2tan theta\ sec^2 theta` `v` `=tan^(2)theta`

 
`int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`

`=[theta\ tan^(2)theta]_0^(pi/4)-int_0^(pi/4)tan^(2)theta\ d theta`

`=(pi/4 xx 1 -0)-int sec^(2)theta-1\ d theta`

`=pi/4-[tan theta-theta]_0^(pi/4)`

`=pi/4-[(1-pi/4)-0]`

`=pi/2-1`

Mechanics, EXT2 M1 2022 HSC 15a

A machine is lifted from the floor of a room using two ropes. The two ropes ensure that the horizontal components of the forces are balanced at all times. It is assumed that at all times the machine moves vertically upwards at a constant velocity.

The machine is located in a room with height `h` metres.

One of the ropes is attached to the point `P` on the machine and to the fixed point `C` on the ceiling of the room. The point `C` is a distance `d` metres to the left of `P`. Let the vertical distance from `P` to the ceiling be `ℓ` metres and let `\theta` be the angle this rope makes with the horizontal.

The other rope is attached to the point `P` and to the fixed point `F` on the floor of the room. The point `F` is a distance `2 d` metres to the right of `P`. Let `\phi` be the angle this rope makes with the horizontal.

Let the tension in the first rope be `T_1` newtons, the tension in the second rope be `T_2` newtons, the mass of the machine be `M` kilograms and the acceleration due to gravity be `g\ text{m s}^(-2)`.
 

  1. By considering horizontal and vertical components of the forces at `P`, show that
  2.            `tan theta=tan phi+(Mg)/(T_(2)cos phi)`   (3 marks)

  3. Hence, or otherwise, show that the point `P` cannot be lifted to a position `{2 h}/{3}` metres above the floor.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   
       

`text{Resolving forces horizontally:}`

`T_1cos\ theta=T_2cos\ phi\ \ text{… (1)}`

`text{Resolving forces vertically:}`

`T_1\ sin\ theta=Mg + T_2\ sin\ phi\ \ text{… (2)}`
 

`text{Divide}\ (2) -: (1):`

`tan\ theta` `=(Mg + T_2\ sin\ phi)/(cos\ phi)`  
  `=(Mg)/(T_2\ cos\ phi)+tan\ phi`  

 


Mean mark (i) 56%.

ii.   `text{Using diagram in part (i):}`

`tan\ theta=l/d,\ \ tan\ phi=(h-l)/(2d)`

`text{Using part (i):}`

`l/d` `=(Mg)/(T_2\ cos\ phi)+tan\ phi`  
  `=(Mg)/(T_2\ cos\ phi)+(h-l)/(2d)`  

 
`text{S}text{ince}\ \ (Mg)/(T_2\ cos\ phi)>0`

`=> l/d` `>(h-l)/(2d)`  
`l` `>(h-l)/2`  
`(3l)/2` `>h/2`  
`l` `>h/3`  
`h-l` `>h-h/3`  
`h-l` `>(2h)/3`  

 
`:.\ text{Point}\ P\ text(cannot be lifted to)\ (2h)/3\ text(metres above floor.)`


♦♦♦ Mean mark (ii) 27%.

BIOLOGY, M6 2019 HSC 19-20 MC

Use the following diagram to answer Questions 19-20.

The diagram shows how CRISPR/Cas9 can be used as a new tool for genetic engineering. This technology has dramatically improved scientists' ability to successfully modify genomes.
 

Question 19

What type of structure must Cas9 be?

  1. Enzyme
  2. mRNA
  3. Ribosome
  4. tRNA

 

Question 20

Scientists have been able to use biotechnology to 'cut and paste' DNA for decades.

Why would the new CRISPR/Cas9 technology have improved the scientists' success in cutting DNA of specific genes?

  1. Cas9 is able to combine with specific DNA.
  2. Cas 9 has an active site that cuts target DNA.
  3. gRNA has the same nucleotides as the target DNA.
  4. gRNA has nucleotides complementary to the target DNA.
Show Answers Only

Q19. `A`

Q20. `D`

Show Worked Solution

Question 19

→  Cas9 performs a ‘cutting mechanism’, a property of DNA helicase, a known enzyme.

→ Cas9 also contains a recognition site which it uses to perform its tasks, another property of enzymes.

`=>A`
 

Question 20

→ Cas9 has gRNA with complementary bases to the target gene, as displayed in the diagram above.

→ The improvement is due to this feature, as scientists can now use multiple gRNA sequences to cut various amounts of genes and various recognition sites.

`=>D`


♦♦♦ Mean mark (Q20) 23%.

BIOLOGY, M6 2019 HSC 15 MC

A germ-line mutation is known to have occurred.

How is it possible that there has been no noticeable change in the phenotype of the offspring?

  1. The mutation occurred in a stretch of RNA.
  2. The mutation occurred in a protein-coding region.
  3. The mutation occurred in a stretch of non-coding DNA.
  4. The mutation did not affect the DNA sequence of any gametes.
Show Answers Only

`C`

Show Worked Solution

→ For the mutation to not be noticeable in the offspring, it must have occurred within junk DNA (i.e. large sections of non-coding DNA with no known purpose).

