Band 5

Probability, 2ADV S3 2025 MET1 8

Consider

\begin{align*}
f(x)=\left\{\begin{array}{cc}
\dfrac{3}{8}(4-3 x) & 0 \leq x \leq \dfrac{4}{3} \\
0 & \text {otherwise }
\end{array}\right.
\end{align*}

The continuous random variable $X$ has probability density function $f(x)$.
Find $k$ such that $\operatorname{Pr}(X>k)=\dfrac{9}{16}$. (3 marks)

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The function $h(x)$ is a transformation of $f(x)$ such that
\begin{align*}
\ \ \ \ h(x)=m f(x)+n
\end{align*}
where $m$ and $n$ are real numbers.
Find $\displaystyle \int_0^{\tfrac{4}{3}} h(x) d x$ in terms of $m$ and $n$. (2 marks)

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Show Answers Only

a. $k=\dfrac{1}{3}$

b. $\displaystyle\int_0^{\frac{4}{3}} h(x)=m+\dfrac{4}{3} n$

Show Worked Solution

a. $\text{Solve}\ \ \operatorname{Pr}(X>k)=\dfrac{9}{16}\ \ \text{for}\ k$ :

$\operatorname{Pr}(X>k)$	$=\displaystyle\int_k^{\frac{4}{3}} \frac{3}{8}(4-3 x) d x$
	$=\displaystyle-\frac{3}{8} \int_k^{\frac{4}{3}}(3 x-4) d x$
	$=-\dfrac{3}{8} \cdot \dfrac{1}{2} \cdot \dfrac{1}{3}\left[(3 x-4)^2\right]_k^{\frac{4}{3}}$
	$=-\dfrac{1}{16}\left[(4-4)^2-(3 k-4)^2\right]$
	$=\dfrac{(3 k-4)^2}{16}$

♦ Mean mark (a) 47%.

$\dfrac{(3 k-4)^2}{16}$	$=\dfrac{9}{16}$
$(3 k-4)^2$	$=9$
$3 k-4$	$= \pm 3$

$3 k=1 \ \ \ \ \text{or}\ \ \ \ 3 k=7$

$k=\dfrac{1}{3} \quad \quad \ \ \ k=\dfrac{7}{3}\ \ \left(\text{No solution as}\ k \in\left[0, \dfrac{4}{3}\right]\right)$

$\therefore\ k=\dfrac{1}{3}$

b. $h(x)=m f(x)+n$

	$\displaystyle\int_0^{\frac{4}{3}} h(x)$	$=\displaystyle \int_0^{\frac{4}{3}}(m f(x)+n) d x$
		$=m \underbrace{\displaystyle \int_0^{\frac{4}{3}} f(x) d x}_{=1\ \ \text{since p.d.f. }}+n \displaystyle\int_0^{\frac{4}{3}} 1\ d x$
		$=m+n[x]_0^{\frac{4}{3}}$
		$=m+\dfrac{4}{3} n$

♦♦ Mean mark (b) 38%.

Trigonometry, 2ADV T3 2025 MET1 3

Let $f(x)=2 \cos (2 x)+1$ over the domain $x \in\left[0, 2 \pi \right]$.

State the range of $f(x)$. (1 mark)

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Solve $f(x)=0$ for $x$. (3 marks)

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Sketch the graph of $y=f(x)$ for $x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]$ on the axes below.
Label the endpoints with their coordinates. (2 marks)

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a. $\text{Range of } f(x):-1 \leqslant y \leqslant 3$

b. $x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

Show Worked Solution

a. $\text{Amplitude}=2 \ \ \text{about} \ \ y=1.$

$\text{Range of } f(x):\ -1 \leqslant y \leqslant 3$

b.	$2 \cos (2 x)+1$	$=0$
	$\cos (2 x)$	$=-\dfrac{1}{2}$

$\text{Base angle}=\dfrac{\pi}{3}$

$2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots$

$2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}$

$x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

♦ Mean mark (c) 49%.

Algebra, MET1 2025 VCAA 9

Consider the functions

$f: R \backslash\{1\} \rightarrow R, f(x)=\dfrac{w^2}{(x-1)^2}$

and

$g: R \rightarrow R, g(x)=(x-w)^2$

where $w \in R$.

If $w=-3$, find the four solutions to $f(x)=g(x)$. (3 marks)

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Consider the case where $w>0$.
1. Find, in terms of $w$, the coordinates of the minimum point of the graph of $y=(x-1)(x-w)$. (2 marks)
  
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2. Hence, or otherwise, find the positive values of $w$ for which $f(x)=g(x)$ has exactly three solutions. (2 marks)
  
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a. $\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):$

$\dfrac{(-3)^2}{(x-1)^2}=(x+3)^2$

$(x+3)^2(x-1)^2=9$

$(x+3)(x-1)= \pm 3$

$\text{Case 1}$

$x^2+2 x-3=3$

$x^2+2 x-6=0$

$x=-1 \pm \sqrt{7}$

$\text{Case 2}$

$x^2+2 x-3$	$=-3$
$x(x+2)$	$=0$
$x=0 \ \ \text{or} \ -2$

b.i. $\text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)$

b.ii. $\text{See worked solutions.}$

Show Worked Solution

a. $\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):$

$\dfrac{(-3)^2}{(x-1)^2}=(x+3)^2$

$(x+3)^2(x-1)^2=9$

$(x+3)(x-1)= \pm 3$

$\text{Case 1}$

$x^2+2 x-3=3$

$x^2+2 x-6=0$

$x=-1 \pm \sqrt{7}$

♦♦ Mean mark (a) 34%.

$\text{Case 2}$

$x^2+2 x-3$	$=-3$
$x(x+2)$	$=0$
$x=0 \ \ \text{or} \ -2$

b.i. $\text{Find minimum TP of} \ \ y=(x-1)(x-w)$

$\text{Axis of quadratic:}\ \ x=\dfrac{w+1}{2},\ \ (w>0)$

$y$	$=\left(\dfrac{w+1}{2}-1\right)\left(\dfrac{w+1}{2}-w\right)$
	$=\left(\dfrac{w-1}{2}\right)\left(\dfrac{1-w}{2}\right)$
	$=-\dfrac{1}{4}(w-1)^2$

$\therefore \ \text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)$

♦♦ Mean mark (b.i) 39%.
♦♦♦ Mean mark (b.ii) 11%.

b.ii. $\text{Solve for \(w\ (w>0)$ where $f(x)=g(x)$:}\)

$\dfrac{w^2}{(x-1)^2}$	$=(x-w)^2$
$(x-w)^2(x-1)^2$	$=w^2$
$[(x-w)(x-1)]^2-w^2$	$=0$

