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BIOLOGY, M8 2025 HSC 25

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

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

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

Show Answers Only

Evaluation Judgment:

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

Supporting Evidence:

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

Conclusion:

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

Show Worked Solution

Evaluation Judgment:

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

Supporting Evidence:

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

Conclusion:

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

BIOLOGY, M6 2025 HSC 23

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

Show Answers Only

Similarities:

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

Differences:

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

Show Worked Solution

Similarities:

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

Differences:

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

Calculus, EXT2 C1 2025 HSC 13a

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

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

Show Answers Only

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

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

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

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

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

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

 

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

Show Worked Solution

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

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

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

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

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

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

 

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

Algebra, STD2 A4 2025 HSC 39

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

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

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

Show Answers Only

\(5.4 \ \text{mg}\)

Show Worked Solution

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

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

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

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

   

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

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

Measurement, STD2 M7 2025 HSC 38

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

Measurement, STD2 M1 2025 HSC 32

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

Financial Maths, STD2 F1 2025 HSC 31

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

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

BIOLOGY, M6 2025 HSC 16 MC

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

The identical offspring are clones of

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

\(A\)

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

Other Options:

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

BIOLOGY, M8 2025 HSC 13 MC

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

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

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

Show Answers Only

\(A\)

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

Other Options:

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

BIOLOGY, M6 2025 HSC 12 MC

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

 

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

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

\(D\)

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

Other Options:

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

BIOLOGY, M7 2025 HSC 11 MC

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

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

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

\(C\)

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

Other Options:

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

BIOLOGY, M7 2025 HSC 10 MC

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

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

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

\(C\)

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

Other Options:

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

BIOLOGY, M5 2025 HSC 8 MC

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

What is the reason for this occurrence?

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

\(D\)

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

Other Options:

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

BIOLOGY, M8 2025 HSC 7 MC

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

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

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

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

Show Answers Only

\(C\)

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

Other Options:

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

BIOLOGY, M5 2025 HSC 4 MC

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

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

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

\(B\)

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

Other Options:

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

Financial Maths, STD1 F2 2025 HSC 25

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

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

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

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

Show Answers Only

\(\text{Option A:}\)

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

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

  
\(\text{Option B:}\)

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

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

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

Show Worked Solution

\(\text{Option A:}\)

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

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

  
\(\text{Option B:}\)

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

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

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

Financial Maths, STD1 F3 2025 HSC 24

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

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

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

Calculate Kat’s monthly repayment.   (5 marks)

Show Answers Only

\($835.31\)

Show Worked Solution

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

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

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

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

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

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

Algebra, STD1 A3 2025 HSC 23

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

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

 
Write TWO linear equations that represent the relationship below.

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

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

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

Show Worked Solution

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

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

Measurement, STD1 M3 2025 HSC 22

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

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

a.    \(h=3.2\ \text{cm}\)

b.    \(39.1\%\)

Show Worked Solution

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

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

\(\text{Using Pythagoras:}\)

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

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

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

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

Measurement, STD1 M3 2025 HSC 20

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

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

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

a.    \(75\ \text{metres}\)

b.    \(106\ \text{metres}\)

c.    \(225^\circ\)

Show Worked Solution

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

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

b.   \(\text{Using Pythagoras:}\)

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

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

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

  
c.    \(\text{Since }AB=BC:\)

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

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

Financial Maths, STD1 F1 2025 HSC 19

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

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

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

  1. Using the table, calculate Alex’s tax payable.   (3 marks)

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

Show Answers Only

a.    \($36\ 320\)

b.    \($2972\)

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

 

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

Measurement, STD1 M4 2025 HSC 18

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

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

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

b.    \(80\ \text{km/h}\)

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

 

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

Algebra, STD1 A2 2025 HSC 16

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

Statistics, STD1 S3 2025 HSC 15

A researcher is using the statistical investigation process to investigate a possible relationship between average number of minutes per day a person spends watching television, and the average number of minutes per day the person spends exercising.

  1. State the statistical question being posed.   (1 mark)

Participants were asked to record the number of minutes they spent watching television each day and the number of minutes they spent exercising each day. The averages for each participant were recorded and graphed, and a line of best fit was included.
 

  1. From the graph, identify the dependent variable.   (1 mark)

  2. Describe the bivariate dataset in terms of its form and direction.   (2 marks)

  3. The points \((0, 70)\) and \((60, 10)\) lie on the line of best fit. By first plotting these points on the graph, find the gradient and the \(y\)-intercept of the line of best fit.   (3 marks)
  4. Explain why it is NOT appropriate to extrapolate the line of best fit to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)
Show Answers Only

a.    \(\text{How does the average daily time spent watching television}\)

\(\text{relate to the average daily time spent exercising?}\)
 

b.    \(\text{Dependent variable: Average minutes per day exercising, or }y.\)
 

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d.   \(\text{Gradient}=-1\)
 

e.    \(\text{The extrapolation of the graph past 70 minutes produces}\)

\(\text{negative average minutes per day exercising (impossible).}\)

Show Worked Solution

a.    \(\text{How does the average daily time spent watching television}\)

\(\text{relate to the average daily time spent exercising?}\)
 

b.    \(\text{Dependent variable:}\)

\(\text{Average minutes per day exercising, or }y.\)
 

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d. 


 

\(y-\text{intercept = 70}\)

\(\text{Gradient}=\dfrac{\text{rise}}{\text{run}}=\dfrac{-60}{60}=-1\)
 

e.    \(\text{The extrapolation of the graph past 70 minutes produces}\)

\(\text{negative average minutes per day exercising (impossible).}\)

Algebra, STD1 A3 2025 HSC 5 MC

A baker makes and sells cakes.

The straight-line graphs represent cost \((C )\) and revenue \((R)\) in dollars, and \(n\) is the number of cakes.
 

What profit will the baker make by selling 6 cakes?

  1. $10
  2. $20
  3. $40
  4. $60
Show Answers Only

\(A\)

Show Worked Solution

\(\text{When }n=6\)

\(\text{Revenue}\) \(=10\times 6=60\)
\(\text{Cost}\)  \(=20+5\times 6=50\)
\(\therefore\ \text{Profit }\) \(=$60-$50=$10\)

  
\(\Rightarrow A\)

Financial Maths, STD2 F4 2025 HSC 36

The graph shows the salvage value of a car over 5 years.

The salvage values are based on the declining-balance method.