`=>C`

BIOLOGY, M8 2019 HSC 12 MC

The glucose tolerance test is used to investigate the control of glucose in the human body. Patients consume 75 g of glucose and their blood glucose is monitored.

Type 2 diabetes is a condition where the cells of the body do not respond adequately to insulin.

Which graph could represent the results of glucose tolerance tests in a non-diabetic person and a person with untreated Type 2 diabetes?
  

Show Answers Only

`D`

Show Worked Solution

→ In a type 2 diabetic, glucose cannot be actively stored as glycogen as fast as a person with a functioning pancreases, leading to a build up of glucose in the bloodstream.

→ Graph D is the only graph that shows a type 2 diabetics BGL above that of a non-diabetic. 

`=>D`


Mean mark 55%.

BIOLOGY, M5 2019 HSC 11 MC

Which of the following is always true of a mutation that produces a dominant allele?

  1. It will be lethal in a population.
  2. It will be expressed in heterozygous individuals.
  3. It will only be expressed in homozygous individuals.
  4. It will spread more quickly through the population than a recessive allele.
Show Answers Only

`B`

Show Worked Solution

→ If a mutation occurs in a dominant allele, this mutation will be expressed in the individual (or in the offspring if it occurs in germline cells).

`=>B`

BIOLOGY, M5 2019 HSC 4 MC

The figure shows a part of a DNA molecule.

How many complete nucleotides are shown in the figure?

  1. 1
  2. 2
  3. 4
  4. 8
Show Answers Only

`D`

Show Worked Solution

→ Nucleotides are made up of one base, one sugar and one phosphate.

→ The diagram therefore shows 8 nucleotides.

`=>D`


Mean mark 56%.

CHEMISTRY, M6 2019 HSC 24

A conductometric titration was undertaken to determine the concentration of a barium hydroxide solution. The solution was added to 250.0 mL of standardised 1.050 × 10 ¯ 3 mol L ¯1 hydrochloric acid solution. The results of the titration are shown in the conductivity graph.
 


 

  1. Explain the shape of the titration curve.   (3 marks)

  2. The equivalence point was reached when a volume of 17.15 mL of barium hydroxide was added.
  3. Calculate the concentration of barium hydroxide (in mol L¯1), and give a relevant chemical equation.  (4 marks)

Show Answers Only

a.   Titration curve shape:

→ The conductivity of a solution of hydrochloric acid is initially very high due to the high concentration of ions in the solution.

→ As barium hydroxide is added, the conductivity decreases because the concentration of \(\ce{H+}\) decreases due to the neutralization reaction with \(\ce{OH-}\) ions.

→ The conductivity reaches a minimum at the equivalence point, which is when all the \(\ce{H+}\) ions have been removed and the solution only contains \(\ce{Ba^2+}\) and \(\ce{Cl-}\) ions.

→ These ions are much less mobile than \(\ce{H+}\) or \(\ce{OH-}\) ions, so the conductivity is lower at the equivalence point.

→ After the equivalence point, the conductivity increases as more barium hydroxide is added, resulting in an increase in the concentration of \(\ce{OH-}\) ions and therefore an increase in conductivity.
 

b.   \(\ce{2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2H2O(l)} \)

\[\ce{Ba(OH)2 }\] `=7.653 xx 10^{-3}\ text{mol L}^{-1}`  
Show Worked Solution

a.   Titration curve shape:

→ The conductivity of a solution of hydrochloric acid is initially very high due to the high concentration of ions in the solution.

→ As barium hydroxide is added, the conductivity decreases because the concentration of \(\ce{H+}\) decreases due to the neutralization reaction with \(\ce{OH-}\) ions.

→ The conductivity reaches a minimum at the equivalence point, which is when all the \(\ce{H+}\) ions have been removed and the solution only contains \(\ce{Ba^2+}\) and \(\ce{Cl-}\) ions.

→ These ions are much less mobile than \(\ce{H+}\) or \(\ce{OH-}\) ions, so the conductivity is lower at the equivalence point.

→ After the equivalence point, the conductivity increases as more barium hydroxide is added, resulting in an increase in the concentration of \(\ce{OH-}\) ions and therefore an increase in conductivity.
 


♦ Mean mark (a) 46%.

b.   \(\ce{2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2H2O(l)} \)

\(\ce{n(HCl) = c \times V = 1.050 \times 10^{-3} \times 250 = 2.625 \times 10^{-4} mol }\)

\(\ce{n(Ba(OH)2) = 0.5 \times n(HCl) = 1.3125 \times 10^{-4} mol}\)

\[\ce{[Ba(OH)2] = \frac{n}{V}}\] `=(1.3125 xx 10^{-4})/(0.01715)`  
  `=7.653 xx 10^{-3}\ text{mol L}^{-1}`  

CHEMISTRY, M7 2019 HSC 23

The following apparatus was used in an experiment to determine the molar enthalpy of combustion of ethanol.
 