$\text{Difference between two squares:}$

$[(x-w)(x-1)-w][(x-w)(x-1)+w]=0$

$\text{Case 1}$

$(x-w)(x-1)-w$	$=0$
$x^2-x-w x+w-w$	$=0$
$x^2-(w+1) x$	$=0$
$x[x-(w+1)]$	$=0$

$x=0 \ \ \text{or} \ \ w+1$

$\text{Case 2}$

$(x-w)(x-1)+w$	$=0$
$x^2-(w+1) x+2 w$	$=0$

$\text{1 solution if} \ \ \Delta=0:$

$\Delta$	$=[-(w+1)]^2-4 \times 1 \times 2 w$
	$=w^2+2 w+1-8 w$
	$=w^2-6 x+1$

$\text{If} \ \ \Delta=0:$

$w$	$=\dfrac{6 \pm \sqrt{(-6)^2-4 \times 1 \times 1}}{2}$
	$=\dfrac{6 \pm \sqrt{32}}{2}$
	$=3+2 \sqrt{2}\ \ (\text{reject}\ w=3-2 \sqrt{2}\ \ \text{as}\ \ w>0)$

$\therefore f(x)=g(x) \ \ \text{has exactly 3 solutions}\ (w>0).$

Calculus, MET1 2025 VCAA 7

Let $f: R \rightarrow R, f(x)=x^3-x^2-16 x-20$.

Verify that $x=5$ is a solution of $f(x)=0$. (1 mark)

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Express $f(x)$ in the form $(x+d)^2(x-5)$, where $d \in R$. (2 marks)

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Consider the graph of $y=f(x)$, as shown below.
Complete the coordinate pairs of all axial intercepts of $y=f(x)$. (1 mark)

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Let $g: R \rightarrow R, g(x)=x+2$.
1. State the coordinates of the stationary point of inflection for the graph of $y=f(x) g(x)$. (1 mark)
  
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2. Write down the values of $x$ for which $f(x) g(x) \geq 0$. (1 mark)
  
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a. $f(5)=5^3-5^2-16 \times 5-20 =0$

b. $f(x)=(x+2)^2(x-5)$

d.i. $(-2,0)$

d.ii. $x \in(-\infty,-2] \cup[5, \infty)$

Show Worked Solution

a. $f(x)=x^3-x^2-16 x-20$

$f(5)=5^3-5^2-16 \times 5-20=125-25-80-20=0$

$\therefore x=5\ \ \text{is a solution of}\ f(x)$.

b. $\text{By long division:}$

$f(x)=\left(x^2+4 x+4\right)(x-5)=(x+2)^2(x-5)$

d.i.	$y$	$=f(x) \cdot g(x)$
		$=(x+2)^2(x-5)(x+2)$
		$=(x+2)^3(x-5)$

$(x+2)^3\ \text{factor}\ \Rightarrow\ \text{SP of inflection at} \ (-2,0)$

♦ Mean mark (d.i) 44%.
♦♦ Mean mark (d.ii) 27%.

d.ii.

$\text{By inspection of graph:}$

$f(x) g(x) \geqslant 0 \ \ \text{when}\ \ x \leqslant-2\ \cup\ x \geqslant 5$

$x \in(-\infty,-2] \cup[5, \infty) \ \text{also correct.}$

BIOLOGY, M8 EQ-Bank 26

The biological structure shown is part of one of the systems in the body.

Name the biological structure. (1 mark)

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Outline how neural pathways allow the hypothalamus to maintain body temperature within normal range. (3 marks)

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a. Nerve cell/neuron

b. Steps involved in system’s response to a stimulus:

Thermoreceptors detect a change in body temperature and send nerve impulses to the hypothalamus.
The hypothalamus (control centre) processes the signal and sends neural signals to effectors.
Effectors respond (e.g. sweat glands, skeletal muscles, blood vessels) to restore temperature to the normal range.

Show Worked Solution

a. Nerve cell/neuron

b. Steps involved in system’s response to a stimulus:

Thermoreceptors detect a change in body temperature and send nerve impulses to the hypothalamus.
The hypothalamus (control centre) processes the signal and sends neural signals to effectors.
Effectors respond (e.g. sweat glands, skeletal muscles, blood vessels) to restore temperature to the normal range.

Graphs, MET1 2025 VCAA 3

Let $f:[0,2 \pi] \rightarrow R, f(x)=2 \cos (2 x)+1$.

State the range of $f$. (1 mark)

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Solve $f(x)=0$ for $x$. (3 marks)

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Sketch the graph of $y=f(x)$ for $x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]$ on the axes below.
Label the endpoints with their coordinates. (2 marks)

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a. $\text{Range of } f(x):-1 \leqslant y \leqslant 3$

b. $x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

Show Worked Solution

a. $\text{Amplitude}=2 \ \ \text{about} \ \ y=1.$

$\text{Range of } f(x):\ -1 \leqslant y \leqslant 3$

b.	$2 \cos (2 x)+1$	$=0$
	$\cos (2 x)$	$=-\dfrac{1}{2}$

$\text{Base angle}=\dfrac{\pi}{3}$

$2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots$

$2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}$

$x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

♦ Mean mark (c) 49%.

Networks, GEN2 2025 VCAA 18

Frances is constructing a home gym. This project requires 12 activities, $A$ to $L$, to be completed.

The activity network below shows each activity and its completion time in days.

This network contains two critical paths.
State the activities that are common to both critical paths. (1 mark)

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Determine the latest start time, in days, for activity $E$. (1 mark)

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Which activity has the longest float time? (1 mark)

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The table below shows five activities that can have their completion time reduced.
It shows the maximum reduction time (days) and the additional cost per day, for each of the five activities.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \textbf{Activity} \ \ & \textbf{Maximum reduction time} & \textbf{Additional cost per day }\\
\textbf{} & \textbf{(days)} \rule[-1ex]{0pt}{0pt} & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} \textit{A} \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{500} \\
\hline
\rule{0pt}{2.5ex} \textit{F} \rule[-1ex]{0pt}{0pt} & \text{4} \rule[-1ex]{0pt}{0pt} & \text{150} \\
\hline
\rule{0pt}{2.5ex} \textit{G} \rule[-1ex]{0pt}{0pt} & \text{4} \rule[-1ex]{0pt}{0pt} & \text{150} \\
\hline
\rule{0pt}{2.5ex} \textit{H} \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{300} \\
\hline
\rule{0pt}{2.5ex} \textit{K} \rule[-1ex]{0pt}{0pt} & \text{1} \rule[-1ex]{0pt}{0pt} & \text{100} \\
\hline
\end{array}

Frances would like to construct the home gym in three days less than was previously possible.
What is the minimum additional amount Frances will need to pay? (1 mark)

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a. $\text{Critical paths: }ADHJL, ADIKL$

b. $\text{LST for Activity}\ E=9 \ \text{days}$

c. $\text{Activity \(F$.}\)

d. $\$1400$

Show Worked Solution

a. $\text{Scan network forwards/backwards:}$

$\text{Critical paths: }ADHJL, ADIKL$

b. $\text{LST for Activity}\ E$

$=20-4-4-3$

$=9 \ \text{days}$

♦ Mean mark (a) 44%.
♦ Mean mark (b) 46%.