By what amount will the car’s value depreciate during the 10th year?   (4 marks)

Show Answers Only

\($1476.40\)

Show Worked Solution

\(\text{Find}\ r:\)

\(\text{When}\ \ n=1, \ S=$44\ 000\ \ \text{(see graph)}\)

\(S\) \(=V_0(1-r)^n\)
\(44\ 000\) \(=55\ 000(1-r)^1\)
\(\dfrac{44\ 000}{55\ 000}\) \(=1-r\)
\(1-r\) \(=0.8\)
\(r\) \(=1-0.8=0.20\)

  
\(\text{Find \(S\) when}\ \ n=9\ \ \text{and}\ \ n=10:\)

\(S_9=55\ 000(1-0.20)^{9}=$7381.97504\)

\(S_{10}=55\ 000(1-0.20)^{10}=$5905.5800\)

\(S_9-S_{10}=$7381.9750-$5905.580=$1476.40\ \text{(nearest cent)}\)
 

\(\therefore\ \text{The car’s value will depreciate by \$1476.40 in the 10th year.}\)

Algebra, STD2 A4 2025 HSC 30

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

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

 
Write TWO linear equations that represent the relationship below.

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

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

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

Show Worked Solution

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

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

Statistics, STD2 S1 2025 HSC 28

The heights of students in a class were recorded.

The results for this class are displayed in the cumulative frequency graph shown.
 

 

The shortest student in this class is 130 cm and the tallest student is 180 cm.

Construct a box-plot for this class in the space below.   (3 marks)
 

Show Answers Only

Show Worked Solution

\(Q_1(7.5 \ \text{students })=135\)

\(Q_3(22.5 \ \text{students })=160\)

\(\text{Median (15 students )}=140\)
 

Financial Maths, STD2 F4 2025 HSC 27

A credit card has an interest-free period of 45 days from and including the date of purchase. Interest is charged on purchases made, compounding daily at a rate of 13.74% per annum, from and including the day following the interest-free period.

Concert tickets were purchased for a total of $392 using this credit card.

Full payment was made on the 68th day from the date of purchase. There were no other purchases on this credit card.

What was the total interest charged when the account was paid in full?   (3 marks)

Show Answers Only

\(\text{Interest charged}\ =\$ 3.41\)

Show Worked Solution

\(\text{Day 1-45: no interest is charged}\)

\(\text{Day 46-68: interest charged (23 days)}\)

\(\text{Daily interest rate}=\dfrac{13.74}{365} \%=\dfrac{13.74}{365 \times 100}\)

\(\text{Amount owing}=392\left(1+\dfrac{13.74}{365 \times 100}\right)^{23}=\$ 395.41\)

\(\text{Interest charged}=395.41-392=\$ 3.41\)

Measurement, STD2 M1 2025 HSC 26

A toy has a curved surface on the top which has been shaded as shown. The toy has a uniform cross-section and a rectangular base.
 

  1. Use two applications of the trapezoidal rule to find an approximate area of the cross-section of the toy.   (2 marks)

  2. The total surface area of the plastic toy is 1300 cm².
  3. What is the approximate area of the curved surface?   (2 marks)

  4. The measurements shown on the diagram are given to the nearest millimetre.
  5. What is the percentage error of the measurement of 10.2 cm? Give your answer correct to 3 significant figures.   (2 marks)

Show Answers Only

a.   \(34.68 \ \text{cm}^2\)

b.   \(582.64 \ \text{cm}^2\)

c.   \(0.490 \%\)

Show Worked Solution

a.    \(\begin{array}{|c|c|c|c|}
\hline\rule{0pt}{2.5ex} \quad x \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 5.1 \quad & \quad 10.2 \quad\\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 6 & 3.8 & 0 \\
\hline
\end{array}\)

\(A\) \(\approx \dfrac{h}{2}\left(y_0+2y_1+y_2\right)\)
  \(\approx \dfrac{5.1}{2}\left(6+2 \times 3.8+0\right)\)
  \(\approx 34.68 \ \text{cm}^2\)

 

b.    \(\text{Toy has 5 sides.}\)

\(\text{Area of base}=10.2 \times 40=408 \ \text{cm}^2\)

\(\text{Area of rectangle}=6.0 \times 40=240 \ \text{cm}^2\)

\(\text{Approximated areas}=2 \times 34.68=69.36 \ \text{cm}^2\)

\(\therefore \ \text{Area of curved surface}\) \(=1300-(408+240+69.36)\)
  \(=582.64 \ \text{cm}^2\)

 

c.   \(\text{Convert cm to mm:}\)

\(10.2 \ \text{cm}\times 10=102 \ \text{mm}\)

\(\text{Absolute error}=\dfrac{1}{2} \times \text{precision}=0.5 \ \text{mm}\)

\(\% \ \text{error}=\dfrac{0.5}{102} \times 100=0.490 \%\)

Vectors, EXT1 V1 2025 HSC 9 MC

The vectors \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) have magnitudes 3, 5 and 7 respectively.
 

Given that \(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}=\underset{\sim}{0}\), what is the size of angle \(\theta\) between \(\underset{\sim}{a}\) and \(\underset{\sim}{b}\) ?

  1. \(\dfrac{\pi}{6}\)
  2. \(\dfrac{\pi}{3}\)
  3. \(\dfrac{2 \pi}{3}\)
  4. \(\dfrac{5 \pi}{6}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Find}\ \theta\ \text{using:}\ \ \cos\,\theta = \dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}} \, \abs{\underset{\sim}{b}}}\)

\(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}=0 \ \Rightarrow\  \underset{\sim}{a}+\underset{\sim}{b}=-\underset{\sim}{c}\)

\(\abs{\underset{\sim}{a}+\underset{\sim}{b}}=\abs{-\underset{\sim}{c}}=7\)

\(\abs{\underset{\sim}{a}+\underset{\sim}{b}}^2\) \(=(\underset{\sim}{a}+\underset{\sim}{b})(\underset{\sim}{a}+\underset{\sim}{b})=49\)
\(49\) \(=\underset{\sim}{a} \cdot \underset{\sim}{a}+2 a \cdot \underset{\sim}{b}+\underset{\sim}{b} \cdot \underset{\sim}{b}\)
\(49\) \(=\abs{\underset{\sim}{a}}^2+2 a \cdot b+\abs{\underset{\sim}{b}}^2\)
\(49\) \(=9+2 \underset{\sim}{a} \cdot \underset{\sim}{b}+25\)
\(2 \underset{\sim}{a} \cdot \underset{\sim}{b}\) \(=15\)
\(\underset{\sim}{a} \cdot \underset{\sim}{b}\) \(=\dfrac{15}{2}\)

\(\cos \theta\) \(=\dfrac{\frac{15}{2}}{3 \times 5}=\dfrac{1}{2}\)
\(\therefore \theta\) \(=\dfrac{\pi}{3}\)

 
\(\Rightarrow B\)

Vectors, EXT1 V1 2025 HSC 14c

The hands of an analogue clock are \(OA\) and \(OB\),

where \(A\) is \(\left(\sin \left(\dfrac{\pi t}{360}\right), \cos \left(\dfrac{\pi t}{360}\right)\right), B\) is \(\left(2 \sin \left(\dfrac{\pi t}{30}\right), 2 \cos \left(\dfrac{\pi t}{30}\right)\right)\),

\(O\) is the origin, and  \(t \geq 0\)  is the number of minutes past midnight.