 

  1. Calculate the experimental molar enthalpy of combustion `(Delta_(c) H)` of ethanol when 0.370 g ethanol was used to raise the water temperature from 18.5°C to 30.0°C.   (4 marks)

  2. Upon replication, the molar enthalpy of combustion obtained in the experiment was consistently much lower than the accepted value.
  3. Explain ONE change that could be made to the experiment that would improve the accuracy of the obtained value.   (2 marks)

Show Answers Only

a.  `-628\ text{kJ mol}^(-1)`

b.   Improvement to experiment accuracy (one of many possible answers):

→ The molar enthalpy of combustion measured in the experiment was not very accurate due to heat being lost to the surroundings.

→ To improve the accuracy, it would be helpful to move the spirit burner closer to the beaker to reduce heat loss.

Show Worked Solution
a.    `text{n(ethanol)}` `= text{m}/text{MM}`
    `=0.370/46.068`
    `=0.008032\ text{mol}`

 
`q=mC DeltaT = 105\ text{g} xx 4.18\ text{J g}^(-1)\ text{K}^(-1) xx (30-18.5)\ text{K} = 5047.35\ text{J}`

`(Delta_(c) H)` `=-q/n`  
  `=- 5047.35/0.008032`  
  `=-628\ 405\ text{J mol}^(-1)`  
  `=-628\ text{kJ mol}^(-1)\ \ text{(3 sig fig)}`  

 

b.   Improvement to experiment accuracy (one of many possible answers):

→ The molar enthalpy of combustion measured in the experiment was not very accurate due to heat being lost to the surroundings.

→ To improve the accuracy, it would be helpful to move the spirit burner closer to the beaker to reduce heat loss.

CHEMISTRY, M7 2019 HSC 21

  1. The structural formula for 2-methylpropan-2-ol is shown in the table.
  2. Draw one structural isomer of this alcohol and state its name.   (2 marks)

     

     

  3. The structural formulae for two compounds are shown below.
     

   
 

  1. Why are these two compounds classed as functional group isomers?   (2 marks)

  2. A chemical test is required to distinguish between the isomers in part (b).
  3. Identify a suitable test and explain the expected observations.   (3 marks)

Show Answers Only

a.   Successful answers should have one of the following:

 

b.  Functional Group isomers

→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.

→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
 

c.   Tollens’ Test:

→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).

→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.

→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.

Show Worked Solution

a.   Successful answers should have one of the following:

b.  Functional Group isomers

→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.

→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
 

c.   Tollens’ Test:

→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).

→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.

→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.


♦♦ Mean mark (c) 39%.

BIOLOGY, M7 2019 HSC 10 MC

A group of islands are separated from each other by large stretches of water. Each island has its own policy on quarantine.

A nearby country is experiencing an outbreak of an infectious disease in its cattle.

An investigation is to be designed to find which of the quarantine policies operating on the islands is the most effective.

Which of the following would be a suitable design feature of the investigation?

  1. The control is the smallest island.
  2. The control is the number of infected cattle.
  3. The independent variable is quarantine policy.
  4. The independent variable is the number of infected cattle.
Show Answers Only

`C`

Show Worked Solution

To measure effectiveness of quarantine policies:

→ The independent variable must be the different quarantine policies being investigated

`=>C`

BIOLOGY, M5 2019 HSC 9 MC

Which of the following is an advantage for animals using internal fertilisation rather than external fertilisation?

  1. It prevents dehydration of gametes.
  2. It involves large numbers of gametes.
  3. It relies on adaptations such as mating rituals.
  4. It allows gametes to combine to form unique offspring.
Show Answers Only

`A`

Show Worked Solution

→ The damp environment utilised in internal fertilisation prevents dehydration of the gametes, which would lead to the death of the cells.

`=>A`


Mean mark 57%.

BIOLOGY, M7 2019 HSC 7 MC

Two types of bacteria were isolated from a patient's throat swab and grown in pure culture on separate agar plates. On each plate there were FOUR different antibiotic discs, `W, X, Y` and `Z`.

The photograph shows the plates seven days later.
 

Which antibiotic should be used to treat the patient?

  1. `W`
  2. `X`
  3. `Y`
  4. `Z`
Show Answers Only

`A`

Show Worked Solution

→ The area of each circle shows the amount of bacterial growth each antibiotic has inhibited.

→ The larger the circle, the more effective that antibiotic is at inhibiting bacterial growth.

→ Antibiotic `W` has the most consistently large disc, and should be used to treat the patient.

`=>A`

BIOLOGY, M6 2019 HSC 3 MC

The diagram shows the impact of birds feeding on a population of beetles over time.
 

Which of the following accounts for the change in the beetle population?

  1. Mutation
  2. Gene flow
  3. Genetic drift
  4. Environmental pressure
Show Answers Only

`D`

Show Worked Solution

→ The diagram shows the process of natural selection.

→ The environmental pressure of a predator eliminates individuals from a population based on phenotype.

`=>D`

CHEMISTRY, M6 2019 HSC 15 MC

What is the concentration of hydroxide ions (in mol L ¯1) in a solution that has a pH of 8.53?