c.	$\begin{array}{\|l\|c\|c\|c\|c\|c\|} \hline \rule{0pt}{2.5ex}\ \ \text{Activity} \ \ \rule[-1ex]{0pt}{0pt}& \ \ B \ \ &\ \ C \ \ & \ \ E \ \ & \ \ F \ \ & \ \ G \ \ \\ \hline \rule{0pt}{2.5ex} \ \ \text{Float} \ \ \rule[-1ex]{0pt}{0pt}& 5 & 1 & 5 & 6 & 1 \\ \hline \end{array}$

$\therefore\ \text{Activity \(F$ has the longest float time.}\)

♦ Mean mark (c) 47%.

d. $\text{Current time = 25 days}$

$\text{New time = 22 days.}$

$\text{Reduce:} \ A(2 \ \text {days}), H(1 \ \text{day}), K(1 \ \text{day})$

$\text{Minimum cost}=2 \times 500+1 \times 300+1 \times 100=\$1400$

♦♦♦ Mean mark (d) 21%.

Networks, GEN2 2025 VCAA 17

A gym is hosting a competition in which competitors will complete activities at eight different stations.

On the network below, the vertices represent the stations. The edges represent the walkways between the stations and the numbers show the distance, in metres, between them.

The gym owner would like to make sure that all the walkways are clear before the competition starts.

The gym owner would like to begin and end the inspection of the walkways at station $A$.

Explain, with reference to the degrees of the vertices, why the gym owner's intended route must involve some repeated edges. (1 mark)

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What is the minimum distance, in metres, that the gym owner will cover when completing the inspection? (1 mark)

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a. $\text{Vertices \(C$ and $E$ are of odd degree.}\)

$\text{The circuit described is Eulerian which requires all vertices to be}$

$\text{of even degree if edges are only to be used once.}$

b. $\text{Minimum distance = 19 m}$

Show Worked Solution

a. $\text{Vertices \(C$ and $E$ are of odd degree.}\)

$\text{The circuit described is Eulerian which requires all vertices to be}$

$\text{of even degree if edges are only to be used once.}$

♦ Mean mark (a) 40%.

b. $\text{Minimum distance}$

$=28+11+23+31+7+7+8+12+22+19+14+14$

$=196 \ \text{m}$

♦♦ Mean mark (b) 27%.

Matrices, GEN2 2025 VCAA 14

An early learning centre runs seven different activities during its 40-day holiday program.

The activities are cooking $(C)$, drama $(D)$, gardening $(G)$, lunch $(L)$, music $(M)$, reading $(R)$ and sport $(S)$.

The timetabled order of the activities for day one of the holiday program is shown below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \text{9 am} \ \ \rule[-1ex]{0pt}{0pt} & \text{10 am} \rule[-1ex]{0pt}{0pt} & \text{11 am} \rule[-1ex]{0pt}{0pt} & \text{12 pm} \rule[-1ex]{0pt}{0pt} & \ \ \text{1 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{2 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{3 pm} \ \ \\
\hline
\rule{0pt}{2.5ex} \textit{C} \rule[-1ex]{0pt}{0pt} & \textit{D} \rule[-1ex]{0pt}{0pt} & \textit{G} \rule[-1ex]{0pt}{0pt} & \textit{L} \rule[-1ex]{0pt}{0pt} & \textit{M} \rule[-1ex]{0pt}{0pt} & \textit{R} \rule[-1ex]{0pt}{0pt} & \textit{S}\\
\hline
\end{array}

The timetabled order of the activities for day one is also shown in matrix $X$ below.

\begin{aligned}
X = & \begin{bmatrix}
C \\
D \\
G \\
L \\
M \\
R\\
S \\
\end{bmatrix}
\end{aligned}

Matrix $P$, shown below, is a permutation matrix used to determine the timetabled order of activities from one day to the next.

\begin{aligned}
P = & \begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{aligned}

A column matrix containing the timetabled order of activities on one day is multiplied by matrix $P$ to determine the timetabled order of activities for the next day.

State the activities that are always held at the same time on each day of the program. (1 mark)

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Determine the timetabled order of the seven activities on day three of the program. (1 mark)

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$P^4$ is an identity matrix.
Explain what this means for the timetabled order of the activities over the 40-day holiday program. (1 mark)

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a. $\text{Gardening, lunch, music.}$

b. $\text{Drama, cooking, gardening, lunch, music, sport, reading.}$

c. $\text{Order of activities rotate on a 4 day cycle.}$

$\text{Over 40 days, there will be 10 cycles of activities}$

Show Worked Solution

a. $\text{Activities held af the same time:}$

$\Rightarrow \ \text{correspond to 1’s in leading diagonal}$

$\text{Gardening, lunch, music.}$

\begin{aligned}X_2=P \times X=\begin{bmatrix}
R \\
S \\
G \\
L \\
M \\
D \\
C
\end{bmatrix}, \quad X_3=P \times X_2=\begin{bmatrix}
D \\
C \\
G \\
L \\
M \\
S \\
R
\end{bmatrix}
\end{aligned}

$\text{Order on day 3:}$

$\text{Drama, cooking, gardening, lunch, music, sport, reading.}$

♦♦ Mean mark (b) 27%.

c. $P^4 \ \Rightarrow \ \text{identity matrix}$

$\text{Order of activities rotate on a 4 day cycle.}$

$\text{Over 40 days, there will be 10 cycles of activities}$

♦♦♦ Mean mark (c) 11%.

Networks, GEN2 2025 VCAA 15

Frances lives in a housing estate.

On the graph below the vertices represent her favourite locations, and the edges represent the roads between them.

Calculate the sum of the degrees of all the vertices in this graph. (1 mark)

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Euler's formula, $v+f=e+2$, holds for this graph.
Complete the formula by writing the appropriate numbers in the boxes below. (1 mark)

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Frances is at the gym. She would like to visit each of the other locations once and end at her home.
What is the mathematical term used to describe this route? (1 mark)

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Using edges from the original graph, construct a spanning tree below. (1 mark)

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a. $\text{Sum of degrees}=2+4+2+3+3=14$

b. $5+4=7+2$

c. $\text{Hamiltonian path}$

d. $\text{Spanning tree (one of many possibilities):}$

Show Worked Solution

a. $\text{Sum of degrees}=2+4+2+3+3=14$

b. $v+f=e+2 \ \ \Rightarrow \ \ 5+4=7+2$

c. $\text{Hamiltonian path}$

♦ Mean mark (c) 53%.

d. $\text{Spanning tree (one of many possibilities):}$

Matrices, GEN2 2025 VCAA 12

The early learning centre contains three rooms, Nursery $(N)$, Toddler $(T)$ and Pre-kinder $(P)$ .

