SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Networks, STD2 N3 2025 HSC 19

The activities and corresponding durations in days for a project are shown in the network diagram.
 

 

  1. Complete the table showing the immediate prerequisites for each activity. Indicate with an \(\text{X}\) any activities without any immediate prerequisites.   (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

  1. Find the critical path for this project AND state the minimum duration for the project.   (2 marks)

  1. The duration of activity \( A \) is increased by 2. Does this affect the critical path for the project? Give a reason for your answer.   (1 mark)

Show Answers Only

a.           

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \text{X} \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & C,D \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & E \\
\hline
\end{array}

  
b.   \(\text{Critical Path:}\ BDEFH\)

\(\text{Minimum Duration}\ =4+5+5+7+5=26\ \text{days}\)
 

c.   \(\text{If duration of activity \(A\) is increased by 2:}\)

\(\text{The critical path remains unchanged (EST of activity \(E\) remains = 9)}\)

Show Worked Solution

a.   

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \text{X} \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & C,D \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & E \\
\hline
\end{array}

  
b.   

\(\text{Critical Path:}\ BDEFH\)

\(\text{Minimum Duration}\ =4+5+5+7+5=26\ \text{days}\)
 

c.   \(\text{If duration of activity \(A\) is increased by 2:}\)

\(\text{The critical path remains unchanged (EST of activity \(E\) remains = 9)}\)

Financial Maths, STD2 F5 2025 HSC 18

A table of future value interest factors for an annuity of $1 is shown.
 

The prize in a lottery is an annuity of $5000 a year for 10 years, invested at 4.5% per annum compounding annually.

What will be the value of the prize at the end of 10 years?   (2 marks)

Show Answers Only

\($61\,440\)

Show Worked Solution

\( r=4.5\%\ \text{annually}\)

\(\text{Compounding periods = 10}\)

\(\text{Annuity factor = 12.288}\)

\(\therefore\ \text{FV (annuity)}\ = 5000 \times 12.288=$61\,440\)

Statistics, STD2 S4 2025 HSC 17

The scatter plot shows a bivariate dataset, where \(x\) is the independent variable and \(y\) is the dependent variable. 
 

 

The points \( (0,14) \) and \( (5,4)\) lie on the line of best fit.

Plot the points \( (0,14) \) and \( (5,4) \) on the graph and hence find the equation of the line of best fit.   (3 marks)

Show Answers Only

\( y=-2 x+14 \)

Show Worked Solution

\(\text{Gradient (LOBF)}\ =\dfrac{y_2-y_1}{x_4-x_1}=\dfrac{4-14}{5-0}=-2 \)

\(\text{Find equation of line}\ \ m=-2 \ \ \text{through}\ (0,14):\)

\( y-y_1\) \(=m(x-x_1)\)  
\( y-14\) \(=-2(x-0)\)  
\( y\) \(=-2 x+14 \)  

Measurement, STD2 M2 2025 HSC 12 MC

A football game is being played in Quito (UTC –5) starting at 3:40 pm on Tuesday.

What is the time in Sydney (UTC +10) when the game starts in Quito?

  1. 10:40 am Tuesday
  2. 8:40 pm Tuesday
  3. 12:40 am Wednesday
  4. 6:40 am Wednesday
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Sydney is 15 hours ahead of Quito.}\)

\(\text{3:40 pm Tuesday + 15 hours}\)

\(\text{= 3:40 am Wednesday + 3 hours}\)

\(\text{= 6:40 am Wednesday (Sydney time)}\)

\(\Rightarrow D\)

Measurement, STD2 M7 2025 HSC 10 MC

An electricity company charges customers 37 cents per kWh for electricity used, \( U\), and pays customers 5 cents per kWh for electricity produced, \(P\).

The electricity company also charges customers a fee of 71 cents per day.

Which formula should be used to calculate a customer's daily cost of electricity, \(C\), in cents?

  1. \( C=71+37 U-5 P \)
  2. \( C=71+37 U+5 P \)
  3. \( C=71-37 U-5 P \)
  4. \( C=71-37 U+5 P \)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Customer pays:}\ 71 + 37 \times U\)

\(\text{Customer receives:}\ 5 \times P\)

\(\text{Daily cost}\ = 71 + 37 \times U-5 \times P\)

\(\Rightarrow A\)

Measurement, STD2 M7 2025 HSC 9 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, STD2 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
\(P(\text{White})\) \(=\dfrac{50}{100}=\dfrac{1}{2}\)
\(P(\text{Red})\) \(=\dfrac{25}{100}=\dfrac{1}{4}\)  
\(P(\text{Yellow})\) \(=\dfrac{15}{100}=\dfrac{3}{20}\)
\(P(\text{Green})\) \(=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}\)

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

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

\(\Rightarrow A\)

Networks, STD1 N1 2025 HSC 6 MC

The network shows the distances, in kilometres, along a series of roads that connect towns.
 

What is the value of the largest weighted edge included in the minimum spanning tree for this network?

  1. 7
  2. 8
  3. 9
  4. 10
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Minimum spanning tree:}\)
 

\(\text{Using Kruskal’s algorithm:}\)

\(\text{Edge 1 = 4, Edge 2/3 = 4, Edge 4 = 6, Edge 5 = 9}\) 

\(\therefore\ \text{The largest weighted edge in the MST = 9.}\)

\(\Rightarrow C\)

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`

Calculus, 2ADV C4 EO-Bank 11

  1. Differeniate \(y=\dfrac{x}{x^2+1}\)  (2 marks)
  1. Hence evaluate \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\)   (2 marks)
Show Answers Only

a.    \(\dfrac{dy}{dx} = \dfrac{1-x^2}{(x^2+1)^2}\)

b.    \(\dfrac{1}{2}\)

Show Worked Solution

a.    Using the quotient rule:

\(\dfrac{dy}{dx} = \dfrac{x^2+1-2x^2}{(x^2+1)^2} = \dfrac{1-x^2}{(x^2+1)^2}\)
 

b.    Using part a: 

 \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\) \(=\left[\dfrac{x}{x^2+1}\right]_0^1\)  
  \(=\dfrac{1}{2} -0\)  
  \(=\dfrac{1}{2}\)  

 

Calculus, 2ADV C4 EO-Bank 10

Given that  `int_0^k ( 2x + 4 )\ dx = 21`, and  `k`  is a constant, find the value of  `k`.   (2 marks)

Show Answers Only

`k = 3`

Show Worked Solution
`int_0^k ( 2x + 4 ) \ dx` `= 21`
`int_0^k ( 2x + 4 ) \ dx` `= [ x^2 + 4x ]_0^k`
  `= [(k^2 + 4k ) – 0 ]`
  `= k^2 + 4k`

 

`=> k^2 + 4k` `=21`
`k^2+4k-21` `= 0`
`(k-3)(k+7)` `= 0`
`k` `=3, -7`
`k` `=3\ text(as ) k >0`