Find the values of \(t\) when the hands are perpendicular for the first and second time after midnight. Give your answers to 3 decimal places.   (3 marks)

Show Answers Only

\(t=16.364, 49.091 \ \text{mins}\).

Show Worked Solution

\(\text{Express \(OA\) and \(OB\) as vectors:}\)

\(\overrightarrow{O A}=\displaystyle \binom{\sin \left(\frac{\pi t}{360}\right)}{\cos \left(\frac{\pi t}{360}\right)}, \quad \overrightarrow{O B}=\displaystyle \binom{2\, \sin \left(\frac{\pi t}{30}\right)}{2\, \cos \left(\frac{\pi t}{30}\right)}\)

\(\text{When hands are parallel,} \ \ \overrightarrow{OA} \cdot \overrightarrow{OB}=0:\)

\(\sin \left(\dfrac{\pi t}{360}\right) \times 2\, \sin \left(\dfrac{\pi t}{30}\right)+\cos \left(\dfrac{\pi t}{360}\right) \times 2\, \cos \left(\dfrac{\pi t}{30}\right)=0\)

\(\cos \left(\dfrac{\pi t}{30}-\dfrac{\pi t}{360}\right)\) \(=0\)
\(\cos \left(\dfrac{11 \pi t}{360}\right)\) \(=0\)

 

\(\dfrac{11 \pi t}{360}=\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\)

\(t=\dfrac{\pi}{2} \times \dfrac{360}{11 \pi}=16.364 \ \text{mins}\)

\(t=\dfrac{3 \pi}{2} \times \dfrac{360}{11 \pi}=49.091 \ \text{mins}\).

Vectors, EXT1 V1 2025 HSC 14b

Points \(A\) and \(B\) lie vertically above the origin. Point \(A\) is higher than point \(B\) such that  \(\dfrac{OA}{O B}=k\), where  \(k>1\).

A particle is projected horizontally from point \(A\) with velocity \(U\ \text{ms}^{-1}\). After \(T\) seconds, another particle is projected horizontally from point \(B\) with velocity \(V\ \text{ms}^{-1}\). The two particles land on the ground in the same place.
 

Show that the ratio \(\dfrac{V}{U}\) depends only on \(k\).   (4 marks)

Show Answers Only

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

Show Worked Solution

\(\text{Let} \ \ OB=h \ \Rightarrow \ OA=k \times OB=k h\)

\({\underset{\sim}{a}}_A=\displaystyle \binom{0}{-g}, \quad {\underset{\sim}{v}}_A=\displaystyle \binom{U}{-g t_1}, \quad {\underset{\sim}{r}}_A=\displaystyle \binom{U t_1}{k h-\frac{1}{2} g t_1^2}\)

 
\(\text{Time of flight for} \ B\left(t_2\right) \neq \text{Time of flight for} \ A\left(t_1\right)\)

\({\underset{\sim}{a}}_B=\displaystyle \binom{0}{-g}, \quad {\underset{\sim}{v}}_B=\displaystyle\binom{V}{-g t_2}, \quad {\underset{\sim}{r}}_B=\displaystyle\binom{V t_2}{h-\frac{1}{2} g t_2^2}\)
 

\(\text{Time of flight for} \ A\left(t_1\right)\) :

\(kh-\dfrac{1}{2} g t_1^2=0 \ \Rightarrow \  t_1^2=\dfrac{2kh}{g} \ \Rightarrow \ t_1=\sqrt{\dfrac{2 kh}{g}}\)

\(\text{Range of} \ A=Ut_1=U \sqrt{\dfrac{2 k h}{g}}\)

 
\(\text{Time of flight for}\ B\left(t_2\right):\)

\(h-\dfrac{1}{2} g t_2^2=0 \ \Rightarrow \ t_2^2=\dfrac{2 h}{g}\ \Rightarrow \  t_2=\sqrt{\dfrac{2 h}{g}}\)

\(\text{Range of} \ B=Vt_2=V \sqrt{\dfrac{2h}{g}}\)

\(\text{Equating ranges:}\)

\(V \sqrt{\dfrac{2h}{g}}\) \(=U \sqrt{\dfrac{2 kh}{g}}\)  
\(\dfrac{V}{U}\) \(=\sqrt{\dfrac{2 kh}{g}} \times \sqrt{\dfrac{g}{2 h}}=\sqrt{k}\)  

 
\(\therefore \dfrac{V}{U} \ \text{ratio depends only on} \ k.\)

Combinatorics, EXT1 A1 2025 HSC 14a

Prove that the product of any seven distinct factors of 60 must be a multiple of 60.   (2 marks)

Show Answers Only

\(\text {Consider the factors of 60:}\)

\(\{1,60\},\{2,30\},\{3,20\},\{4,15\},\{5,12\},\{6,10\}\)

\(\text{Select 1 factor from each pair}\)

\(\Rightarrow 6 \ \text{numbers}\)

\(\text{The 7th number chosen must complete a pair whose product}=60.\)

\(\therefore \ \text{By PHP, the product of any 7 distinct numbers is a multiple of 60.}\)

Show Worked Solution

\(\text {Consider the factors of 60:}\)

\(\{1,60\},\{2,30\},\{3,20\},\{4,15\},\{5,12\},\{6,10\}\)

\(\text{Select 1 factor from each pair}\)

\(\Rightarrow 6 \ \text{factors}\)

\(\text{The 7th number chosen must complete a pair whose product}=60.\)

\(\therefore \ \text{By PHP, the product of any 7 distinct factors is a multiple of 60.}\)

Statistics, EXT1 S1 2025 HSC 13d

A bag contains counters, some of which are green.

One hundred trials of an experiment are run. In each trial, one counter is selected from the bag at random and its colour noted. The counter is returned to the bag after each trial.