From one year to the next, children can move between rooms, stay in the same room, or may leave $(L)$ the centre. The following transition matrix, $M$, shows the expected proportion of children who will move between categories or stay in the same category from one year to the next.

\begin{aligned}
& \quad \quad \quad \quad \quad \textit{this year} \\
& \quad \quad \ \ \ \ N \quad \quad \ \ T \quad \quad P \quad \ L \\
M&=\begin{bmatrix}
0.25 & 0 & 0 & 0 \\
0.625 & 0.25 & 0 & 0 \\
0 & 0.625 & 0.1 & 0 \\
0.125 & 0.125 & 0.9 & 1
\end{bmatrix}\begin{array}{l}
N\\
T \\
P \\
P
\end{array}\quad \textit{next year}
\end{aligned}

The number of children expected to be in each of the four categories, from one year to the next, can be calculated using the matrix recurrence relation
1. $S_{n+1}=M S_n$
where $S_n$ represents the expected number of children in each of the four categories at the start of year $n$.
The state matrix $S_{2024}$, shown below, gives the number of children in each category at the start of 2024.
1. 1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
Find $S_{2023}$. (1 mark)

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From the start of 2025, new children commenced in the early learning centre at the start of each year.
A new matrix recurrence relation for determining the expected number of children in each of the four categories from one year to the next is
1. 1. \begin{align*}
    S_{n+1}=M S_n+B
    \end{align*}
where
1. 1. \begin{align*}B=\left[\begin{array}{c}12 \\5 \\10 \\0\end{array}\right] \begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
gives the number of new children enrolled in each room of the early learning centre at the start of each year.
Given the state matrix
1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
find the expected total number of children to be enrolled in the early learning centre at the start of 2026. Round your answer to the nearest whole number. (1 mark)

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$S_{2023}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}$

b. $\text{Children enrolled (2026) =50}$

Show Worked Solution

a. $S_{n+1}=M S_n \ \Rightarrow \ S_n=M^{-1} S_{n+1}$

$S_{2023}=\begin{bmatrix}0.25 & 0 & 0 & 0 \\ 0.625 & 0.25 & 0 & 0 \\ 0 & 0.625 & 0.1 & 0 \\ 0.125 & 0.125 & 0.9 & 1\end{bmatrix}^{-1}\begin{bmatrix}4 \\ 15 \\ 15 \\ 27\end{bmatrix}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}$

♦ Mean mark (a) 53%.

$S_{2025}=M S_{2024} + B=\begin{bmatrix}13 \\ 11.25 \\ 20.87 \\ 42.87\end{bmatrix}$

$S_{2026}=M S_{2025} + B=\begin{bmatrix}15.25 \\ 15.93 \\ 19.11 \\ 64.69\end{bmatrix}$

$\text{Children enrolled (2026)\(=15.25+15.93+19.11=50$ (nearest whole)}\)

♦♦♦ Mean mark (b) 13%.

Financial Maths, STD2 F4 2025 GEN1 7

Declan is a filmmaker and content creator.

He has taken out a reducing balance loan to fund a new production.

Interest is calculated monthly and Declan makes monthly repayments.

Three rows of the amortisation table for Declan’s loan are shown below.

\begin{array}{|c|c|c|c|c|}
\hline
\hline \rule{0pt}{2.5ex}\quad \textbf{Payment} \quad & \quad\textbf{Payment} \quad & \quad\textbf{Interest} \quad& \textbf{Principal} & \quad\textbf{Balance}\quad\\
\textbf{number} & \textbf{(\$)} \rule[-1ex]{0pt}{0pt}& \textbf{(\$)} & \quad\textbf{reduction (\$)} \quad& \textbf{(\$)}\\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 0.00 & 0.00 & 0.00 & 850\,000.00 \\
\hline \hline \rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & 15\,730.88 & 2975.00 & 12\,755.88 & 837\,244.12 \\
\hline \hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & 2930.35 & 12\,800.53 & 824\,443.59 \\
\hline
\end{array}

What amount, in dollars, did Declan borrow? (1 mark)
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Why is the interest associated with payment 2 lower than the interest associated with payment 1? (1 mark)
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The interest rate on Declan’s loan is 4.2% per annum, compounding monthly.
Using the values in the table, complete the table below.
Round all values to the nearest cent. (2 marks)

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

Show Answers Only

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

Show Worked Solution

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Monthly interest}=\dfrac{4.2}{12}=0.35\%$

$\text{Calculating missing values in the table:}$

$\text{Interest}=\dfrac{0.35}{100} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

♦ Mean mark (c) 40%.

Matrices, GEN2 2025 VCAA 11

An early learning centre contains three rooms, Nursery $(N)$, Toddler $(T)$ and Pre-kinder $(P)$.

The Nursery and Toddler rooms each have capacity for eight children and the Pre-kinder room has capacity for 20 children, as shown in matrix $C$ below.

\begin{align*}
C=\left[\begin{array}{c}
8 \\
8 \\
20
\end{array}\right] \begin{aligned}
& N \\
& T \\
& P
\end{aligned}
\end{align*}

Matrix $E$ shows enrolment numbers for each room for one week, Monday to Friday.

\begin{aligned}
& \quad \ \ \ Mon \quad Tue\quad Wed \ \ \ Thu \ \ \ Fri \\
E&=\begin{bmatrix}
6 & \quad 8 & \quad 8 & \quad 8 & \quad 5 \\
7 & \quad 8 & \quad 7 & \quad 8 & \quad 6 \\
18 & \ \ \ 18 &\ \ \ 17 &\ \ \ 15 &\ \ \ 13
\end{bmatrix}\begin{array}{l}
N \\
T \\
P
\end{array}\\
\end{aligned}

State the order of matrix $E$. (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
The following matrix multiplication has been completed to determine a new matrix, $W$.
1. \begin{align*}
  \begin{bmatrix}
  1 & 1 & 1
  \end{bmatrix} \times E=W
  \end{align*}
What information does matrix $W$ provide regarding enrolments at the early learning centre? (1 mark)

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It has been decided that the capacity of the Nursery room will be increased by 25% and the capacity of the Toddler room will be increased by 50%. The capacity of the Pre-kinder room will be reduced by 10%.
The new capacities for the three rooms $(C_{\text {new}})$ can be determined from the matrix product
1. $C_{\text {new }}=F C$
where $F$ is a diagonal matrix.
Write down the matrix $F$. (1 mark)
--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Order of matrix} \ \ E=3 \times 5$

b. $\text { Total enrolments for each day of the week.}$

$F=\begin{bmatrix}1.25 & 0 & 0 \\ 0 & 1.5 & 0 \\ 0 & 0 & 0.9\end{bmatrix}$

Show Worked Solution

a. $\text{Order of matrix} \ \ E=3 \times 5$

b. $\text { Total enrolments for each day of the week.}$

$F=\begin{bmatrix}1.25 & 0 & 0 \\ 0 & 1.5 & 0 \\ 0 & 0 & 0.9\end{bmatrix}$

♦ Mean mark (c) 44%.