Calculus, 2ADV C1 2014 HSC 13c v1

The displacement of a particle moving along the  `x`-axis is given by

 `x =2t -3/sqrt(t+1)`,

where  `x`  is the displacement from the origin in metres,  `t`  is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.    (2 marks)

  2. What value does the velocity approach as  `t`  increases indefinitely?    (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2`
Show Worked Solution
a.    `x` `=2t -3/sqrt(t+1)`
    `=2t -3(t+1)^(-1/2)`

 

`dot x` `= 2\ -3(-1/2) (t+1)^(-3/2)`
  `= 2 + 3/(2(t+1)^(3/2))`

 

`ddot x` `= -(9/4)(t+1)^(-5/2)`
  `= – 9/(4sqrt((t+1)^5))`

 
`text(S)text(ince)\ \ t >= 0,`

`=> 1/sqrt((t+1)^5) > 0`

`=> – 9/(4sqrt((t+1)^5)) < 0`
 

`:.\ text(Acceleration is always negative.)`

 

b.    `text(Velocity)\ (dot x) = 2 + 3/(2(t+1)^(3/2))`

 
`text(As)\ t -> oo,\ 3/(2(t+1)^(3/2)) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 2`

Calculus, 2ADV C1 EO-Bank 11

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = (t^2+4)/sqrt(3t+1)`  metres per second

Find the acceleration of the particle when  `t = 2`.

Express your answer as an exact value in its simplest form.  (3 marks)

Show Answers Only

` (16sqrt(7))/49\ \ text(ms)^(−2)`

Show Worked Solution

`v = (t^2+4)/sqrt(3t+1)`

`alpha` `= (dv)/(dt)`

`text(Using quotient rule:)`

`u=t^2+4,`     `v=(3t+1)^(1/2)`  
`u^{′} = 2t,`     `v^{′} = 3/2 (3t+1)^(-1/2)`  
     
`alpha` `= (u^{′} v-v^{′} u)/v^2`
  `= (2t (3t+1)^(1/2)-3/2(t^2+4) (3t+1)^(-1/2))/(3t+1)`

 
`text(When)\ \ t = 2,`

`alpha` `= (4(7)^(1/2)-12(7)^(-1/2))/(7)`
  `= (4sqrt(7))/7 -12/(7sqrt(7))`
  `= (28sqrt(7))/49-(12sqrt(7))/49`
  `= (16sqrt(7))/49\ \ text(ms)^(−2)`

Calculus, 2ADV C1 EO-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = t^3-4t^2 +5t + 6`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 3`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

Show Answers Only

i.    `8\ text(ms)^(−1)`

ii.   `1, 5/3\ text(s)`

Show Worked Solution

i.   `x =t^3-4t^2 +5t + 6` 

`v = (dx)/(dt) = 3t^2-8t +5`

 
`text(When)\ t = 3,`

`v` `= 3 xx 3^2-8 · 3 +5`
  `= 8\ text(ms)^(−1)`

 

ii.   `text(Particle is stationary when)\ \ v = 0`

`3t^2-8t +5` `= 0`
`3t^2-3t-5t+5` `= 0`
`3t(t-1)-5(t-1)` `= 0`
`(t-1)(3t-5)` `= 0`
`t` `= 1, 5/3\ text(s)`

Functions, 2ADV F2 2025 HSC 30

The parabola with equation  \(y=(x-1)(x-5)\)  is translated both horizontally to the right and vertically up by \(k\) units, where \(k\) is positive.

The translated parabola passes through the point \((6,11)\).

Find the value of \(k\).   (3 marks)

Show Answers Only

\(k=6\)

Show Worked Solution

\(y=(x-1)(x-5)\)

\(\text{Translate \(k\) units to the right:}\)

\(y \rightarrow y^{\prime}=(x-k-1)(x-k-5)\)

\(\text{Translate \(k\) units vertically up:}\)

\(y^{\prime} \rightarrow y^{\prime \prime}=(x-k-1)(x-k-5)+k\)

\(y^{\prime \prime} \ \text{passes through}\ (6,11):\)

\(11=(6-k-1)(6-k-5)+k\)

\(11=(5-k)(1-k)+k\)

\(11=5-6 k+k^2+k\)

\(0=k^2-5 k-6\)

\(0=(k-6)(k+1)\)

\(\therefore k=6 \quad(k>0)\)

Trigonometry, 2ADV T1 2025 HSC 28

A farmer wants to use a straight fence to divide a circular paddock of radius 10 metres into two segments. The smaller segment is \(\dfrac{1}{4}\) of the paddock and is shaded in the diagram. The fence subtends an angle of \(\theta\) radians at the centre of the circle as shown.
 

  1. Show that  \(\theta=\sin \theta+\dfrac{\pi}{2}\).   (2 marks)

  2. The graph of  \(y=\sin \theta+\dfrac{\pi}{2}\) is shown.
     
  4. Use the graph and the result in part (a) to estimate the arc length of the smaller segment to the nearest metre.   (2 marks)

Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.   \(\text{Arc length} \ \approx 23 \ \text{metres.}\)

Show Worked Solution

a.    \(\text{Area of paddock} =\pi \times 10^2=100 \pi\)

\(\text{Area of segment} =\dfrac{1}{4} \times 100 \pi =25 \pi\)

\(\text{Area of sector} =\dfrac{\theta}{2 \pi} \times 100 \pi=50 \theta\)

\(\text{Area of triangle} =\dfrac{1}{2} ab \, \sin C=50 \, \sin \theta\)

\(\text{Equating sector areas:}\)

\(25 \pi\) \(=50 \theta-50\, \sin \theta\)
\(50 \theta\) \(=25 \pi+50 \, \sin \theta\)
\(\theta\) \(=\dfrac{\pi}{2}+\sin \theta\)

 
b.    \(\text{Find where} \ \ \theta=\sin \theta+\dfrac{\pi}{2}\)

\(\text {Intersection occurs where: }\)

\(y=\theta \ \ \text{intersects with} \ \ y=\sin \theta+\dfrac{\pi}{2}\)

\(\text{At intersection (from graph):} \ \ \theta \approx 2.3 \ \text{radians}\)

\(\text{Arc length} \ \approx 10 \times 2.3 \approx 23 \ \text{metres.}\)

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 C4 2025 HSC 25

  1. Show that  \(\dfrac{d}{d x}(\sin x-x\, \cos x)=x\, \sin x\).   (2 marks)
  2. Hence, find the value of  \(\displaystyle\int_0^{2025 \pi} x\, \sin x \, dx\).   (2 marks)

  3. The regions bounded by the \(x\)-axis and the graph of  \(y=x\, \sin x\)  for  \(x \geq 0\)  are shown.
     

  1. Let  \(A_n=\displaystyle \int_{(n-1) \pi}^{n \pi} x\, \sin x \,dx\),  where \(n\) is a positive integer.
  2. It can be shown that  \(\left|A_n\right|=(2 n-1) \pi\).  (Do NOT prove this.)
  3. Find the exact total area of the regions bounded by the curve  \(y=x \sin x\), and the \(x\)-axis between  \(x=0\)  and  \(x=2025 \pi\).   (2 marks)  
Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.   \(2025 \pi\)

c.   \(4\,100\,625 \pi \ \text{units}^2\)