  1. `3.0 xx10^(-9)`
  2. `3.4 xx10^(-6)`
  3. `5.5`
  4. `3.0 xx10^(5)`
Show Answers Only

`B`

Show Worked Solution
`text{pOH}` `=14-text{pH}`  
  `=14-8.53`  
  `=5.47`  

 

\(\ce{[OH-]}\) `=10^(text{–pH})`  
  `=10^(-5.47)`  
  `=3.4 xx10^(-6)`  

 
`=>B`

BIOLOGY, M8 2022 HSC 27

The incidence of cervical cancer in the Australian population is shown in the graph from 1985-2015. Public health campaigns commenced during this period to reduce the incidence of cervical cancer. This included a national screening program to detect pre-cancerous cells and a vaccination program against human papillomavirus (HPV) which causes most cervical cancers.

Evaluate the success of these campaigns in reducing the incidence of cervical cancer in Australian women. Include reference to the data in your answer.  (6 marks)

Show Answers Only

In 1991, a national screening program for women aged 20-69 years was introduced to try and lower the incidence of cervical cancer.

→ Screening programs work by detecting pre-cancerous cells so they can be treated before they spread.

→ The graph above shows that the incidence of cervical cancer almost halved from 1990 to 2002, showing that the screening program was most likely successful in reducing incidence.
 

In 2007 and 2013, HPV vaccine programs for girls and boys respectively were introduced.

→ HPV is a virus that is responsible for most cervical cancers. A national vaccine rollout may help to reduce the incidence of HPV, thus reducing cervical cancers.

→ The graph only shows the incidence of cervical cancer per 100 000 women, so based on this graph alone, it is impossible to determine if the boys program was successful.

→ After the girls rollout in 2007, the incidence of cervical cancer has not shown noticeable fluctuations.

→ Extending the data up until the present day should be undertaken for a more accurate analysis of the programs’ effectiveness.

Show Worked Solution

In 1991, a national screening program for women aged 20-69 years was introduced to try and lower the incidence of cervical cancer.

→ Screening programs work by detecting pre-cancerous cells so they can be treated before they spread.

→ The graph above shows that the incidence of cervical cancer almost halved from 1990 to 2002, showing that the screening program was most likely successful in reducing incidence.
 

In 2007 and 2013, HPV vaccine programs for girls and boys respectively were introduced.

→ HPV is a virus that is responsible for most cervical cancers. A national vaccine rollout may help to reduce the incidence of HPV, thus reducing cervical cancers.

→ The graph only shows the incidence of cervical cancer per 100 000 women, so based on this graph alone, it is impossible to determine if the boys program was successful.

→ After the girls rollout in 2007, the incidence of cervical cancer has not shown noticeable fluctuations.

→ Extending the data up until the present day should be undertaken for a more accurate analysis of the programs’ effectiveness.


Mean mark 61%.

BIOLOGY, M7 2022 HSC 26

Jelly Bush honey has been used by Aboriginal Peoples to treat cuts, sores and burns. Recent studies have shown that Jelly Bush honey has a very high level of methylglyoxal which is known to help fight infection.

A scientist wants to test the effectiveness of Jelly Bush honey using agar plates.

Design a safe procedure that the scientist could use in a laboratory to investigate the effectiveness of Jelly Bush honey as a pharmaceutical to inhibit bacterial growth.  (5 marks)

Show Answers Only

STEP 1: Sterilise 5 agar plates to remove contamination.

STEP 2: Set up the 5 agar plates as 1 blank (control), 1 with just bacteria* (control), 1 with bacteria and Jelly Bush honey, 1 with pure methylglyoxal and 1 with bacteria and a known antibiotic which will inhibit the bacterial growth.

→ *Bacteria must be a pathogen encountered during infection that is also native to Australia.

STEP 3: Store all agar plates in an incubator at 37°C and 1 atm for 5 days.

STEP 4: Record the amount of bacterial growth in each agar plate in comparison to the pure bacteria plate.

STEP 5: Repeat 3 times to ensure no other variables are responsible for inhibiting bacterial growth.
 

Experiment summary:

→ This experiment is a valid way of determining Jelly Bush honey as a pharmaceutical as it tests it against pure methylglyoxal (and 2 controls).

→ This experiment should show evidence whether methylglyoxal or some other factor(s) within the honey contribute to its ability to fight infection.

Show Worked Solution

STEP 1: Sterilise 5 agar plates to remove contamination.

STEP 2: Set up the 5 agar plates as 1 blank (control), 1 with just bacteria* (control), 1 with bacteria and Jelly Bush honey, 1 with pure methylglyoxal and 1 with bacteria and a known antibiotic which will inhibit the bacterial growth.

→ *Bacteria must be a pathogen encountered during infection that is also native to Australia.

STEP 3: Store all agar plates in an incubator at 37°C and 1 atm for 5 days.

STEP 4: Record the amount of bacterial growth in each agar plate in comparison to the pure bacteria plate.

STEP 5: Repeat 3 times to ensure no other variables are responsible for inhibiting bacterial growth.
 

Experiment summary:

→ This experiment is a valid way of determining Jelly Bush honey as a pharmaceutical as it tests it against pure methylglyoxal (and 2 controls).

→ This experiment should show evidence whether methylglyoxal or some other factor(s) within the honey contribute to its ability to fight infection.


♦ Mean mark 54%.

CHEMISTRY, M8 2019 HSC 14 MC

A molecule, `text{C}_(4) text{H}_(9) text{Cl}`, is analysed. The \(\ce{^1H NMR} \) spectrum of this molecule is shown.
 