Financial Maths, GEN2 2025 VCAA 9

Declan takes out a new loan of $50 000 to promote his new film.

Interest on this loan compounds weekly and Declan makes weekly repayments of $75.

With these weekly repayments of $75, suppose the balance of Declan’s loan does not change over time.
Determine the weekly interest rate. (1 mark)

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With these weekly repayments of $75, suppose the balance of Declan's loan now reduces over time.
The balance of the loan, in dollars, after $n$ weeks, $L_n$, can be determined using a recurrence relation of the form
1. $L_0=50\,000, \quad L_{n+1}=R L_n-75$
Assume there are exactly 52 weeks in a year
After one year, Declan owes $49 565.34
Determine
1. the per annum interest rate, compounding weekly, as a percentage, rounded to two decimal places. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---
2. the value of $R$ rounded to four decimal places. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Weekly interest rate}=0.15 \%$

b.i. $\text{Interest rate}=6.96 \% \ \text{(2 d.p)}$

b.ii. $R=1.0013$

Show Worked Solution

a. $\text{Weekly interest rate}=\dfrac{75}{50\,000} \times 100=0.15 \%$

b.i. $L_0=50\,000 \quad L_{n+1}=R L_n-75$

$\text{When} \ \ N=52,\ \text{balance}=\$ 49\,565.34$

$\text{Solve for \(I$ (by CAS):}\)

$N$	$=52$
$I(\%)$	$=\boldsymbol{6.96000}$
$PV$	$=50\,000$
$PMT$	$=-75$
$FV$	$=-49\,565.34$
$PY$	$=CY=52$

$\text{Interest rate}=6.96 \% \ \text{(2 d.p)}$

b.ii. $R=1+\dfrac{6.96}{100 \times 52}=1.0013$

♦ Mean mark (a) 40%.
♦ Mean mark (b.i) 40%
♦♦ Mean mark (b.ii) 30%

Financial Maths, GEN2 2025 VCAA 8

Declan depreciates the value of his lighting equipment using flat rate depreciation.

The graph below shows the value, in dollars, of the lighting equipment, $V_n$, after $n$ years.

The value of the lighting equipment could be modelled by either a recurrence relation or a rule.
1. Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$. (1 mark)
  
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2. Write a rule for $V_n$ in terms of $n$. (1 mark)
  
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What is the annual flat rate depreciation percentage applied to the lighting equipment? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---

Show Answers Only

a.i. $V_0=40\,000 \quad V_{n+1}=V_n-8000$

a.ii. $V_n=40\,000-8000 \times n$

b. $\text{Depreciation}=20\%$

Show Worked Solution

a.i. $\text{Recurrence relation:}$

$V_0=40\,000 \quad V_{n+1}=V_n-8000$

a.ii. $V_n=40\,000-8000 \times n$

b. $\text{Depreciation %}=\dfrac{8000}{40\,000} \times 100=20\%$

♦ Mean mark (a.ii) 50%.

Financial Maths, GEN2 2025 VCAA 7

Declan is a filmmaker and content creator.

He has taken out a reducing balance loan to fund a new production.

Interest is calculated monthly and Declan makes monthly repayments.

Three rows of the amortisation table for Declan’s loan are shown below.

What amount, in dollars, did Declan borrow? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
Why is the interest associated with payment 2 lower than the interest associated with payment 1? (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
The interest rate on Declan’s loan is 4.2% per annum, compounding monthly.
Using the values in the table, complete the table below.
Round all values to the nearest cent. (1 mark)

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

The last payment required to fully repay the loan is $15 730.71, correct to the nearest cent.
How many payments of $15 730.88 did Declan make before this final payment? (1 mark)

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

Show Answers Only

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

d. $\text{59 payments made before the final payment.}$

Show Worked Solution

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

♦ Mean mark (c) 40%.

d. $\text{Solve for \(N$ (by CAS):}\)

$N$	$=\boldsymbol{59.99 \ldots}$
$I(\%)$	$=4.2$
$PV$	$=850\,000$
$PMT$	$=-15\,730.88$
$FV$	$=0$
$PY$	$=CY=12$

$\therefore \ \text{59 payments made before the final payment.}$

♦♦♦ Mean mark (d) 21%.

Data Analysis, GEN2 2025 VCAA 5

The table below shows sale price and the number of days on the market before sale, days, for a sample of 10 apartments sold in a particular suburb.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\ \ \textbf{Sale price(\$)} \ \ \rule[-1ex]{0pt}{0pt}& \quad \quad \textbf{Days}\quad \quad \\
\hline \rule{0pt}{2.5ex}950\,000 \rule[-1ex]{0pt}{0pt}& 15 \\
\hline \rule{0pt}{2.5ex}925\,000 \rule[-1ex]{0pt}{0pt}\rule[-1ex]{0pt}{0pt}& 18 \\
\hline \rule{0pt}{2.5ex}900\,000 & 23 \\
\hline \rule{0pt}{2.5ex} 900\,000 \rule[-1ex]{0pt}{0pt}& 24 \\
\hline \rule{0pt}{2.5ex}905\,000 \rule[-1ex]{0pt}{0pt}& 26 \\
\hline \rule{0pt}{2.5ex}750\,000 \rule[-1ex]{0pt}{0pt}& 28 \\
\hline \rule{0pt}{2.5ex}680\,000 \rule[-1ex]{0pt}{0pt}& 31 \\
\hline \rule{0pt}{2.5ex}800\,000 \rule[-1ex]{0pt}{0pt}& 35 \\
\hline \rule{0pt}{2.5ex}590\,000 \rule[-1ex]{0pt}{0pt}& 46 \\
\hline \rule{0pt}{2.5ex}600\,000 \rule[-1ex]{0pt}{0pt}& 65 \\
\hline
\end{array}

Use the data in the table above to find the equation of the least squares line.
Write your answers in the boxes below, rounding both values to four significant figures. (2 marks)

--- 0 WORK AREA LINES (style=lined) ---
sale price$=\begin{array}{|c|}\hline \rule{0pt}{5 ex} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \hline \end{array} \times \begin{array}{|c|}\hline \rule{0pt}{5 ex}\quad \quad \quad \quad \quad \quad\\ \hline \end{array} \times$days
For this data, Pearson's correlation coefficient is $r=-0.866$, rounded to three decimal places.
Explain the meaning of the coefficient of determination, as a whole percentage, in the context given in this question. (1 mark)

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

Show Answers Only

a. $\textit{sale price}=1\,050\,000-8050 \times \textit{days} \ \text{(4 sig. fig.)}$

b. $\text{75% of the variation in sale price can be explained by the variation in days.}$

Show Worked Solution

a. $\text{By calculator:}$

$\textit{sale price}=1\,050\,000-8050 \times \textit{days} \ \text{(4 sig. fig.)}$

♦ Mean mark (a) 50%.

b. $(-0.866)^2=0.7499 \ldots \approx 75 \%$

$\text{75% of the variation in sale price can be explained by the variation in days.}$

♦ Mean mark (b) 44%.