Show Worked Solution
a.     \(\dfrac{d}{dx}(\sin x-x\, \cos x)\) \(=\dfrac{d}{dx} \sin x-\dfrac{d}{dx} x\, \cos x\)
    \(=\cos x+x\, \sin x-\cos x\)
    \(=x\, \sin x\)

 

b.     \(\displaystyle\int_0^{2025 \pi} x\, \sin x\) \(=\Big[\sin x-x\, \cos x\Big]_0^{2025 \pi}\)
    \(=\Big[(\sin (2025\pi)-2025 \pi \times \cos (2025 \pi))-0\Big]\)
    \(=0-2025 \pi \times -1\)
    \(=2025 \pi\)

 

c.    \(\text{Area}=\displaystyle \int_0^\pi x\, \sin x \, dx+\left|\int_\pi^{2 \pi} x\, \sin x \, dx\right|+\cdots+\int_{2024 \pi}^{2025 \pi} x\, \sin x \, dx\)

\(\text{Using}\ \ \left|A_n\right|=(2n-1) \pi:\)

\(A_1=\pi, A_2=3 \pi, A_3=5 \pi, \ldots, A_{2025}=4049 \pi\)

\begin{aligned}
\rule{0pt}{2.5ex} \text{Area}& =\underbrace{\pi+3 \pi+5 \pi+\ldots+4049 \pi}_{\text {AP where } a=\pi, \ l=4049 \pi, \ n=2025} \\
\rule{0pt}{4.5ex} & =\frac{2025}{2}(\pi+4049 \pi) \\
\rule{0pt}{3.5ex} & =4\,100\,625 \pi \ \text{units }^2
\end{aligned}

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}:\)

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

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

Trigonometry, 2ADV T2 2025 HSC 22

Prove that

\(\dfrac{\sin ^4 \theta+\cos ^4 \theta}{\sin ^2 \theta\, \cos ^2 \theta}+2=\sec ^2 \theta\, \operatorname{cosec}^2 \theta\).   (2 marks)

Show Answers Only

\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{Note: RHS } =\sec ^2 \theta\, \operatorname{cosec}^2 \theta=\dfrac{1}{\sin ^2 \theta\, \cos ^2 \theta}\)

\(\text{RHS}\) \(=\dfrac{\sin ^4 \theta+\cos ^4 \theta}{\sin ^2 \theta \cos ^2 \theta}+2\)
  \(=\dfrac{\sin ^4 \theta+\cos ^4 \theta+2 \sin ^2 \theta \cos ^2 \theta}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\dfrac{\left(\sin ^2 \theta+\cos ^2 \theta\right)^2}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\dfrac{1}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\sec ^2 \theta\, \operatorname{cosec} ^2 \theta\)

Financial Maths, 2ADV M1 2025 HSC 20

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

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

\(\text{Lin’s investment:}\)

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

\(\text{Yemi’s investment:}\)

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

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

Functions, 2ADV F1 2025 HSC 18

Find the range of \(g(f(x))\),  given  \(f(x)=\dfrac{3}{x-1}\)  and  \(g(x)=x+5\).   (2 marks)

Show Answers Only

\(\text{Range} \ g(f(x)): \text{All} \  y, \  y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Show Worked Solution

\(f(x)=\dfrac{3}{x-1}, \ \ g(x)=x+5\)

\(g(f(x))=\dfrac{3}{x-1}+5 \Rightarrow \ \text{vertical translation +5 of} \ f(x)\)

\(\text{Range} \ f(x):\ \text{All} \ y, \ y \neq 0\)

\(\text{Range} \ g(f(x)): \text{All} \  y, \  y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Financial Maths, 2ADV M1 2025 HSC 17

A borrower obtains a reducing-balance loan of $800 000 to buy a house.

Interest is charged at 0.5% monthly, compounded monthly.

On the last day of each month, interest is added to the balance owing on the loan and then the monthly repayment of $5740 is made.

Let \(\$ A_n\) be the balance owing on the loan at the end of \(n\) months.

  1. Show that  \(A_2=800\,000(1.005)^2-5740(1.005)-5740\).   (2 marks)

  2. Show that  \(A_n=1\,148\,000-348\,000(1.005)^n\).   (3 marks)

  3. After how many months will the balance owing on the loan first be less than $400 000?   (2 marks)

Show Answers Only

a.   \(\text{See Worked Solutions.}\)

b.   \(\text{See Worked Solutions.}\)

c.   \(\text{154 months}\)

Show Worked Solution
a.     \(A_1\) \(=800\,000(1.005)-5740\)
  \(A_2\) \(=A_1 \times (1.005)-5470\)
    \(=[800\,000(1.005)-5740](1.005)-5740\)
    \(=800\,000(1.005)^2-5740(1.005)-5470\)

 

b.     \(A_3\) \(=800\,000(1.005)^3-5740(1.005)^2-5740(1.005)-5740\)
  \(A_n\) \(=800\,000(1.005)^n-5740(1.005)^{n-1} \ldots -5740(1.005)-5740\)
    \(=800\,000(1.005)^n-5740 \underbrace{\left(1.005+1.005^2+\cdots+1.005^{n-1}\right)}_{\text {GP where} \  a=1 ,  r=1.005}\)
    \(=800\,000(1.005)^n-5740\left(\dfrac{a\left(r^n-1\right)}{r-1}\right)\)
    \(=800\,000(1.005)^n-5740\left(\dfrac{1.005^n-1}{0.005}\right)\)
    \(=800\,000(1.005)^n-1\,148\,000\left(1.005^n-1\right)\)
    \(=1\,148\,000-1\,148\,000(1.005)^n+800\,000(1.005)^n\)
    \(=1\,148\,000-348\,000(1.005)^n\)

 

c.    \(\text{Find} \  n \ \text{such that} \ A_n < 400\,000:\)

\(1\,148\,000-348\,000(1.005)^n\) \(< 400\,000\)
\(348\,000(1.005)^n\) \(>1\,148\,000-400\,000\)
\(1.005^n\) \(>\dfrac{748\,000}{348\,000}\)
\(n \times \ln 1.005\) \(>\ln \left(\frac{187}{87}\right)\)
\(n\) \(>\dfrac{\ln \left(\frac{187}{87}\right)}{\ln 1.005}\)
\(n\) \(>153.42\)

 
\(\therefore \ \text{Balance owing is less than \$400 000 after 154 months.}\)

Calculus, 2ADV C3 2025 HSC 16

Consider the function  \(f(x)=\dfrac{x^2}{e^x}\).