Let \(X\) be the random variable representing the number of times that a green counter is selected.

Given that  \(E(X)=20\)  and  \(P(X \geq k)=0.0668\), find the value of \(k\). Use of a standard normal approximation table is allowed.   (4 marks)

Show Answers Only

\(k=26\)

Show Worked Solution

\(E(X)=20, n=100\)

\(E(X)=np \ \ \Rightarrow \ \ 20=100 p\ \ \Rightarrow \ \ p=\dfrac{1}{5}\)

\(X \sim B(n, p) \sim B\left(100, \dfrac{1}{5}\right)\) 

\(P(X \geqslant k)=0.0668\) 

\(1-0.0668=0.9332\)

\(\text{Using Normal Dist Table (ref = 0.9332):}\)

\(P(X \geqslant k)=P(z \geqslant 1.5)\)

\(\operatorname{Var}(X)=np(1-p)=20\left(1-\dfrac{1}{5}\right)=16\)

\(\sigma(X)=\sqrt{16}=4\)
 

\(\text{Using}\ \ z=\dfrac{x-\mu}{\sigma}:\)

\(\dfrac{k-20}{4}\) \(=1.5\)
\(k-20\) \(=6\)
\(k\) \(=26\)

Combinatorics, EXT1 A1 2025 HSC 13b

Eight guests are to be seated at a round table. If two of these guests refuse to sit next to each other, how many seating arrangements are possible?   (2 marks)

Show Answers Only

\(\text{Total combinations}\ = 3600\)

Show Worked Solution

\(\text{Strategy 1}\)

\(\text{If no restrictions:}\)

\(\text{Total combinations\(=7!\)}\)

\(\text{If two people must sit together:}\)

\(\text{Combinations\(=6!2!\)}\)

\(\text{If two people refuse to sit together:}\)

\(\text{Combinations\(=7!-6!2!=3600\)}\)
 

\(\text{Strategy 2}\)

\(\text{Sit one of the refusers in any seat.}\)

\(\text{Possible seats for other refuser = 5}\)

\(\text{Combinations for other 6 people}\ =6!\)

\(\text{Total combinations}\ = 5 \times 6! = 3600\)

Vectors, EXT1 V1 2025 HSC 8 MC

Points \(A\) and \(B\) have non-zero, non-parallel position vectors \(\underset{\sim}{a}\) and \(\underset{\sim}{b}\) respectively.

Point \(C\) has position vector  \(\underset{\sim}{c}=3 \underset{\sim}{a}-2 \underset{\sim}{b}\).

The points \(A, B\) and \(C\) lie on the same line.

Which of the following must be true?

  1. Point \(A\) always lies between Points \(B\) and \(C\).
  2. Point \(B\) always lies between Points \(A\) and \(C\).
  3. Point \(C\) always lies between Points \(A\) and \(B\).
  4. The order of the points cannot be determined.
Show Answers Only

\(A\)

Show Worked Solution

\(\underset{\sim}{c}=3 \underset{\sim}{a}-2 \underset{\sim}{b}\ \ldots\  (1)\)

\(\text{Since \(A, B\) and \(C\) are collinear:}\)

\(\overrightarrow{A C}\) \(=\lambda \overrightarrow{A B}\)
\(\underset{\sim}{c}-\underset{\sim}{a}\) \(=\lambda(\underset{\sim}{b}-\underset{\sim}{a})\)
\(\underset{\sim}{c}\) \(=\underset{\sim}{a}+\lambda(\underset{\sim}{b}-\underset{\sim}{a})\)
\(\lambda\) \(=-2\ \text{(see (1) above)}\)

 

\(\Rightarrow A\)

Measurement, STD1 M5 2025 HSC 26

A scale of \(1 : 50\) is used to draw a rectangular area on a 2 mm grid as shown. The actual rectangular area is to be tiled.

The tiles cost $150 per square metre and the tiler orders 15% extra tiles to allow for cutting and breakage.

The tiler charges $90 per hour and will take 20 hours to complete the tiling.

Calculate the total cost of the tiles and tiling. Give your answer to the nearest dollar.   (4 marks)

Show Answers Only

\($6105.60\)

Show Worked Solution

\(\text{Using scale where each grid is 2mm × 2 mm:}\)

\(2\ \text{mm}\ =50 \times 2 = 100\ \text{mm in actual length}\)

\(\text{100 mm = 10 cm}\)

\(\Rightarrow \ \text{i.e. each grid represents 10 cm in actual length.}\)

\(\text{Width}\ = 52 \times 10 = 520\ \text{cm}\ = 5.2\ \text{m}\)

\(\text{Height}\ = 48 \times 10 = 480\ \text{cm}\ = 4.8\ \text{m}\)

\(\text{Area}=5.2 \times 4.8=24.96\ \text{m}^2\)

\(\text{Cost of tiles}=24.96\times $150\times 1.15=$4305.60\)

\(\text{Labour Cost}=90\times 20=$1800\)
 

\(\therefore\ \text{Total Cost}=$4305.60+$1800=$6105.60\)

Calculus, 2ADV C3 2025 HSC 9 MC

The diagram shows the graph of  \(y=f^{\prime}(x)\).
 

Given  \(f(1)=6\), which interval includes the best estimate for \(f(1.1)\) ?

  1. \([6.2,6.4)\)
  2. \([6.0,6.2)\)
  3. \([5.8,6.0)\)
  4. \([5.6,5.8)\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Gradient of \(f(x)\)  at  \(x=1\)  is 2 (see graph).}\)

\(\text{Gradient of \(f(x)\)  at  \(x=1.1\)  is slightly below 2 (see graph).}\)

\(\text{As \(x\) increases 0.1 (from 1.0 to 1.1), \(y\) will increase less than 0.2 units.}\)

\(\therefore f(1.1) \in [6.0,6.2)\)

\(\Rightarrow B\)

Networks, STD2 N3 2025 HSC 22

A network of pipes with one cut is shown. The number on each edge gives the capacity of that pipe in L/min.
 