Data Analysis, GEN2 2025 VCAA 4

The scatterplot below shows the sale price of a home, in dollars, against the distance of the home from the city centre of Melbourne, in kilometres, distance from city centre.

The sample consists of three‑bedroom homes sold between 2016 and 2018

The equation of the least squares line for the data in the scatterplot is

sale price$=1\,765\,353-35\,054 \times$distance from city centre

The coefficient of determination is 0.0806

Identify the explanatory variable in the least squares equation. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Calculate the value of the correlation coefficient $r$. Round your answer to three decimal places. (1 mark)

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Use the equation of the least squares line to predict the sale price for a three-bedroom home, located in the city centre of Melbourne, sold between 2016 and 2018. (1 mark)

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Jocelyn wants to sell her three-bedroom home located two kilometres from the city centre of Melbourne.
Would the predicted sale price be an example of interpolation or extrapolation?
Briefly explain your answer. (1 mark)

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Describe the linear association between sale price and distance from city centre in terms of its strength and direction. Answer in the table below. (2 marks)

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

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

A residual plot associated with the least squares line is shown below.
It is missing one point.

The residual associated with the home that is furthest from the city centre of Melbourne is missing from the residual plot. The home is 15.5 km from the city centre and sold for $1 250 000.
1. Show that the value of the missing residual is 27 984. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
2. Plot the residual from part i by placing an $\text{X}$ on the residual plot above. (1 mark)
  
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Show Answers Only

a. $\text{Distance from city centre}$

b. $r=-0.284$

c. $\text{sale price}=1\,765\,353$

d. $\text{Extrapolation as 2 km lies outside the explanatory variable data range.}$

e. $\text{Strength: weak. Direction: negative}$

f.i. $\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016$

$\text{Actual sale price}=1\,250\,000$

$\text{Residual}=1\,250\,000-1\,222\,016=27\,984$

f.ii.

Show Worked Solution

a. $\text{Distance from city centre}$

b. $\text{Since slope is negative}$

$r=-\sqrt{0.0806}=-0.284$

♦♦♦ Mean mark (b) 21%.

c. $\text{Find sale price when distance fran city centre}=0:$

$\text{sale price}=1\,765\,353$

d. $\text{Extrapolation as 2 km lies outside the explanatory variable data range.}$

$\text{(Note: interpolation/extrapolation should be referenced to the}$

$\text{explanatory variable range).}$

♦♦ Mean mark (d) 27%.

e. $\text{Strength: weak}$

$\text{Direction: negative}$

f.i. $\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016$

$\text{Actual sale price}=1\,250\,000$

$\text{Residual}=1\,250\,000-1\,222\,016=27\,984$

♦ Mean mark (f.i) 41%.
♦ Mean mark (f.ii) 45%

f.ii.

Data Analysis, GEN2 2025 VCAA 1

Table 1, below, shows the prices in dollars, price, for a sample of 20 homes sold in an inner Melbourne suburb during 2017.

The type of home sold is either an apartment or a house.

Table 1

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \ \textit{Price(\$)}\quad \ \rule[-1ex]{0pt}{0pt}& \textit{Type} \\
\hline \rule{0pt}{2.5ex}350\,000 \rule[-1ex]{0pt}{0pt}& \ \ \text{apartment}\ \ \\
\hline \rule{0pt}{2.5ex}490\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}500\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}620\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}720\,000 \rule[-1ex]{0pt}{0pt}\rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}830\,000 & \text{apartment}\\
\hline \rule{0pt}{2.5ex}875\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}995\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,100\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,520\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}800\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}840\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,010\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,263\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,398\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,460\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline
\end{array}

Find the median, in dollars, of the variable price. (1 mark)

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State whether the variable type is numerical, nominal or ordinal. (1 mark)

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1. Complete the table below by finding the standard deviation, to the nearest whole number, for the sale price of apartments in the sample. (1 mark)
  
  --- 0 WORK AREA LINES (style=lined) ---
2. Table 2
  \begin{array}{|c|c|}
  \hline \rule{0pt}{2.5ex} \quad \ \ \textbf{Type of home} \quad \ \ & \quad \textbf{Standard deviation of} \quad \\
  & \rule[-1ex]{0pt}{0pt}\textbf{sale price (\$)}\\
  \hline \rule{0pt}{2.5ex}\text{house} \rule[-1ex]{0pt}{0pt}& 300\,911 \\
  \hline \rule{0pt}{2.5ex}\text{apartment} \rule[-1ex]{0pt}{0pt}& \\
  \hline
  \end{array}
3. Using the information in Table 2, comment on the relative spread in the distribution of the sale prices of houses compared with apartments in this sample. (1 mark)
  
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Table 3, below, shows the percentage of houses and apartments with prices in the given ranges. Some information is missing.
Use the data from Table 1 to complete Table 3.
--- 0 WORK AREA LINES (style=lined) ---
Table 3

Show Answers Only

a. $\text{Median}=920\,000$

b. $\text{Variable is nominal}$

c.i. $\text{Std deviation = \$346 466}$

c.ii. $\text {Apartment sale prices have a higher spread (standard deviation) than the}$

$\text{spread of house sale prices.}$

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Show Worked Solution

a. $\text{Median}=\dfrac{10^{\text{th}}+11^{\text{th}}}{2}=920\,000$

b. $\text{Type is qualitative and cannot be ordered}$

$\Rightarrow \ \text{Variable is nominal}$

c.i. $\text{By calculator,}$

$\text{Std deviation = \$346 466}$

c.ii. $\text {Apartment sale prices have a higher spread (standard deviation) than the}$

$\text{spread of house sale prices.}$

♦ Mean mark (c.ii) 47%.

Networks, STD2 N3 EQ-Bank 38

The precedence table below shows the 12 activities required to complete a project. The duration in days and immediate predecessors are shown.

\begin{array}{|c|c|c|}
\hline \rule{0pt}{2.5ex}\ \ \textbf{Activity} \ \ & \ \ \textbf{Duration} \ \ & \textbf{Immediate predecessors} \\
\hline \rule{0pt}{2.5ex}A \rule[-1ex]{0pt}{0pt}& 4 & - \\
\hline \rule{0pt}{2.5ex}B \rule[-1ex]{0pt}{0pt}& 6 & A \\
\hline \rule{0pt}{2.5ex}C \rule[-1ex]{0pt}{0pt}& 8 & A \\
\hline \rule{0pt}{2.5ex}D \rule[-1ex]{0pt}{0pt}& 3 & A \\
\hline \rule{0pt}{2.5ex}E \rule[-1ex]{0pt}{0pt}& 9 & B\\
\hline \rule{0pt}{2.5ex}F \rule[-1ex]{0pt}{0pt}& 6 & C \\
\hline \rule{0pt}{2.5ex}G \rule[-1ex]{0pt}{0pt}& 7 & B, D, F \\
\hline \rule{0pt}{2.5ex}H \rule[-1ex]{0pt}{0pt}& 12 & C \\
\hline \rule{0pt}{2.5ex}I \rule[-1ex]{0pt}{0pt}& 6 & G, H \\
\hline \rule{0pt}{2.5ex}J \rule[-1ex]{0pt}{0pt}& 4 & E, I \\
\hline \rule{0pt}{2.5ex}K \rule[-1ex]{0pt}{0pt}& 3 & G, H \\
\hline \rule{0pt}{2.5ex}L \rule[-1ex]{0pt}{0pt}& 9 & J \\
\hline
\end{array}

The project is to be completed in minimum time.