  1. Find the stationary points of the function and determine their nature.   (4 marks)

  2. A partially completed graph of  \(f(x)=\dfrac{x^2}{e^x}\)  is shown.
  3. Use your answer from part (a) to complete the graph.   (1 mark)


     

Show Answers Only

a.   \(\text{MIN at}\ \ (0,0)\)

\(\text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)

b.   
     

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{x^2}{e^x}\)
  \(f^{\prime}(x)\) \(=2 x \cdot e^x-e^x \cdot x^2\)
    \(=\dfrac{x e^x(2-x)}{e^{2 x}}\)
    \(=\dfrac{x(2-x)}{e^x}\)

 
\(\text{Find} \ x\ \text{when} \ \ f^{\prime}(x)=0:\)

\(x(2-x)=0\)

\(x=0 \ \text {or} \ 2\)

\(\text{When} \ \ x=0 \ \Rightarrow \ f(0)=0\)

\(\text {When}\ \  x=2 \ \Rightarrow \ f(2)=\dfrac{4}{e^2}\)

\(\text {Checking nature of SP’s:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline x & -1 & 0 & 1 & 2 & 3 \\
\hline f^{\prime}(x) & -3 e & 0 & \dfrac{1}{e} & 0 & -\dfrac{3}{e^3} \\
\hline
\end{array}

\(\therefore \ \text{MIN at}\ \ (0,0)\)

\(\quad \ \ \text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)
 

b.
     

Statistics, STD2 S4 2025 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.    \(\text{At} \ \  x=120\text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

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

Statistics, 2ADV S2 2025 HSC 14

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.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

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.    \(\text{34.9 minutes}\)
 

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

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

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

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

Algebra, STD2 A4 2025 HSC 20

The graph of a quadratic function represented by the equation  \(h=t^2-8 t+12\)  is shown.
 

  1. Find the values of \(t\) and \(h\) at the turning point of the graph.   (2 marks)

  2. The graph shows  \(h=12\)  when  \(t=0\).
  3. What is the other value of \(t\) for which  \(h=12\)?   (1 mark)

Show Answers Only

a.   \(\text{Turning point at} \ \ (4,-4)\)

b.   \(t=8\)

Show Worked Solution

a.    \(\text{Axis of quadratic occurs when}\ \ t= \dfrac{2+6}{2} = 4\)

\(\text{At} \ \ t=4:\)

\(h=4^2-8 \times 4+12=-4\)

\(\therefore \ \text{Turning point at} \ \ (4,-4)\)
 

b.    \(\text {When} \ \ h=12:\)

\(t^2-8 t+12\) \(=12\)
\(t(t-8)\) \(=0\)

 
\(\therefore \ \text{Other value:} \ \ t=8\)

Statistics, 2ADV S3 2025 HSC 8 MC

The minimum daily temperature, in degrees, of a town each year follows a normal distribution with its mean equal to its standard deviation. The minimum daily temperature was recorded over one year.

What percentage of the recorded minimum daily temperatures was above zero degrees?

  1. 16%
  2. 50%
  3. 68%
  4. 84%
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Consider a possible example:}\)

\(\text{Let mean min daily temperature = 8°C}\)

\(\text{Std dev = 8°C}\)

\(z\text{-score (0°C)}\ =-1\)

\(\text{Percentage above 0°C} = 50+34=84\%\)

\(\Rightarrow D\)

Probability, 2ADV S1 2025 HSC 7 MC

A ten-sided die has faces numbered 1 to 10 .

The die is constructed so that the probability of obtaining the number 1 is greater than the probability of obtaining any of the other numbers. The numbers 2 to 10 are equally likely to occur.

When the die is rolled 153 times, a 1 is obtained 72 times.

By using the relative frequency of rolling a 1, which of the following is the best estimate for the probability of rolling a 10 ?

  1. \(\dfrac{1}{17}\)
  2. \(\dfrac{1}{11}\)
  3. \(\dfrac{1}{10}\)
  4. \(\dfrac{1}{9}\)
Show Answers Only

\(A\)

Show Worked Solution

\(P(1) = \dfrac{72}{153}=\dfrac{8}{17} \)

\(\text{Let}\ \ p=P(2)=P(3) = … =P(10) \)

\(\dfrac{8}{17}+9p\) \(=1\)  
\(9p\) \(=1-\dfrac{8}{17}\)  
\(p\) \(=\dfrac{1}{17}\)  

 
\(\Rightarrow A\)

Calculus, 2ADV C1 EO-Bank 2

  1.  Find the equations of the tangents to the curve  `y = x^2-5x+6`  at the points where the curve cuts the `x`-axis.  (2 marks)

  2.  Where do the tangents intersect?  (2 marks)

Show Answers Only
  1. `y = −x+2`
    `y = x-3`
  2. `(5/2, −1/2)`
Show Worked Solution
a.   `y` `= x^2-5x+6`
  `= (x-2)(x-3)`

 
`text(Cuts)\ xtext(-axis at)\ \ x = 2\ \ text(or)\ \ x = 3`
 

`(dy)/(dx) = 2x-5`

 
`text(At)\ \ x = 2 \ => \ (dy)/(dx) = -1`

`T_1\ text(has)\ \ m = −1,\ text{through (2, 0)}`

`y -0` `= -1(x-2)`
`y` `= -x+2`

  

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 1`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y -0` `= 1(x-3)`
`y` `= x -3`

 

b.   `text(Intersection occurs when:)`

`-x+2` `= x-3`
`2x` `= 5`
`x` `= 5/2`

  

`y = 5/2 – 3 = −1/2`

`:.\ text(Intersection at)\ \ (5/2, −1/2)`

Calculus, 2ADV C1 EO-Bank 1

Find the equation of the tangent to the curve  \(y=e^{x^2+3x}\)  at the point where \(x=1\).  (2 marks)

Show Answers Only

`y = 5e^4x-4e^4`

Show Worked Solution
\(y\) \(=e^{x^2+3x}\)
`(dy)/(dx)` \(=(2x+3)e^{x^2+3x}\)

 
`text(When)\ x = 1,\ \ (dy)/(dx) = 5e^4`

`text(Equation of tangent through)\ (1, e^4)`

`y-e^4` `= 5e^4(x – 1)`
`y` `= 5e^4x-4e^4`

Calculus, 2ADV C1 EO-Bank 14 v1

Evaluate `f^{′}(1)`, where `f(x) = x^2 / sqrt(2x + 3)`. (4 marks)

Show Answers Only

`9 / (5sqrt5)`

Show Worked Solution

`f(x) = x^2(2x + 3)^(-1/2)`

`f^{′}(x)` `= 2x(2x + 3)^(-1/2) + x^2(-1/2)(2x + 3)^(-3/2)(2)`

`= (2x)/(sqrt(2x + 3)) – (x^2)/(2x + 3)^(3/2)`

`= [2x(2x + 3) – x^2] / (2x + 3)^(3/2)`

`= (3x^2 + 6x) / (2x + 3)^(3/2)`

`f^{′}(1)` `= (3(1)^2 + 6(1)) / (2(1) + 3)^(3/2)`

`= 9 / (5sqrt5)`

Calculus, 2ADV C1 EO-Bank 6

Use the definition of the derivative, `f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}`  to find  `f^{\prime}(x)`  if  `f(x)=x-3x^2`.   (2 marks)

Show Answers Only

`f(x)=x-3x^2`

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Show Worked Solution

`f(x)=x-3x^2`

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Calculus, 2ADV C1 2013 HSC 11b v1

Evaluate  `lim_(x->1) ((x-1)(x+2)^2)/(x^2+x-2)`.   (2 marks)

Show Answers Only

 `3`

Show Worked Solution

`lim_(x ->1) ((x-1)(x+2)^2)/(x^2+x-2)`

COMMENT: This question has been simplified as students no longer need to factorise the difference between 2 cubes (`x^3-2^3`).