  1. What is the capacity of the cut shown?   (1 mark)

  2. The diagram shows a possible flow for this network of pipes.
     

    1. What is the value of \(x\)? Give a reason for your answer.   (2 marks)

    2. Which of the pipes in the flow are at full capacity?   (1 mark)

    3. The maximum flow for this network is 50 L/min.
    4. Which path of pipes could have an increase in flow of 2 L/min to achieve the maximum flow?   (1 mark)

Show Answers Only

a.    \(\text{Capacity} =62\)

b.i.   \(x=30\) 

b.ii. \(DE, DG, CF \ \text{and} \ FG\)

b.iii.  \(ACEG\)

Show Worked Solution

a.    \(\text{Capacity} =26+24+12=62\)
 

b.i.   \(\text{Inflow into} \ C\) \(=\text{Outflow from} \ C\)
  \(x\) \(=5+13+12\)
    \(=30\)

 

b.ii. \(DE, DG, CF \ \text{and} \ FG\)
 

b.iii.  \(ACEG\)

Statistics, STD2 S4 2025 HSC 25

In a research study, participants were asked to record the number of minutes they spent watching television and the number of minutes they spent exercising each day over a period of 3 months. The averages for each participant were recorded and graphed.
 

 

  1. Describe the bivariate dataset in terms of its form and direction.   (2 marks)
  2. Form:  ..................................................................
  3. Direction:  ............................................................

The equation of the least-squares regression line for this dataset is

\(y=64.3-0.7 x\)

  1. Interpret the values of the slope and \(y\)-intercept of the regression line in the context of this dataset.   (2 marks)

  2. Jo spends an average of 42 minutes per day watching television.
  3. Use the equation of the regression line to determine how many minutes on average Jo is expected to exercise each day.   (1 mark)

  4. Explain why it is NOT appropriate to extrapolate the regression line to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)

Show Answers Only

a.    \(\text{Form: Linear. Direction: Negative}\)

Show Worked Solution

a.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

b.    \(\text{Slope}=-0.7\)

\(\text{This means that for each added minute of watching television per day, a participant, on average,}\)

\(\text{will exercise for 0.7 minutes less.}\)

\(y \text{-intercept}=64.3\)

\(\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}\)
 

c.    \(\text{At} \ \ x=42:\)

\(y=64.3-0.7 \times 42=34.9\)

\(\therefore \ \text{Jo is expected to exercise for 34.9 minutes}\)
 

d.    \(\text{At} \ \  x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

\(\text{The model predicts a negative value for time spent exercising, which is not possible.}\)

Measurement, STD2 M7 2025 HSC 24

The population of snails in a garden is approximately 90.

One night Bobbie collected 18 snails from the garden. He tagged each snail and released it back into the garden.

The next night 20 snails were captured from the garden.

Approximately how many of the snails in the second sample are expected to have a tag?   (2 marks)

Show Answers Only

\(\text{4 snails}\)

Show Worked Solution

\(\text{Snail population}=90\)

\(\text{After 1st night tagging:}\)

\(\text{% snails with tag}=\dfrac{18}{90} = 20\% \)

\(\text{Expected snails with tags \((s)\):}\)

\(s=20\% \times 20 = 4\)

Financial Maths, STD2 F4 2025 HSC 23

Company \(\text{A}\) and Company \(\text{B}\) both issue an annual dividend per share as shown in the table.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} {Company} \rule[-1ex]{0pt}{0pt} & {Current \ share \ price} \rule[-1ex]{0pt}{0pt} & {Annual \ dividend}\\
 {}  & {} & { per \ share}\\
\hline
\rule{0pt}{2.5ex} \text{A} \rule[-1ex]{0pt}{0pt} & \text{\$25.43} \rule[-1ex]{0pt}{0pt} & \text{\$4.92}\\
\hline
\rule{0pt}{2.5ex} \text{B} \rule[-1ex]{0pt}{0pt} & \text{\$2.13} \rule[-1ex]{0pt}{0pt} & \text{45c}\\
\hline
\end{array}
Based on the dividend yield, which company would be better to invest in?   (2 marks)

Show Answers Only

\(\text{Div yield (A)} =\dfrac{4.92}{25.43}=0.1934 = 19.3\% \)

\(\text{Div yield (B)} =\dfrac{0.45}{2.13}=0.211 = 21.1\% \)

\(\text{Since Div yield (B) > Div yield (A), it is better to invest in company B.}\)

Show Worked Solution

\(\text{Div yield (A)} =\dfrac{4.92}{25.43}=0.1934 = 19.3\% \)

\(\text{Div yield (B)} =\dfrac{0.45}{2.13}=0.211 = 21.1\% \)

\(\text{Since Div yield (B) > Div yield (A), it is better to invest in company B.}\)

Measurement, STD2 M6 2025 HSC 14 MC

Points \( M \) and \( P \) are the same distance from a third point \(R\).

The bearing of \( M \) from \( R \) is 017° and the bearing of \( P \) from \( R \) is 107°.

Which of the following best describes the bearing of \(P\) from \(M\)?

  1. Between 000° and 090°
  2. Exactly 090°
  3. Between 090° and 180°
  4. Exactly 180°
Show Answers Only

\(C\)

Show Worked Solution

\(\angle MRP = 107-17=90^{\circ}\)

\(\angle RMP = \angle MPR = 45^{\circ}\ \text{(equilateral triangle)}\)

\(\text{Bearing of \(P\) from \(M\)}\ = 180-28=152^{\circ}\)

\(\Rightarrow C\)

Algebra, STD2 A4 2025 HSC 11 MC

The thickness of the skin of a spherical balloon varies inversely with the surface area of the balloon.

What would be the effect on the thickness of the skin if the radius of the balloon is doubled?

  1. Divided by 2
  2. Multiplied by 2
  3. Divided by 4
  4. Multiplied by 4
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Thickness}\ (T) \propto \dfrac{1}{\text{S.A. balloon}}\)

\(\text{Since}\ \ \text{S.A.}\ = 4\pi\,r^2:\)

\(T \propto\ \dfrac{1}{r^2}\)

\(\text{If \(r\) is doubled, \(T\) will be reduced to}\ \dfrac{1}{4}T.\)

\(\Rightarrow C\)

Measurement, STD1 M5 2025 HSC 10 MC

The ratio of the dimensions of a model car to the dimensions of an actual car is \(1:64\). The actual car has a length of 4.9 m.

What is the length of the model car in cm, correct to 1 decimal place?

  1. 3.1
  2. 7.7
  3. 13.1
  4. 59.1
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Actual length}=4.9\ \text{m}\ =490\ \text{cm}\)

\(\therefore\ \text{Model car length}\ =490\times\dfrac{1}{64}=7.65625\approx 7.7\ \text{cm}\)

 \(\Rightarrow B\)

Probability, STD1 S2 2025 HSC 8 MC

A spinner made up of 4 colours is spun 100 times. The frequency histogram shows the results.
 

Which of these spinners is most likely to give the results shown?
 