What is the structural formula of this molecule?
 


 


 


 

Show Answers Only

`A`

Show Worked Solution

→ The \(\ce{^1H NMR} \) spectrum exhibits a single signal, which suggests that all the hydrogens in the sample are present in the same chemical environment.

→ Therefore, the answer A, where all the hydrogens are chemically equivalent is correct.

`=>A`

BIOLOGY, M8 2022 HSC 31a

Studies have shown that lung cancer can be linked to environmental causes.

In one historical study across 29 health districts in Japan, non-smoking married women aged 40 and above were followed up for 14 years (1966-79) and annual mortality rates for lung cancer were assessed according to the smoking habits of their husbands. They were compared to women who smoked.

The results are shown.

  1. Evaluate the method used in this epidemiological study.   (4 marks)

  2. Justify conclusions that could be drawn from the results of the study.   (3 marks)

Show Answers Only

i.    Sample size:

→ Large sample (36 302 average amount per group).

→ Widespread (29 different health districts) which means results will be more accurate.

 
Other factors to consider:

→ The study was conducted over an extended period of 14 years, which will again lead to more accurate results.

→ The study also has a control group (non-smokers with non-smoker husbands) which provides a base for the study.

→ Overall, it can be concluded that this study is a valid way of showing the relationship between first and second hand smoking and mortality in women.
 

ii.   Conclusions of study:

→ The control group (non-smoker women with non-smoker husbands) shows that 8.7 women per 100 000 have died to lung cancer.

→ This could be due to friends or other family members who smoke or exposure to smoke in the workplace.

→ Other possible contributing factors in Japan, such as poor air quality can have an effect but be outside the scope of the study.

→ The study also shows that non-smoker women who have smoker husbands have almost double the mortality of women in the control group, and women who smoke themselves with smoker husbands have triple the mortality rate.

→ With this information, it can be concluded that both smoking and exposure to second hand smoke, increases a woman’s chance of death by lung cancer.

Show Worked Solution

i.    Sample size:

→ Large sample (36 302 average amount per group).

→ Widespread (29 different health districts) which means results will be more accurate.

 
Other factors to consider:

→ The study was conducted over an extended period of 14 years, which will again lead to more accurate results.

→ The study also has a control group (non-smokers with non-smoker husbands) which provides a base for the study.

→ Overall, it can be concluded that this study is a valid way of showing the relationship between first and second hand smoking and mortality in women.
 


♦ Mean mark (i) 44%.
ii.   Conclusions of study:

→ The control group (non-smoker women with non-smoker husbands) shows that 8.7 women per 100 000 have died to lung cancer.

→ This could be due to friends or other family members who smoke or exposure to smoke in the workplace.

→ Other possible contributing factors in Japan, such as poor air quality can have an effect but be outside the scope of the study.

→ The study also shows that non-smoker women who have smoker husbands have almost double the mortality of women in the control group, and women who smoke themselves with smoker husbands have triple the mortality rate.

→ With this information, it can be concluded that both smoking and exposure to second hand smoke, increases a woman’s chance of death by lung cancer.

ENGINEERING, PPT 2020 HSC 23b

A diagram of the front steering system of a bicycle is shown.
 


 

  1. A rider leans forward so that 55% of their mass is exerted as a force, as shown in the diagram. A turning moment of 132.19 Nm is generated about the front axle.
  2. What is the mass of the rider?   (3 marks)
  1. In overcoming the resistance due to gravity, drag and rolling resistance, a rider applies a force of 500 N on the pedals. This allows the rider to cover, in 5 seconds, a distance of 20 metres from the starting point.
  2. How much power was produced by the rider?  (2 marks)
Show Answers Only

i.   `95\ text{kg}`

ii.  `2000\ text{W}`

Show Worked Solution

i.  
         

`sin20°` `= d/740`  
`d` `= 740 xx sin20°`  
  `= 253.09\ text{mm}`  

 

`M` `= Fxxd`  
`F` `= 132.19/0.25309`  
  `= 522.5\ text{N}`  

 

`55\text{%} xx F_(text{total})` `= 522.5\ text{N}`  
`F_(text{total})` `= 522.5/0.55`  
  `= 950\ text{N}`  

 

`:. M_(text{rider})` `= 950/10`  
  `= 95\ text{kg}`  

♦ Mean mark 47%.

ii.   `F=500\ text{N}, \ s=20\ text{m},\ t=5\ text{sec}`

`P` `= Fxxv`  
  `= F xx s/t`  
  `= 500 xx 20/5`  
  `= 2000\ text{W}`  

ENGINEERING, PPT 2020 HSC 23a

The diagrams show two bicycles, one from 1900 and one from 2020.


The basic components of the bicycle frame have not changed for 120 years. However, manufacturing and materials technologies have evolved significantly during this period.

Compare TWO technologies used in the production of modern bicycle frames to those used in earlier times.

Show Answers Only

Successful answers should include two of the following:

→ Lightweight composite materials (such as carbon fibre) have replaced heavy steel tubing in modern bike frames due to their high strength to weight ratio.

→ Seamless tubing is now used instead of rolling a flat sheet into a circular shape and welding it together.