`=lim_(x->1) ( (x -1)(x+2)^2)/( (x-1)(x+2)`

`=lim_(x->1) (x+2)`

`=3`

Calculus, 2ADV C4 EO-Bank 4 MC SJ

 Let  `f^(')(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

  1. `f(x)=2sqrt(2x-3)`
  2. `f(x)=sqrt(2x-3)-2`
  3. `f(x)=2sqrt(2x-3)-2`
  4. `f(x)=sqrt(2x-3)+2`
Show Answers Only

`=>C`

Show Worked Solution
`f^{‘}(x)` `=2/(sqrt(2x-3))`  
`f(x)` `=2 int(2x-3)^{- 1/2}`  
  `=2*1/2*2(2x-3)^{1/2}+c`  
  `=2sqrt(2x-3)+c`  

 
`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3)-2`

`=>C`

Calculus, 2ADV C1 EO-Bank 1 MC v1

The derivative of  \((n^2-1) x^{3n-2}\)  can be expressed as

  1. \(3(n-1)(n^2-1) x^{3n-2}\)
  2. \(3(n-1)(n^2-1) x^{3(n-1)}\)
  3. \((3n-2) (n^2-1) x^{3(n-1)}\)
  4. \((3n-2) (n^2-1) x^{3n-2}\)
Show Answers Only

\(C\)

Show Worked Solution
\(y\) \(=(n^2-1) x^{3n-2}\)  
\(y^{′}\) \(=(3n-2) (n^2-1) x^{3n-2-1)}\)  
  \(=(3n-2) (n^2-1) x^{3(n-1)}\)  

 
\(\Rightarrow C\)

BIOLOGY, M7 2021 VCE 11

Two students designed an experiment to investigate antibiotic resistance in Escherichia coli bacteria. They began with an E. coli culture. The following procedure was conducted in a filtered air chamber using aseptic techniques:

  • On Day 0, spread 1 mL of E. coli culture onto a nutrient agar plate containing \(0 \ \mu \text{g} / \text{mL}\) (micrograms per millilitre) of the antibiotic ampicillin. Spread \(1 \ \text{mL}\) of the \(E. coli \) culture onto a separate nutrient agar plate containing \(1 \ \mu \text{g} / \mathrm{mL}\) of ampicillin. Cover each plate with an airtight lid.
  • On Day 1, transfer a sample of bacteria from one of the Day 0 plates to one of the Day 1 plates containing \(2 \ \mu \text{g} / \text{mL}\) of ampicillin. Transfer a sample of bacteria from the other Day 0 plate to the other Day 1 plate, which also contains \(2 \ \mu \text{g} / \text{mL}\) of ampicillin. Cover as before.
  • On Day 2, transfer a sample of bacteria from one of the Day 1 plates to one of the Day 2 plates containing \(4 \ \mu \text{g} / \text{mL}\) of ampicillin. Transfer a sample of bacteria from the other Day 1 plate to the other Day 2 plate, which also contains \(4 \ \mu \text{g} / \text{mL}\) of ampicillin. Cover as before.
  • On Day 3, transfer a sample of bacteria from one of the Day 2 plates to one of the Day 3 plates containing \(6 \ \mu \text{g} / \text{mL}\) of ampicillin. Transfer a sample of bacteria from the other Day 2 plate to the other Day 3 plate, which also contains \(6 \ \mu \text{g} / \text{mL}\) of ampicillin. Cover and seal the plates.

All plates were incubated at 37 °C for each 24-hour period. Used plates were refrigerated until the end of the experiment. They were then photographed to compare the amount of bacterial growth and disposed of safely.

The students drew a diagram (Figure 1) to help explain the experimental design and to show their predicted results in each condition at the end of each day.
 

  1. Identify any two controlled variables for this experiment.   (2 marks)
  1. Write a suitable hypothesis for this experiment.   (2 marks)
The refrigerated plates kept from Days 0,1,2 and 3 of the experiment were photographed. The diagrams in Figure 2 represent the bacterial growth seen in the photographs.
   

 

  1.  i. Analyse the results of the experiment shown in Figure 2.   (3 marks)
  2. ii. Explain whether the results of the students' experiment shown in Figure 2 support the predicted results shown in Figure 1.   (2 marks)
Show Answers Only

a.    Examples of suitable responses included two of the following:

  • volume of E. coli first applied to plates
  • type of nutrient agar used
  • size of agar plates used
  • batch or source of ampicillin
  • duration of incubation
  • temperature of incubation
  • exposure time to each concentration of ampicillin.

b.    Hypothesis:

  • If E. coli is exposed to low OR 1 µg/ml concentration of ampicillin, it will exhibit increased growth when exposed to higher levels of ampicillin.

c.i.  Experiment analysis:

  • No ampicillin: More E. coli growth on Day 0.
  • 2 µg/mL ampicillin: No effect on bacterial growth.
  • Days 2/3: Fewer colonies for both groups.
  • Higher ampicillin (4 µg/mL, 6 µg/mL): Significant effect, killed many E. coli.
  • More growth than predicted on all plates.
  • By end: More ampicillin-resistant bacteria on experimental plate.
  • 1 µg/mL ampicillin: Initial exposure led to greater number of resistant E. coli.

c.ii.  If the experimental results did not support the predicted results:

  • E. coli was still present on both plates on Day 3.
  • This shows exposing E. coli to 1 µg/ml ampicillin does not increase its resistance to ampicillin.
  • Resistant E. coli grew regardless of initial exposure to ampicillin.

Any two of the following, if the experimental results did support the predicted results:

  • The number of colonies in both groups decreased over time.
  • More E. coli developed ampicillin resistance in the experimental group.
  • There were more colonies in the experimental group than in the control group on Day 3.
  • There were more colonies than expected as ampicillin concentration increased.
  • No growth was expected at the end of Day 3 for the control group.
Show Worked Solution

a.    Examples of suitable responses included two of the following:

  • volume of E. coli first applied to plates
  • type of nutrient agar used
  • size of agar plates used
  • batch or source of ampicillin
  • duration of incubation
  • temperature of incubation
  • exposure time to each concentration of ampicillin.

b.    Hypothesis:

  • If E. coli is exposed to low OR 1 µg/ml concentration of ampicillin, it will exhibit increased growth when exposed to higher levels of ampicillin.