Sketch the network, identifying each activity and its duration, including any dummy activities. (3 marks)
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Determine the critical path of the network. (1 mark)
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Show Answers Only

a. $\text{Sketch the network:}$

b. $\text {Critical Path:}\ ACFGIJL$

Show Worked Solution

a. $\text{Sketch the network:}$

b. $\text{Critical path only requires the forward scanning in the above network.}$

$\text {Critical Path:}\ ACFGIJL$

Networks, GEN1 2025 VCAA 40 MC

The precedence table below shows the 12 activities required to complete a project. The duration in days and immediate predecessors are shown.

The project is to be completed in minimum time.

The float time, in days, of Activity $B$ is

Show Answers Only

$C$

Show Worked Solution

$\text{The precedence table produces the network:}$

$\text {Critical Path:}\ ACFGIJL$

$\text{Float time of \(B=$ LST of $E-$EST of $B-$duration $B=18-4-6=8$}\)

$\Rightarrow C$

♦♦ Mean mark 39%.

Networks, GEN1 2025 VCAA 38 MC

The diagram below shows the network of roads that Sofia can use to travel between home and school.

The numbers on the roads show the length, in metres, of each section of road.

Using this network of roads, the shortest distance that Sofia can take to travel from home to school, in metres, is

3700
3800
3900
5100

Show Answers Only

$A$

Show Worked Solution

$\text{Shortest path}=900+500+400+800+1100=3700$

$\Rightarrow A$

♦ Mean mark 47%.

Financial Maths, GEN1 2025 VCAA 24 MC

Dennis invests $800 000 in an annuity.

The annuity earns interest at 4.8% per annum, compounding monthly.

After the interest is credited to the annuity, Dennis receives a monthly payment of $6000.

After some years, the balance of the annuity is $521118.96

For the remainder of the annuity, Dennis receives a monthly payment of $4767.66.

The total number of years for which Dennis receives payments is closest to

Show Answers Only

$A$

Show Worked Solution

$\text{Calculate time until annuity value is}\ \$521\,118.96.$

$\text{Solve for} \ N \ \text {(by CAS):}$

$N$	$=\boldsymbol{84}$
$I(\%)$	$=4.8$
$PV$	$=800\,000$
$PMT$	$=-6000$
$FV$	$=521\,118.96$
$PY$	$=CY=12$

♦ Mean mark 52%.

$\text{After 84 months, calculate time until annuity = 0.}$

$\text{Solve for} \ N \ \text {(by CAS):}$

$N$	$=\boldsymbol{144}$
$I(\%)$	$=4.8$
$PV$	$=521\,118.96$
$PMT$	$=-4767.66$
$FV$	$=0$
$PY$	$=CY=12$

$\therefore \ \text {Length of annuity}=\dfrac{84+144}{12}=19 \text { years.}$

$\Rightarrow A$

Financial Maths, GEN1 2025 VCAA 23 MC

Virat invested $5000 into an account that earned interest compounding fortnightly.

The effective annual interest rate for Virat's investment was 4.51%.

Assume that there are exactly 26 fortnights in one year.

After five years, the amount of interest earned by Virat was closest to

$1128
$1234
$1262
$1264

Show Answers Only

$B$

Show Worked Solution

$\text{Find nominal interest rate (by CAS):}$

$\operatorname{nom}(4.51,26)=4.415 \%$

$\text{Solve for} \ FV \ \text {(by CAS):}$

$N$	$=5 \times 26=130$
$I(\%)$	$=4.415$
$PV$	$=-5000$
$PMT$	$=0$
$FV$	$=\boldsymbol{6233.89}$
$PY$	$=CY=26$

$\therefore \text{Interest}=6233.89-5000=\$1233.89$

$\Rightarrow B$

♦ Mean mark 41%.

Financial Maths, GEN1 2025 VCAA 19 MC

The table below compares the value of an asset at various times using the flat rate and reducing balance methods of depreciation.

\begin{array}{|l|c|c|}
\hline & \rule{0pt}{2.5ex}\ \ \quad \quad \textbf{Flat rate (\$)} \quad \ \ \quad \rule[-1ex]{0pt}{0pt}& \textbf{Reducing balance (\$)} \\
\hline \rule{0pt}{2.5ex}\text{Original value} \rule[-1ex]{0pt}{0pt}& 60\,000.00 & 60\,000.00 \\
\hline \rule{0pt}{2.5ex}\text{Value after 1 year} \rule[-1ex]{0pt}{0pt}& 56\,000.00 & 55\,200.00 \\
\hline \rule{0pt}{2.5ex}\text{Value after 2 years} \rule[-1ex]{0pt}{0pt}& 52\,000.00 & 50\,784.00 \\
\hline \rule{0pt}{2.5ex}\text{Value after 3 years} \quad \rule[-1ex]{0pt}{0pt}& 48\,000.00 & 46\,721.28 \\
\hline
\end{array}

After how many years will the value using flat rate depreciation first be lower than the value using reducing balance depreciation?

Show Answers Only

$B$

Show Worked Solution

$\text {Flat rate: value decreases by \$4000 each year.}$

$\text{Value}\ =6000-4000 \times n$

$\text {Reducing balance:}\ \ r=\dfrac{55\,200}{60\,000}=0.92$

$\text{Value }=6000 \times(0.92)^n$

$\text{Solve (by CAS):}\ 6000-4000 \times n=6000 \times 0.92^n$

$n=5.582 \ldots$

$\therefore \ \text{Flat rate value is lower after 6 years.}$

$\Rightarrow B$

♦ Mean mark 49%.

Financial Maths, GEN1 2025 VCAA 18 MC

A recurrence relation is of the form

$u_0=a, \quad u_{n+1}=R u_n+d$

If $a>0, R=0.5$ and $d=0$, the sequence generated will be

arithmetic and increasing.
arithmetic and decreasing.
geometric and increasing.
geometric and decreasing.

Show Answers Only

$D$

Show Worked Solution

$\text{Given \(\ d=0\ \ \Rightarrow$ series is geometric.}\)

$\text{Given \(\ 0<R<1\ \ \Rightarrow$ sequence is decreasing.}\)

$\Rightarrow D$

♦ Mean mark 52%.

Financial Maths, GEN1 2025 VCAA 17 MC

Dani invests $4000 for three years.