→ Advanced understanding of aerodynamics has led to bike frames being designed to better reduce wind resistance.

Show Worked Solution

Successful answers should include two of the following:

→ Lightweight composite materials (such as carbon fibre) have replaced heavy steel tubing in modern bike frames due to their high strength to weight ratio.

→ Seamless tubing is now used instead of rolling a flat sheet into a circular shape and welding it together.

→ Advanced understanding of aerodynamics has led to bike frames being designed to better reduce wind resistance.

ENGINEERING, AE 2020 HSC 22c

A periodic aircraft maintenance inspection is to be carried out.

Identify and justify an appropriate test that could be used to check the airworthiness of the landing gear.  (4 marks)

Show Answers Only

Ultrasonic testing:

 → Can determine whether there are internal cracks/flaws within the landing gear.

 → This is appropriate as the equipment is portable, the testing is non-destructive and can be done on site.
 

Other appropriate tests could include:

Dye penetrant or fluorescent inspection tests

 → Reveals surface cracks/flaws and is non-destructive.

Magnetic particle/eddy current inspection

 → Non-destructive and reveals surface and shallow flaws through discrepancies in electric and magnetic fields.

Visual inspection

 → For any obvious faults/damages/cracks.

Radiography

 → Used to find deep internal flaws, non-destructive.

Show Worked Solution

Ultrasonic testing:

 → Can determine whether there are internal cracks/flaws within the landing gear.

 → This is appropriate as the equipment is portable, the testing is non-destructive and can be done on site.
 

Other appropriate tests could include:

Dye penetrant or fluorescent inspection tests

 → Reveals surface cracks/flaws and is non-destructive.

Magnetic particle/eddy current inspection

 → Non-destructive and reveals surface and shallow flaws through discrepancies in electric and magnetic fields.

Visual inspection

 → For any obvious faults/damages/cracks.

Radiography

 → Used to find deep internal flaws, non-destructive.

ENGINEERING, AE 2020 HSC 22b

Describe the use of different types of materials for the wings of aircraft over the last 100 years.  (3 marks)

Show Answers Only

→ The skin of aircraft wings were originally made from fabric.

→ Sheet metal alloys were then used to replace the fabric as they provided strength to the skins’ structure that the fabric did not.

→ These sheet metals primarily included aluminium, titanium and steel alloys.

→ More recently composite materials have been used for their high strength to weight ratios.

→ These include carbon fibre epoxy resin, GLARE and other fibre reinforced metal laminates.
 

Other answers could include:

→ Material developments regarding the frame of the wing where timber was originally used, followed by steel, aluminium/titanium alloys and modern day composite materials.

Show Worked Solution

→ The skin of aircraft wings were originally made from fabric.

→ Sheet metal alloys were then used to replace the fabric as they provided strength to the skins’ structure that the fabric did not.

→ These sheet metals primarily included aluminium, titanium and steel alloys.

→ More recently composite materials have been used for their high strength to weight ratios.

→ These include carbon fibre epoxy resin, GLARE and other fibre reinforced metal laminates.
 

Other answers could include:

→ Material developments regarding the frame of the wing where timber was originally used, followed by steel, aluminium/titanium alloys and modern day composite materials.

ENGINEERING, AE 2020 HSC 22a

Outline how ONE innovation in aeronautical engineering has contributed to improved performance of modern aircraft.  (2 marks)

Show Answers Only

Successful answers should include one of the following:

Retrofitting of winglets 

    → Reduces drag and turbulences, improves fuel efficiency.

Wing shape 

    → Designed to reduce drag and in turn increase fuel efficiency.

Turbofan engines 

    → Have replaced turbojets in modern aircraft due to increased thrust and fuel efficiency.

Materials such as Carbon Fibre Epoxy Composite

    → Used in modern aircraft due to their high specific strength, enabling planes to carry more fuel and/or passengers.

Show Worked Solution

Successful answers should include one of the following:

Retrofitting of winglets 

    → Reduces drag and turbulences, improves fuel efficiency.

Wing shape 

    → Designed to reduce drag and in turn increase fuel efficiency.

Turbofan engines 

    → Have replaced turbojets in modern aircraft due to increased thrust and fuel efficiency.

Materials such as Carbon Fibre Epoxy Composite

    → Used in modern aircraft due to their high specific strength, enabling planes to carry more fuel and/or passengers.

BIOLOGY, M7 2022 HSC 30

Malaria is a disease transmitted by a mosquito vector. There has been no effective vaccine developed.

Refer to the maps below to answer Question 30.
 

   

   

Discuss possible reasons for the differences in the distribution of malaria and its vector. Include detailed reference to Map 1 and Map 2.  (7 marks)

Show Answers Only

→ Malaria is a disease which is transmitted via a female mosquito.

→ The plasmodium responsible for the disease can multiply inside a mosquito.

→ When it bites an individual, it enters the bloodstream via the mosquito’s saliva.

→ The presence of infected individuals and the malaria vector (mosquito) in the same regional area is an obvious overlap between the two maps.

 
Regions with mosquito vector but no infected individuals

→ Areas such as Japan, Western Europe and North America have the mosquito vector responsible for malaria but no infected individuals.