♦♦♦ Mean mark (b) 40%.

c.i.  Experiment analysis:

  • No ampicillin: More E. coli growth on Day 0.
  • 2 µg/mL ampicillin: No effect on bacterial growth.
  • Days 2/3: Fewer colonies for both groups.
  • Higher ampicillin (4 µg/mL, 6 µg/mL): Significant effect, killed many E. coli.
  • More growth than predicted on all plates.
  • By end: More ampicillin-resistant bacteria on experimental plate.
  • 1 µg/mL ampicillin: Initial exposure led to greater number of resistant E. coli.

♦♦♦ Mean mark (c)(i) 6%.

c.ii.  If the experimental results did not support the predicted results:

  • E. coli was still present on both plates on Day 3.
  • This shows exposing E. coli to 1 µg/ml ampicillin does not increase its resistance to ampicillin.
  • Resistant E. coli grew regardless of initial exposure to ampicillin.

Any two of the following, if the experimental results did support the predicted results:

  • The number of colonies in both groups decreased over time.
  • More E. coli developed ampicillin resistance in the experimental group.
  • There were more colonies in the experimental group than in the control group on Day 3.
  • There were more colonies than expected as ampicillin concentration increased.
  • No growth was expected at the end of Day 3 for the control group.

♦♦ Mean mark (c)(ii) 50%.

CHEMISTRY, M1 EQ-Bank 10 MC

250,000 tonnes of iron ore contains 3.2% w/w of magnetite, \(\ce{Fe3O4}\).

What is the mass of magnetite?

  1. \(\text{8000 kg}\)
  2. \(8.0 \times 10^{6}\ \text{kg}\)
  3. \(8.0 \times 10^{9}\ \text{kg}\)
  4. \(8.0 \times 10^{12}\ \text{kg}\)
Show Answers Only

\(B\)

Show Worked Solution
  • Converting tonnes to kilograms:
  •     \(250\ 000\ \text{t} = 250 \ 000\ 000\ \text{kg}\)
  • Calculating the mass of magnetite:
  •     \(0.032 \times 2.5 \times 10^8\ \text{kg} = 8.0 \times 10^6\ \text{kg}\)

\(\Rightarrow B\)

CHEMISTRY, M1 EQ-Bank 6 MC

Elements \(\ce{A}\) and \(\ce{B}\) are in the same period of the Periodic Table. Element \(\ce{A}\) has 1 electron in its outer shell, and element \(\ce{B}\) has 6 electrons in its outer shell.

What is the likely formula of the compound they form together?

  1. \(\ce{A2B}\)
  2. \(\ce{AB2}\)
  3. \(\ce{A2B6}\)
  4. \(\ce{A6B2}\)
Show Answers Only

\(A\)

Show Worked Solution
  • \(\ce{A}\) (1 valence electron) → Group 1 → forms \(\ce{A^+}\).
  • \(\ce{B}\) (6 valence electrons) → Group 16 → forms \(\ce{B^2-}\) ions.
  • Hence they will form the compound with the formula \(\ce{A2B}\).

\(\Rightarrow A\)

CHEMISTRY, M1 EQ-Bank 4 MC

The formula for sodium thiosulfate is \(\ce{Na2S2O3}\).

What is the formula for aluminium thiosulfate?

  1. \(\ce{AlS2O3}\)
  2. \(\ce{Al2(S2O3)3}\)
  3. \(\ce{Al3S2O3}\)
  4. \(\ce{Al(S2O3)2}\)
Show Answers Only

\(B\)

Show Worked Solution
  • From the sodium thiosulfate ion we can identify that it contains a sodium ion \(\ce{Na^+}\) and the thiosulfate ion \(\ce{S2O3^2-}\).
  • The aluminium ion is \(\ce{Al^3+}\).
  • Hence the correct formula for aluminium thiosulfate ion is \(\ce{Al2(S2O3)3}\).

\(\Rightarrow B\)

CHEMISTRY, M1 EQ-Bank 3 MC

What is the correct systematic (IUPAC) name for \(\ce{Ca(NO3)2}\)?

  1. Calcium dinitrogen trioxide
  2. Calcium dinitrate
  3. Calcium nitrate
  4. Calcium\(\text{(I)}\) dinitrate
Show Answers Only

\(C\)

Show Worked Solution
  • For all ionic compounds, the names is written with the metal/postive ion and non-metal/negative ion.
  • The metal (cation) is written first with the name unchanged.
  • The non-metal (anion) is written second and the name is modified to end in either -ide, -ate, or -ite. For this compound, the name of the polyatomic ion is nitrate.

\(\Rightarrow C\)

CHEMISTRY, M1 EQ-Bank 15

A student is given a mixture containing sodium chloride crystals, sand, and iron filings.

The table below shows some physical properties of these substances.

\begin{array} {|c|c|c|}
\hline \text{Substance} &\text{Formula} & \text{Melting point }(^{\circ}C) & \text{Solubility in water} & \text{Magnetic} & \text{Density (g cm}^{-3})\\
\hline \text{Sodium chloride} & \ce{NaCl} & 801 & \text{Soluble} & \text{No} & 2.2 \\
\hline \text{Sand (silicon dioxide)} & \ce{SiO2} & 1710 & \text{Insoluble} & \text{No} & 2.6 \\
\hline \text{Iron filings} & \ce{Fe} & 1538 & \text{Insoluble} & \text{Yes} & 7.9 \\
\hline \end{array}

  1. Explain whether you would expect this mixture to be homogeneous or heterogeneous, and explain your reasoning.   (3 marks)
  1. Draw a flow diagram that shows how this mixture could be separated into pure samples of each substance.   (3 marks)
Show Answers Only

a.    The mixture is a heterogeneous mixture.

  • This is because the components retain their individual physical properties and can be seen as separate phases (solid particles with different appearances).
  • They do not form a uniform composition.

b.    
       

Show Worked Solution

a.    The mixture is a heterogeneous mixture.

  • This is because the components retain their individual physical properties and can be seen as separate phases (solid particles with different appearances).
  • They do not form a uniform composition.

b.    
       

  • The flow chart should outline steps that can be completed in a school science lab.

CHEMISTRY, M7 EQ-Bank 5 MC

Within a homologous series such as the alkanes, boiling points

  1. are all approximately the same.
  2. increase as the molar masses increase.
  3. are independent of intermolecular forces.
  4. decrease as the molecules become larger.
Show Answers Only

`B`

Show Worked Solution
  • Within a homologous series, boiling points increase as molecules get larger due to an increase in dispersion forces.