Show Answers Only

\(A\)

Show Worked Solution
\(P(\text{White})\) \(=\dfrac{50}{100}=\dfrac{1}{2}\)
\(P(\text{Red})\) \(=\dfrac{25}{100}=\dfrac{1}{4}\)  
\(P(\text{Yellow})\) \(=\dfrac{15}{100}=\dfrac{3}{20}\)
\(P(\text{Green})\) \(=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}\)

 
\(\text{Eliminate Options B and D as white}\ \neq \dfrac{1}{2}\ \text{of spinner.}\)

\(\text{Eliminate Option C as red}\ \neq \dfrac{1}{4}\ \text{of spinner.}\)

\(\Rightarrow A\)

Measurement, STD2 M7 2025 HSC 7 MC

There are 960 students at a high school.

Using stratified sampling, 240 students from the whole school are to be chosen for a survey.

If there are 200 students in Year 12, how many Year 12 students should be chosen?

  1. 4
  2. 5
  3. 50
  4. 60
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Stratified sampling}\ \ \Rightarrow\ \ \text{Fraction of Y12 in school = same fraction in sample}\)

\(\text{Yr 12 students in school}\ =\dfrac{200}{960} = \dfrac{5}{24}\)

\(\text{Number in sample}\ = \dfrac{5}{24} \times 240 = 50\)

\(\Rightarrow C\)

Calculus, 2ADV C4 2011 HSC 4d v1

  1. Differentiate  `y=sqrt(16 -x^2)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, find  `int (8x)/sqrt(16 -x^2)\ dx`.    (2 marks)

Show Answers Only
  1. `- x/sqrt(16\ -x^2)`
  2. `-8 sqrt(16\ -x^2) + C`
Show Worked Solution
IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(16-x^2)^(1/2)`.
a.    `y` `= sqrt(16 -x^2)`
    `= (16 -x^2)^(1/2)`

 

`dy/dx` `=1/2 xx (16 -x^2)^(-1/2) xx d/dx (16 -x^2)`
  `= 1/2 xx (16 -x^2)^(-1/2) xx -2x`
  `= – x/sqrt(16 -x^2)`

 

b.    `int (8x)/sqrt(16 – x^2)\ dx` `= -8 int (-x)/sqrt(16 -x^2)\ dx`
    `= -8 (sqrt(16 -x^2)) + C`
    `= -8 sqrt(16 -x^2) + C`

Trigonometry, 2ADV T1 2025 HSC 29

The point \(T\) is the peak of a mountain and the point \(O\) is directly below the mountain's peak. The point \(Y\) is due east of \(O\) and the angle of elevation of \(T\) from \(Y\) is 60°. The point \(F\) is 4 km south-west of \(Y\). The points \(O, Y\) and \(F\) are on level ground. The angle of elevation of \(T\) from \(F\) is 45°.
 

  1. Let the height of the mountain be \(h\).
  2. Show that  \(O Y=\dfrac{h}{\sqrt{3}}\).   (1 mark)

  3. Hence, or otherwise, find the value of \(h\), correct to 2 decimal places.   (3 marks)

  4. Find the bearing of point \(O\) from point \(F\), correct to the nearest degree.   (3 marks)

Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.   \(h=3.03 \ \text{km}\)

c.   \(\text {Bearing of \(O\) from \(F\)}=021^{\circ} \)

Show Worked Solution

a.    \(\text{In}\ \triangle TOY:\)

\(\tan 60^{\circ}\) \(=\dfrac{h}{OY}\)  
\(OY\) \(=\dfrac{h}{\tan 60^{\circ}}\) \(=\dfrac{h}{\sqrt{3}}\)

 

b.   \(\text{In}\ \triangle TOF:\)

\(\tan 45^{\circ}=\dfrac{h}{OF} \ \Rightarrow \ OF=h\)

\(\text{Since \(Y\) is due east of \(O\) and \(F\) is south-west of \(Y\):}\)

\(\angle OYF =45^{\circ}\)
 

\(\text{Using cosine rule in} \ \triangle OYF:\)

\(OY^2+YF^2-2 \times OY \times YF\, \cos 45^{\circ}\) \(=OF^2\)
\(\left(\dfrac{h}{\sqrt{3}}\right)^2+4^2-2 \times \dfrac{h}{\sqrt{3}} \times 4 \times \dfrac{1}{\sqrt{2}}\) \(=h^2\)
\(\dfrac{h^2}{3}+16-\dfrac{8}{\sqrt{6}} h\) \(=h^2\)
\(\dfrac{2}{3} h^2+\dfrac{8}{\sqrt{6}} h-16\) \(=0\)

 

\(h\)

 \(=\dfrac{\dfrac{-8}{\sqrt{6}}+\sqrt{\frac{64}{6}+4 \times \frac{2}{3} \times 16}}{2 \times \frac{2}{3}}\ \ \ (h>0)\)
  \(=3.0277 \ldots\)
  \(=3.03 \ \text{km (to 2 d.p.)}\)

 

c.    \(\text {Using sine rule in} \ \ \triangle OYF:\)

\(\dfrac{\sin\angle FOY}{4}\) \(=\dfrac{\sin 45^{\circ}}{3.03}\)  
\(\sin \angle F O Y\) \(=\dfrac{4 \times \sin 45^{\circ}}{3.03}=0.93347\)  
\(\angle FOY\) \(=180-\sin^{-1}(0.93347)=180-68.98 \ldots = 111^{\circ} \ \text{(angle is obtuse)}\)  

 
\(\angle FOS^{\prime}=111-90=21^{\circ}\)

\(\angle \text{N}^{\prime}FO=21^{\circ}(\text {alternate})\)

\(\therefore \ \text {Bearing of \(O\) from \(F\)}=021^{\circ} \)

Calculus, 2ADV C4 2025 HSC 27

The shaded region is bounded by the graph  \(y=\left(\dfrac{1}{2}\right)^x\), the coordinate axes and  \(x=2\).
 