This can be due to:

→ Accessibility to medical treatment such as pharmaceuticals which can treat infected individuals.

→ Better mosquito control eg effective repellent.

→ Better quarantine of infected individuals.

 
Regions with no mosquito vector but infected individuals

→ Areas such as China, sections of Northern and Southern Africa and areas of the North-Western coast of South America have cases of malaria but no mosquito vectors in the area.

This can be due to:

→ Migration of infected individuals to the areas stated above from surrounding areas where both the vector and disease are present.

Show Worked Solution

→ Malaria is a disease which is transmitted via a female mosquito.

→ The plasmodium responsible for the disease can multiply inside a mosquito.

→ When it bites an individual, it enters the bloodstream via the mosquito’s saliva.

→ The presence of infected individuals and the malaria vector (mosquito) in the same regional area is an obvious overlap between the two maps.

 
Regions with mosquito vector but no infected individuals

→ Areas such as Japan, Western Europe and North America have the mosquito vector responsible for malaria but no infected individuals.

This can be due to:

→ Accessibility to medical treatment such as pharmaceuticals which can treat infected individuals.

→ Better mosquito control eg effective repellent.

→ Better quarantine of infected individuals.

 
Regions with no mosquito vector but infected individuals

→ Areas such as China, sections of Northern and Southern Africa and areas of the North-Western coast of South America have cases of malaria but no mosquito vectors in the area.

This can be due to:

→ Migration of infected individuals to the areas stated above from surrounding areas where both the vector and disease are present.


Mean mark 56%.

BIOLOGY, M6 2022 HSC 29

Bt cotton has been genetically engineered to produce an insecticide that kills cotton bollworm. It was introduced to a cotton-producing nation in 2002.

The graphs show trends of national cotton yield, % Bt cotton grown, total insecticide use, insecticide use to control bollworms and insecticide use to control another insect pest (hemiptera).
 

 

  1. Explain ONE reason why cotton yield changed between 2002 and 2013.  (2 marks)

  2. To what extent do the data support the use of Bt cotton as a method of disease control in cotton?  (5 marks)

Show Answers Only

a.   Cotton yield increase between 2002-2013:

→ The cotton yield increased because the pesticide produced by Bt cotton will decrease the likelihood of the plant contracting Bollworm, which kills the plant.

→ This increases the percentage of healthy cotton, increasing the yield of cotton that is harvested.
  

b.   Use of Bt cotton for disease control:

→ Bt cotton is effective in producing plants resistant to Bollworm.

→ The amount of pesticide used for Bollworm dropped to less than 500 tonnes by 2008, and approaching zero by 2013.

→ This saves farmers money which can be spent on other measures effective for disease control, such as better hygiene in the workplace.

→ The top graph also shows an increase in cotton yield, further supporting the statement above.
 

→ Bt cotton however is not effective as a method of disease control against Hemipteran.

→ The amount of pesticide used for Hemipteran increased by 8000 tonnes from 2002 to 2013. This increase has taken the total pesticide used in 2013 to above that used in 2002.

→ While Bt cotton may be effective against Bollworm, other pests such as Hemipteran can now thrive and cause problems to cotton farms.

→ More studies with other cotton pest species need to be done in a similar style to prove if Bt cotton would be a well-rounded and effective method of disease control.

Show Worked Solution

a.   Cotton yield increase between 2002-2013:

→ The cotton yield increased because the pesticide produced by Bt cotton will decrease the likelihood of the plant contracting Bollworm, which kills the plant.

→ This increases the percentage of healthy cotton, increasing the yield of cotton that is harvested.
  

b.   Use of Bt cotton for disease control:

→ Bt cotton is effective in producing plants resistant to Bollworm.

→ The amount of pesticide used for Bollworm dropped to less than 500 tonnes by 2008, and approaching zero by 2013.

→ This saves farmers money which can be spent on other measures effective for disease control, such as better hygiene in the workplace.

→ The top graph also shows an increase in cotton yield, further supporting the statement above.
 

→ Bt cotton however is not effective as a method of disease control against Hemipteran.

→ The amount of pesticide used for Hemipteran increased by 8000 tonnes from 2002 to 2013. This increase has taken the total pesticide used in 2013 to above that used in 2002.

→ While Bt cotton may be effective against Bollworm, other pests such as Hemipteran can now thrive and cause problems to cotton farms.

→ More studies with other cotton pest species need to be done in a similar style to prove if Bt cotton would be a well-rounded and effective method of disease control.


♦ Mean mark (b) 47%.

BIOLOGY, M5 2022 HSC 28a

Soon after the discovery of the structure of DNA, scientists began investigating how DNA replicated. Three models were proposed.
 

Identify which of the models above is currently accepted.   (1 mark)

Show Answers Only

→ Model 2 is correct.

→ Each replicated copy contains 1 original strand and 1 new strand.

Show Worked Solution

→ Model 2 is correct.

→ Each replicated copy contains 1 original strand and 1 new strand.