`=>B`

CHEMISTRY, M1 EQ-Bank 5

  1. Rank the penetrating ability of alpha, beta, and gamma radiation from lowest to highest.   (1 mark)
  1. Write a nuclear equation to show the \(\alpha\)-decay of \(\ce{^222_86Rn}\).   (1 mark)
  1. Write a nuclear equation to show the \(\beta\)-decay of \(\ce{^210_82Pb}\).   (1 mark)
  1. Explain why the radioactive decay stops when \(\ce{^206_82Pb}\) is formed.   (2 marks)
Show Answers Only

a.    \(\alpha < \beta < \gamma\)

  • Alpha is stopped by paper, beta by aluminium, gamma requires lead/concrete.
     

b.    \(\ce{^222_86Rn -> ^218_84Po + ^4_2He}\)
 

c.    \(\ce{^210_82Pb -> ^210_83Bi + ^0_-1e^-}\)
 

d.    \(\ce{^206_82Pb}\) is a stable isotope.

  • It has the correct proton-to-neutron ratio, so the nucleus no longer needs to emit radiation to become more stable.

  • Therefore, the decay chain ends once \(\ce{^206Pb}\) is formed.

Show Worked Solution

a.    \(\alpha < \beta < \gamma\)

  • Alpha is stopped by paper, beta by aluminium, gamma requires lead/concrete.
     

b.    \(\ce{^222_86Rn -> ^218_84Po + ^4_2He}\)
 

c.    \(\ce{^210_82Pb -> ^210_83Bi + ^0_-1e^-}\)
 

d.    \(\ce{^206_82Pb}\) is a stable isotope.

  • It has the correct proton-to-neutron ratio, so the nucleus no longer needs to emit radiation to become more stable.

  • Therefore, the decay chain ends once \(\ce{^206Pb}\) is formed.

CHEMISTRY, M1 EQ-Bank 4

Compare and contrast two isotopes of hydrogen.   (2 marks)

Show Answers Only
  • Two of hydrogens isotopes are hydrogen-1 (protium) and hydrogen-2 (deuterium).
  • The atoms have the same number of protons but a different number of neutrons in the nucleus. Hydrogen-1 has 1 proton and 0 neutrons, while hydrogen-2 has 1 proton and 1 neutron.
Show Worked Solution
  • Two of hydrogens isotopes are hydrogen-1 (protium) and hydrogen-2 (deuterium).
  • The atoms have the same number of protons but a different number of neutrons in the nucleus. Hydrogen-1 has 1 proton and 0 neutrons, while hydrogen-2 has 1 proton and 1 neutron.

CHEMISTRY, M1 EQ-Bank 2 MC

Which scientist developed the quantum mechanical model of the atom, describing electrons as existing in orbitals `(s, p, d, f)` defined by regions of space with the highest probability of finding an electron?

  1. Niels Bohr
  2. J. J. Thomson
  3. Erwin Schrödinger
  4. John Dalton
Show Answers Only

\(C\)

Show Worked Solution
  • Schrödinger used wave mechanics to describe electrons as occupying orbitals `(s, p, d, f)`, regions of space with the highest probability of finding an electron rather than fixed paths.

\(\Rightarrow C\)

CHEMISTRY, M1 EQ-Bank 3

Thorium exists in several isotopic forms. The existence of these isotopes can be shown by placing a thorium sample in a mass spectrometer, in which atoms are vaporised, electrically charged, and the ratio of the mass/charge for each is compared.

A mass spectrogram of thorium is shown below. The mass number is displayed on the x-axis and the % abundance on the y-axis.
 

  1. What is the \(Z\) for thorium?   (1 mark)
  1. Write the three isotopes of thorium.   (1 mark)
  1. Using the mass spectrogram, calculate the relative atomic weight of natural thorium. Show your working.   (2 marks)
  1. Thorium-232 decays to radium-228. Write the nuclear equation to show the radioactive decay of thorium-232.   (2 marks)
Show Answers Only

a.    \(Z=90\)

b.    \(\ce{^228_90Th, ^230_90Th, ^232_90Th}\).

c.    \(232.0\)

d.    \(\ce{^232_90Th -> ^228_88Ra + ^4_2He}\)

Show Worked Solution

a.    \(Z = 90\) (atomic number)
 

b.    \(\ce{^228_90Th, ^230_90Th, ^232_90Th}\).
 

c.     \(M\) \(=\dfrac{(228 \times 0.02) + (230 \times 0.8) + (232 \times 99.18)}{100}\)
    \(=231.98\)
    \(=232.0\ \text{(4 sig.fig)}\)

 

d.    Thorium-232 undergoes alpha decay:

\(\ce{^232_90Th -> ^228_88Ra + ^4_2He}\)

v1 Measurement, STD2 M7 SM-Bank 15 MC

A brand of cereal is sold in two different boxes.

What is the difference in the price per kilogram of these two boxes?

  1. $0.80
  2. $1.20
  3. $1.50
  4. $2.40
Show Answers Only

`B`

Show Worked Solution

`text(Convert to price per kg:)`

`text(Box 1)`

`400\ text(g) = $3.60`

`1\ text(kg) = 3.60\ xx\ 2.5 = $9.00`

 

`text(Box 2)`

`1.5\ text(kg) = $15.30`

`1\ text(kg) = 15.30\ -:\ 1.5 = $10.20`

 

`:.\ \text(Difference per kg)`

`= 10.20 – 9.00`

`= $1.20`
 

`=> B`

v1 Measurement, STD2 M7 2023 HSC 5 MC

Four mobile data packs are shown, each with the data included and its cost.

Which one represents the best value?

Show Answers Only

`C`

Show Worked Solution

`text{Calculate cost per GB of each option}`

`text{Option}\ A:\ $22.50 -: 10= $2.25\ text{per GB}`

`text{Option}\ B:\ $25.60 -: 12= $2.133\ldots\ \text{per GB}`

`text{Option}\ C:\ $27.30 -: 14= $1.95\ \text{per GB}`

`text{Option}\ D:\ $30.00 -: 15= $2.00\ \text{per GB}`

`=>C`

CHEMISTRY, M1 EQ-Bank 2

Write the electronic configuration for the following substances, using spdf notation:

  1. Calcium atom   (1 mark)
  1. Fluoride ion   (1 mark)
Show Answers Only

a.    \(\ce{Ca}\): \(1s^2\) \(2s^2\) \(2p^6\) \(3s^2\) \(3p^6\) \(4s^2\)

b.    \(\ce{F^-}\): \(1s^2\) \(2s^2\) \(2p^6\)

Show Worked Solution

a.    \(\ce{Ca}\): \(1s^2\) \(2s^2\) \(2p^6\) \(3s^2\) \(3p^6\) \(4s^2\)
 

b.    \(\ce{F^-}\): \(1s^2\) \(2s^2\) \(2p^6\)

CHEMISTRY, M1 EQ-Bank 1

The element tellurium is a brittle, silver-grey metalloid used in solar panels and thermoelectric devices.