  1. Use two applications of the trapezoidal rule to estimate the area of the shaded region.   (2 marks)

  2. Show that the exact area of the shaded region is  \(\dfrac{3}{4 \ln 2}\).   (2 marks)

  3. Using your answers from part (a) and part (b), deduce  \(e<2 \sqrt{2}\).   (2 marks)

Show Answers Only

a.   \(A\approx \dfrac{9}{8}\ \text{units}^2\)
 

b.     \(\text{Area}\) \(=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x\)
    \(=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2\)
    \(=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}\)
    \(=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}\)
    \(=\dfrac{3}{4 \ln 2}\)

 

c.    \(\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}\)

\(\text{between (0,1) and \((2,\dfrac{1}{4})\)}\)

\(\Rightarrow \ \text{Area using trap rule > Actual area}\)

\(\dfrac{9}{8}\) \(>\dfrac{3}{4 \ln 2}\)  
\(36\, \ln 2\) \(>24\)  
\(\ln 2\) \(>\dfrac{2}{3}\)  
\(e^{\small \dfrac{2}{3}}\) \(>2\)  
\(e\) \(>2^{\small \dfrac{3}{2}}\)  
\(e\) \(>2 \sqrt{2}\)  
Show Worked Solution

a.  

\begin{array}{|c|c|c|c|}
\hline \ \ x \ \  & \ \ 0 \ \  & \ \ 1 \ \  & \ \ 2 \ \  \\
\hline y & 1 & \dfrac{1}{2} & \dfrac{1}{4} \\
\hline
\end{array}

\(A\) \(\approx \dfrac{h}{2}\left[1 \times 1+2 \times \dfrac{1}{2}+1 \times \dfrac{1}{4}\right]\)
  \(\approx \dfrac{1}{2}\left(\dfrac{9}{4}\right)\)
  \(\approx \dfrac{9}{8}\ \text{units}^2\)
 
b.     \(\text{Area}\) \(=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x\)
    \(=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2\)
    \(=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}\)
    \(=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}\)
    \(=\dfrac{3}{4 \ln 2}\)

  

c.    \(\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}\)

\(\text{between (0,1) and \((2,\dfrac{1}{4})\)}\)

\(\Rightarrow \ \text{Area using trap rule > Actual area}\)

\(\dfrac{9}{8}\) \(>\dfrac{3}{4 \ln 2}\)  
\(36\, \ln 2\) \(>24\)  
\(\ln 2\) \(>\dfrac{2}{3}\)  
\(e^{\small \dfrac{2}{3}}\) \(>2\)  
\(e\) \(>2^{\small \dfrac{3}{2}}\)  
\(e\) \(>2 \sqrt{2}\)  

Calculus, 2ADV C3 2025 HSC 26

A piece of wire is 100 cm long. Some of the wire is to be used to make a circle of radius \(r\) cm. The remainder of the wire is used to make an equilateral triangle of side length \(x\) cm.

  1. Show that the combined area of the circle and equilateral triangle is given by
  2. \(A(x)=\dfrac{1}{4}\left(\sqrt{3} x^2+\dfrac{(100-3 x)^2}{\pi}\right)\).   (2 marks)

  3. By considering the quadratic function in part (a), show that the maximum value of \(A(x)\) occurs when all the wire is used for the circle.   (3 marks) 
Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.    \(\text{See Worked Solutions}\)

Show Worked Solution

a.    \(\text{Area of equilateral triangle:}\)

\(A_{\Delta}=\dfrac{1}{2} a b\, \sin C=\dfrac{1}{2} \times x^2 \times \sin 60^{\circ}=\dfrac{\sqrt{3} x^2}{4}\)
 

\(\text{Wire remaining to make circle}=100-3 x\)

\(\text {Find radius of circle:}\)

\(2 \pi r=100-3 x \ \Rightarrow \ r=\dfrac{100-3 x}{2 \pi}\)

\(\text{Area of circle: }\)

\(A_{\text {circ }}=\pi r^2=\pi \times\left(\dfrac{100-3 x}{2 \pi}\right)^2=\dfrac{(100-3 x)^2}{4 \pi}\)

\(\text{Total Area}\) \(=\dfrac{\sqrt{3} x^2}{4}+\dfrac{(100-3 x)^2}{4 \pi}\)
  \(=\dfrac{1}{4}\left(\sqrt{3} x^2+\dfrac{(100-3 x)^2}{\pi}\right)\)

 

b.   \(\text{Note strategy clue in question: “By considering the quadratic..”}\)

\(\text {Expanding} \ A(x):\)

 \(A(x)\)  \(=\dfrac{1}{4}\left(\sqrt{3} x^2+\dfrac{10\,000}{\pi}-\dfrac{600 x}{\pi}+\dfrac{9 x^2}{\pi}\right)\)
   \(=\dfrac{1}{4}\left(\sqrt{3}+\dfrac{9}{\pi}\right) x^2-\dfrac{150}{\pi} x+\dfrac{2500}{\pi}\)

 
\(\text{Consider limits on} \ x:\)

\(3 x \leqslant 100 \ \Rightarrow \ x \in\left[0,33 \frac{1}{3}\right]\)

\(\text{Consider vertex of concave up parabola}\ A(x):\)

\(x=-\dfrac{b}{2 a}=\dfrac{150}{\pi} \ ÷ \ \dfrac{1}{2}\left(\sqrt{3}+\dfrac{9}{\pi}\right) \approx 20.8\)
 

\(\text{By symmetry of the quadratic for} \ x \in\left[0,33 \dfrac{1}{3}\right],\)

\(A(x)_{\text{max}} \ \text{occurs at} \ \ x=0.\)

\(\text{i.e. when the wire is all used in the circle.}\)

Calculus, 2ADV C3 2025 HSC 24

The graphs of  \(y=e\, \ln x\)  and  \(y=a x^2+c\)  are shown. The line  \(y=x\)  is a tangent to both graphs at their point of intersection.
 

Find the values of \(a\) and \(c\).   (4 marks)

Show Answers Only

\(a=\dfrac{1}{2e}, \ c=\dfrac{1}{2} e\)

Show Worked Solution

\begin{array}{ll}
y=e\, \ln x\ \ldots \ (1) & y=a x^2+c\ \ldots \ (2) \\
y^{\prime}=\dfrac{e}{x} & y^{\prime}=2 a x
\end{array}

\(\text{At point of tangency, gradient}=1.\)

\(\text{Find \(x\) such that:}\)

   \(\dfrac{e}{x}=1 \ \Rightarrow \ x=e\)

\(\text{At} \ \ x=e, \ y^{\prime}=2 a x=1:\)

   \(2ae=1 \ \Rightarrow \ a=\dfrac{1}{2e}\)
 

\(\text{Find point of tangency using (1):}\)

   \(y=e\, \ln e=e\)

\(\text{Point of tangency at} \ (e,e).\)
 

\(\text{Since} \ (e, e) \text{ lies on (2):}\)

   \(e=\dfrac{1}{2e} \times e^2+c\)

   \(c=\dfrac{1}{2} e\)