BIOLOGY, M8 2022 HSC 24

Birth defects in humans can be caused by chromosomal abnormalities.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Maternal} \quad & \textit{Prevalence of}\\
\textit{age} & \textit{chromosomal}\\
\text{(years)} & \textit{abnormalities}\\
\text{} & \rule[-1ex]{0pt}{0pt}\text{(per 1000 births)}\\
\hline \rule{0pt}{2.5ex}20 \rule[-1ex]{0pt}{0pt}& 1.5 \\
\hline \rule{0pt}{2.5ex}30 \rule[-1ex]{0pt}{0pt}& 3 \\
\hline \rule{0pt}{2.5ex}35 \rule[-1ex]{0pt}{0pt}& 8 \\
\hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 22 \\
\hline \rule{0pt}{2.5ex}45 \rule[-1ex]{0pt}{0pt}& 38 \\
\hline
\end{array}

  1. Draw a suitable graph of the data provided in the table. Include a suitable line of best fit.  (3 marks)
     

  1. Outline the trend shown in the data.  (2 marks)

  2. Explain the cause of a type of chromosomal mutation.  (3 marks)

Show Answers Only

a. 

 

b.   Data trend:

→ After 20 years of age, the older a woman is when she gives birth, the chance of her child having a chromosomal abnormality increases exponentially.
 

c.   Cause of chromosomal mutation:

→ Aneuploidy is a chromosomal mutation where the child contains either an extra or a deficit of a chromosome.

→ This is a result of non-disjunction, where in meiosis, chromosomes fail to separate.

→ This can produce a gamete with an extra chromosome (trisomy, 2N+1) or one missing a certain chromosome (monosomy, 2N-1).

→ Down Syndrome (Trisomy 21) is the result of a child containing an extra chromosome 21.
 

Other answers could include

→ Polyploidy, duplications of whole sets of chromosomes (XN).

→ Rearrangement, deletion, insertion, translocation or duplications of sections of a chromosome.

Show Worked Solution

a. 

b.   Data trend:

→ After 20 years of age, the older a woman is when she gives birth, the chance of her child having a chromosomal abnormality increases exponentially.


♦ Mean mark (a) 54%.
Mean mark (b) 56%.

c.   Cause of chromosomal mutation:

→ Aneuploidy is a chromosomal mutation where the child contains either an extra or a deficit of a chromosome.

→ This is a result of non-disjunction, where in meiosis, chromosomes fail to separate.

→ This can produce a gamete with an extra chromosome (trisomy, 2N+1) or one missing a certain chromosome (monosomy, 2N-1).

→ Down Syndrome (Trisomy 21) is the result of a child containing an extra chromosome 21.
 

Other answers could include

→ Polyploidy, duplications of whole sets of chromosomes (XN).

→ Rearrangement, deletion, insertion, translocation or duplications of sections of a chromosome.


♦♦ Mean mark (c) 32%.

BIOLOGY, M6 2022 HSC 23

  1. Outline the process of artificial pollination.  (2 marks)

  2. Explain a possible outcome of the use of artificial pollination on subsequent populations.  (2 marks)

Show Answers Only

a.   Artificial pollination Process

→ This is the process of manually pollinating plants with pollen of known characteristics in order to obtain offspring with similar qualities.

→ First, anthers of the recipient plant are cut off in order to remove chance of self pollination.

→ Then, pollen is manually transferred to the stigma of the recipient plant by hand using a brush, or using machinery.
   

b.   The use of the same pollen in order to achieve desirable plants will end up:-

→ decreasing the variability amongst the species

→ decreasing the genetic diversity.

Show Worked Solution

a.   Artificial pollination Process

→ This is the process of manually pollinating plants with pollen of known characteristics in order to obtain offspring with similar qualities.

→ First, anthers of the recipient plant are cut off in order to remove chance of self pollination.

→ Then, pollen is manually transferred to the stigma of the recipient plant by hand using a brush, or using machinery.
   

b.   The use of the same pollen in order to achieve desirable plants will end up:-

→ decreasing the variability amongst the species

→ decreasing the genetic diversity.


♦ Mean mark (b) 49%.

BIOLOGY, M5 2022 HSC 22

Eggplant fruit comes in three colours: dark purple, white and violet. A genetic cross between the dark purple and white eggplants will always result in the violet phenotype.

What phenotypic ratio would you expect to see when two violet offspring are crossed? Show your working. (3 marks)

Show Answers Only

→ Violet is a mix between dark purple and white.

→ Thus proving this characteristic follows a co-dominance inheritance

→ Therefore, violet eggplants have a Pp genotype.

→ `P : Pp xx Pp`

\begin{array} {|c|c|c|}\hline  & P & p \\ \hline P & PP & Pp \\ \hline p & Pp & pp \\  \hline \end{array}

 
Genotype: PP – Pp – pp  (1 – 2 – 1)     

Phenotype: Dark Purple – Violet – White  (1 – 2 – 1)

Show Worked Solution

→ Violet is a mix between dark purple and white.

→ Thus proving this characteristic follows a co-dominance inheritance

→ Therefore, violet eggplants have a Pp genotype.

→ `P : Pp xx Pp`

\begin{array} {|c|c|c|}\hline  & P & p \\ \hline P & PP & Pp \\ \hline p & Pp & pp \\  \hline \end{array}

 
Genotype: PP – Pp – pp  (1 – 2 – 1)     

Phenotype: Dark Purple – Violet – White  (1 – 2 – 1)