  1. Tellurium has two naturally occurring stable isotopes that contribute to its average atomic mass. One of these isotopes, \(\ce{^126Te}\), has a percentage abundance of 45.2%. Calculate the mass number of the other isotope.   (2 marks)
  1. One unstable isotope of tellurium is \(\ce{^130Te}\). Write a nuclear equation for the decay of this isotope when it undergoes beta decay.   (1 marks)
Show Answers Only

a.    \(\ce{^129Te}\)

b.    \(\ce{^130_52Te -> ^130_53I + ^0_-1e}\)

Show Worked Solution

a.    Let the mass number of the other isotope of Tellurium be \(x\). Therefore:

\((0.452 \times 126) + (0.548x)\)  \(=127.6\)  
\(0.548x\) \(=70.648\)  
\(x\) \(=128.9 \approx 129\)  
     
  • The mass number of the other stable isotope of Tellurium is 129.

b.    \(\ce{^130_52Te -> ^130_53I + ^0_-1e}\)

CHEMISTRY, M1 EQ-Bank 6 MC

Which of the following elements reacts most vigorously with water?

  1. Lithium 
  2. Sodium 
  3. Potassium 
  4. Calcium 
Show Answers Only

\(C\)

Show Worked Solution
  • Reactivity increases down Group 1 (alkali metals), so \(\ce{K}\) is more reactive than \(\ce{Li}\) and \(\ce{Na}\).
  • Potassium reacts violently with water, producing enough heat to ignite the hydrogen gas with a lilac flame.
  • Calcium reacts more slowly than potassium despite being a reactive metal as Potassium’s outer electron is further from the nucleus, making it easier to lose and more reactive.

\(\Rightarrow C\)

v1 Algebra, STD2 A2 2014 HSC 22 MC

Maya’s hybrid uses fuel at the rate of 7.4 L per 100 km for highway driving and 10.2 L per 100 km for urban driving.

She drove a total of 720 km, of which 120 km were urban driving.

Approximately how much fuel did Maya’s car use on the journey?

  1. 55 L
  2. 57 L
  3. 60 L
  4. 65 L
Show Answers Only

`B`

Show Worked Solution

`text(Fuel used in urban driving)`

`= 120/100 xx 10.2\ text(L) = 12.24\ text(L)`

`text(Fuel used on highway)`

`= 600/100 xx 7.4\ text(L) = 44.4\ text(L)`

`:.\ text(Total Fuel)` `= 12.24 + 44.4`
  `= 56.64\ text(L) ~~ \approx 57\ text(L)`

`=> B`

CHEMISTRY, M1 EQ-Bank 13

Using your knowledge of the atomic radii for the 2nd and 3rd period elements.

  1. Explain the general trend in atomic radius across Period 2.   (3 marks)
  1. Explain why the elements of Period 3 have larger atomic radii than the corresponding elements of Period 2.   (2 marks)
Show Answers Only

a.   Across Period 2, atomic radius decreases from lithium to neon:

  • This occurs because the number of protons (nuclear charge) increases across the period, while electrons are added to the same energy level.
  • As a result, the increased nuclear attraction pulls the electron cloud closer to the nucleus.
  • This leads to reduced atomic radius despite more electrons being present.
  • Therefore, there is a clear trend of decreasing atomic size from left to right across Period 2.

b.   Period 3 atomic radii are larger than corresponding Period 2 element:

  • Period 3 elements have larger atomic radii than Period 2 elements because they have one additional electron shell.
  • This occurs because Period 3 elements have electrons in the n=3 shell, while Period 2 elements only fill up to n=2.
  • Period 3 elements have more protons that create a greater nuclear attractive force with the orbiting electrons but the extra electron shell outweighs the effect of increased nuclear charge.
  • As a result, the outermost electrons in Period 3 are further from the nucleus, causing increased atomic size despite having more protons.
Show Worked Solution

a.   Across Period 2, atomic radius decreases from lithium to neon:

  • This occurs because the number of protons (nuclear charge) increases across the period, while electrons are added to the same energy level.
  • As a result, the increased nuclear attraction pulls the electron cloud closer to the nucleus.
  • This leads to reduced atomic radius despite more electrons being present.
  • Therefore, there is a clear trend of decreasing atomic size from left to right across Period 2.

b.   Period 3 atomic radii are larger than corresponding Period 2 element:

  • Period 3 elements have larger atomic radii than Period 2 elements because they have one additional electron shell.
  • This occurs because Period 3 elements have electrons in the n=3 shell, while Period 2 elements only fill up to n=2.
  • Period 3 elements have more protons that create a greater nuclear attractive force with the orbiting electrons but the extra electron shell outweighs the effect of increased nuclear charge.
  • As a result, the outermost electrons in Period 3 are further from the nucleus, causing increased atomic size despite having more protons.

CHEMISTRY, M1 EQ-Bank 11

Explain the relationship between electronegativity and atomic radius with non-metal reactivity down Group 17 (the halogens) of the Periodic Table.   (4 marks)

Show Answers Only
  • Atomic radius increases down Group 17 because each element gains more electron shells.
  • This leads to weaker nuclear attraction for incoming electrons, due to greater distance from the nucleus and increased shielding by inner shells.
  • As a result, electronegativity decreases down the group, and consequently decreasing reactivity down Group 17
  • For instance, fluorine (small radius, high electronegativity) is most reactive while in contrast, iodine (large radius, low electronegativity) is least reactive.
  • Therefore, smaller atoms with higher electronegativity are more reactive non-metals.
Show Worked Solution
  • Atomic radius increases down Group 17 because each element gains more electron shells.
  • This leads to weaker nuclear attraction for incoming electrons, due to greater distance from the nucleus and increased shielding by inner shells.
  • As a result, electronegativity decreases down the group, and consequently decreasing reactivity down Group 17
  • For instance, fluorine (small radius, high electronegativity) is most reactive while in contrast, iodine (large radius, low electronegativity) is least reactive.
  • Therefore, smaller atoms with higher electronegativity are more reactive non-metals.

CHEMISTRY, M1 EQ-Bank 4 MC

The graph shows the trend in a particular property across the second period of elements.
 

Property A increases steadily from left to right across the period.

Property A is most likely:

  1. Atomic radius
  2. Electronegativity
  3. Metallic character
  4. Reactivity of metals
Show Answers Only

\(B\)

Show Worked Solution
  • Across a period, electronegativity increases as nuclear charge increases and atomic radius decreases.
  • Metallic character/reactivity of metals decreases across the period, not increases.

\(\Rightarrow B\)

CHEMISTRY, M1 EQ-Bank 3 MC

Which is the most electronegative element?

  1. Caesium
  2. Tellurium
  3. Chlorine
  4. Thallium
Show Answers Only

\(C\)

Show Worked Solution
  • Electronegativity is the tendency of an atom to attract electrons and the most electronegative elements are found in the top right corner of the periodic table (excluding noble gases).
  • Hence chlorine is the most electronegative element from the choices.

\(\Rightarrow C\)