Statistics, STD2 S5 2025 HSC 40

  1. In a flock of 12 600 sheep, the ratio of males to females is \(1:20\).
  2. The weights of the male sheep are normally distributed with a mean of 76.2 kg and a standard deviation of 6.8 kg.
  3. In the flock, 15 of the male sheep each weigh more than \(x\) kg. 
  4. Find the value of \(x\).   (4 marks)

  5. The weights of the female sheep are also normally distributed but have a smaller mean and smaller standard deviation than the weights of male sheeр.
  6. Explain whether it could be expected that 300 of the females from the flock each weigh more than \(x\) kg, where \(x\) is the value found in part (a).   (1 mark)

Show Answers Only

a.   \(x=89.8 \ \text{kg}\)

b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that  300 females weigh > 89.8 kg.}\)

Show Worked Solution

a.    \(12\,600 \ \text{sheep} \ \Rightarrow \ \text{male : female}=1:20\)

\(\text{Number of sheep in “1 part”} = \dfrac{12\,600}{21}=600\)

\(\Rightarrow \ \text{male : female}=600:12\,000\)

\(\text{male sheep} \ \%=\dfrac{15}{600}=0.025\%\)

\(z \text{-score }(0.025 \%)=2\)

\(\text{Using } \ z=\dfrac{x-\bar{x}}{s}, \ \text{find} \ \ x_m:\)

\(2\) \(=\dfrac{x_m-76.2}{6.8}\)  
\(x_m\) \(=76.2+2 \times 6.8=89.8 \ \text{kg}\)  

 
b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that  300 females weigh > 89.8 kg.}\)

Statistics, 2ADV S3 2025 HSC 23

  1. In a flock of 12 600 sheep, the ratio of males to females is \(1:20\).
  2. The weights of the male sheep are normally distributed with a mean of 76.2 kg and a standard deviation of 6.8 kg.
  3. In the flock, 15 of the male sheep each weigh more than \(x\) kg. 
  4. Find the value of \(x\).   (4 marks)

  5. The weights of the female sheep are also normally distributed but have a smaller mean and smaller standard deviation than the weights of male sheeр.
  6. Explain whether it could be expected that 300 of the females from the flock each weigh more than \(x\) kg, where \(x\) is the value found in part (a).   (1 mark)

Show Answers Only

a.   \(x=89.8 \ \text{kg}\)

b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that  300 females weigh > 89.8 kg.}\)

Show Worked Solution

a.    \(12\,600 \ \text{sheep} \ \Rightarrow \ \text{male : female}=1:20\)

\(\text{Number of sheet in “1 part”} = \dfrac{12\,600}{21}=600\)

\(\Rightarrow \ \text{male : female}=600:12\,000\)

\(\text{male sheep} \ \%=\dfrac{15}{600}=0.025\%\)

\(z \text{-score }(0.025 \%)=2\)

\(\text{Using } \ z=\dfrac{x-\bar{x}}{s}, \ \text{find} \ \ x_m:\)

\(2\) \(=\dfrac{x_m-76.2}{6.8}\)  
\(x_m\) \(=76.2+2 \times 6.8=89.8 \ \text{kg}\)  

 
b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that  300 females weigh > 89.8 kg.}\)

Statistics, 2ADV S3 2025 HSC 21

A continuous random variable \(X\) has a probability density function given by

\(f(x)= \begin{cases}\ 0 & \quad x<1 \\ \dfrac{1}{x} & \quad 1 \leq x \leq e \\ \ 0 & \quad x>e\end{cases}\)

  1. Find the mode of the given probability density function. Justify your answer.   (2 marks)

  2. Calculate the value of the 25th percentile \(\left(Q_1\right)\) of this distribution. Give your answer correct to 3 decimal places.   (3 marks)

Show Answers Only

a.    
     

\(\text{Graph is monotonically decreasing.}\)

\(\text{Mode:} \ \ x=1\)

b.   \(Q_1=1.284 \)

Show Worked Solution

a.    
         

\(\text{Graph is monotonically decreasing.}\)

\(f(x)_{\text{max}}\ \text{occurs on interval when}\ \ x=1.\)

\(\text{Mode:} \ \ x=1\)
 

b.    \(\text{Find \(k\) such that} \ P(X<k)=0.25:\)

\(\displaystyle \int_1^{Q_1} \frac{1}{x}\) \(=0.25\)
\(\ln Q_1-\ln 1\) \(=0.25\)
\(Q_1\) \(=e^{0.25}\)
  \(=1.284 \ \text{(3 d.p.)}\)

Financial Maths, STD2 F5 2025 HSC 34

The table shows future value interest factors for an annuity of $1.

Lin invests a lump sum of $21 000 for 7 years at an interest rate of 6% per annum, compounding monthly.

Yemi wants to achieve the same future value as Lin by using an annuity. Yemi plans to deposit a fixed amount into an investment account at the end of each month for 7 years. The investment account pays 6% per annum, compounding monthly.

Using the table provided, determine how much Yemi needs to deposit each month.   (3 marks)

Show Answers Only

\($306.78\)

Show Worked Solution

\(r=\dfrac{0.06}{12}=0.005, \ n=12 \times 7=84\)

\(\text{Lin’s investment:}\)

\(F V=21\,000(1+0.005)^{84}=31\,927.76\)
 

\(\text{Yemi’s investment:}\)

\(\text{Annuity factor:} \ 104.07393\)

\(\text{Annuity} \times 104.07393\) \(=$31\,927.76\)
\(\text{Annuity}\) \(=\dfrac{31\,927.76}{104.07393}=$306.78\)

Probability, 2ADV S1 2025 HSC 19

Three girls, Amara, Bala and Cassie, have nominated themselves for the local soccer team. Exactly one of the girls will be selected. The chances of their selection are in the ratio \(1:2:3\), respectively.

The probability that the team wins when:

  • Amara is selected is 0.5
  • Bala is selected is 0.4
  • Cassie is selected is 0.2.

Given that the team wins, find the probability that Amara was selected.   (3 marks)

Show Answers Only

\(P(A |W)=\dfrac{5}{19}\)

Show Worked Solution

\(P(A |W)=\dfrac{P(A \cap W)}{P(W)}\)

\(P(W)=\dfrac{1}{6} \times \dfrac{1}{2}+\dfrac{1}{3} \times \dfrac{2}{5}+\dfrac{1}{2} \times \dfrac{1}{5}=\dfrac{19}{60}\)

\(P(A \cap W)=\dfrac{1}{6} \times \dfrac{1}{2}=\dfrac{1}{12}\)

\(\therefore P(A |W)=\dfrac{1}{12} \times \dfrac{60}{19}=\dfrac{5}{19}\)