The account earns simple interest at 4% per annum.

The balance in the account after three years can be calculated using

$4000 \times 1.04^3$
$3(1.04 \times 4000)$
$4000+(0.04 \times 4000)^3$
$4000+3(0.04 \times 4000)$

Show Answers Only

$D$

Show Worked Solution

$\text {Account balance}$	$=P+\dfrac{P R T}{100}$
	$=4000+\dfrac{4000 \times 4 \times 3}{100}$
	$=4000+3(0.04 \times 4000)$

$\Rightarrow D$

♦ Mean mark 53%.

Data Analysis, GEN1 2025 VCAA 16 MC

The seasonal index for the number of meat pie sales in winter is 1.75

To correct for seasonality, the actual number of meat pie sales for winter should be reduced, to the nearest whole percentage, by

Show Answers Only

$B$

Show Worked Solution

$\text{Seasonal index}$	$=\dfrac{\text {actual}}{\text {deseasonalised}}$
$\text{deseasonalised}$	$=\dfrac{\text{actual}}{\text{seasonal index}}=\dfrac{\text{actual}}{1.75}=\text{actual} \times 0.57$

$\therefore \ \text{Reduce actual figure by 43% to adjust for seasonality.}$

$\Rightarrow B$

♦ Mean mark 50%.

Data Analysis, GEN1 2025 VCAA 4 MC

The Association of Southeast Asian Nations (ASEAN) has 10 member nations.

The population density, in people per km², for each of these nations in 2024 is displayed in the histogram below. The histogram has a logarithmic (base 10) scale.

Singapore is a member of ASEAN.

In 2024, Singapore's population was 6 028 460 and its total area was 720 km².

In which labelled column does the value of $\log _{10}$(population density) for Singapore lie?

$\text{A}$
$\text{B}$
$\text{C}$
$\text{D}$

Show Answers Only

$D$

Show Worked Solution

$\text{Population density}\ = \dfrac{6\,028\,460}{720}=8372.86…\ $

$\log_{10}(8372.86)=3.92… $

$\Rightarrow D$

♦ Mean mark 52%.

Financial Maths, STD2 EQ-Bank 34

Yuki works as a musician and receives royalties from streaming services. A spreadsheet is used to calculate her quarterly royalty earnings.

Total royalty earnings = Number of streams $\times$ Royalty rate per stream

A spreadsheet showing Yuki's quarterly royalty earnings is shown.

In the following quarter, Yuki's total royalty earnings were $4200. The royalty rate per stream remained at $0.004. Calculate the number of streams Yuki received that quarter. (2 marks)

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

Show Answers Only

$1\,050\,000$

Show Worked Solution

$\text{Total royalty earnings (B8)} = $4200$

$\text{Royalty rate per stream (B5)}=$0.004$

$\text{Let the number of streams (B4)}=N$

$\text{Total royalty earnings} = \text{Number of streams} \times\text{Royalty rate per stream}$

$\text{B8}$	$=\text{B4}^*\text{B5}$
$4200$	$=N\times 0.004$
$N$	$=\dfrac{4200}{0.004}$
	$=1\,050\,000$

$\text{Yuki had 1 050 000 streams in the next quarter}$

Financial Maths, STD2 EQ-Bank 29

Priya works as a sales representative and earns a base wage plus commission. A spreadsheet is used to calculate her weekly earnings.

Total weekly earnings = Base wage + Total sales $\times$ Commission rate

A spreadsheet showing Priya's weekly earnings is shown.

Write down the formula used in cell B9, using appropriate grid references. (1 mark)

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In the following week, Priya earned total weekly earnings of $1437.50. Her base wage and commission rate remained unchanged. Calculate Priya's total sales for that week. (2 marks)

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Priya's employer increases her commission rate to 4.2% but keeps her base wage the same. If Priya makes $22 000 in sales, calculate her new total weekly earnings. (2 marks)

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a. $=\text{B4}+\text{B5}^*\text{B6}/100$

b. $$22\, 500$

c. $$1574$

Show Worked Solution

a. $\text{Total weekly earnings} = \text{Base wage}+\text{Total sales} \times \text{Commission rate}$

$\therefore\ \text{Formula:}\ =\text{B4}+\text{B5}^*\text{B6}/100$

b. $\text{Total weekly earnings} = \text{Base wage}+\text{Total sales} \times \text{Commission rate}$

$\text{Let the Total sales}=S$

$1437.50$	$=650+S\times \dfrac{3.5}{100}$
$787.50$	$=S\times \dfrac{3.5}{100}$
$S$	$=\dfrac{787.50\times 100}{3.5}=$22\,500$

$\text{The amount of Priya’s total sales was \$22 500.}$

c. $\text{Base wage}=650$

$\text{Total sales}=$22\,000$

$\text{New commission rate}=4.2\%$

$\text{Total weekly earnings}$	$=650+22\,000\times \dfrac{4.2}{100}$
	$=650+924=$1574$

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.

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

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

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a. $34.68 \ \text{cm}^2$

b. $582.64 \ \text{cm}^2$

Show Worked Solution

a. $\text{Solution 1}$

$A$	$=\dfrac{5.1}{2}(6.0+3.8) + \dfrac{5.1}{2}(3.8+0) $
	$=34.68\ \text{cm}^2$

$\text{Solution 2}$

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

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

b. $\text{Toy has 5 sides.}$

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

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

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

$\therefore \ \text{Area of curved surface}$

$=1300-(408+240+69.36)=582.64 \ \text{cm}^2$

♦ Mean mark (b) 48%.

Financial Maths, STD2 EQ-Bank 31

Chen earns an annual salary of $72 800. He is entitled to four weeks annual leave with 17.5% leave loading. A spreadsheet is used to calculate his total holiday pay.

Total holiday pay = 4 × weekly wage + 4 × weekly wage × 17.5%

A spreadsheet showing Chen's holiday pay calculation is shown.

What value from the question should be entered in cell B5? (1 mark)

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Write down the formula used in cell B9 to calculate the weekly wage, using appropriate grid references. (2 marks)

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Verify the amount of Chen's total holiday pay using calculations. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---

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a. $4$

b. $=\text{B4}/52$

c. $\text{Total holiday pay}=\text{weekly wage}\times 4 +\ \text{weekly wage}\times 4\ \times 17.5\%$

$\text{Total holiday pay}=1400\times 4+1400\times 4\times\dfrac{17.5}{100}=5600+980=$6580$

Show Worked Solution

a. $\text{Chen gets 4 weeks leave }\rightarrow\ 4$

b. $\text{Weekly wage }=\dfrac{\text{Annual salary}}{52}$

$\text{Formula: }=\text{B4}/52$

c. $\text{Total holiday pay}=\text{weekly wage}\times 4 +\ \text{weekly wage}\times 4\ \times 17.5\%$

$\text{Total holiday pay}=1400\times 4+1400\times 4\times\dfrac{17.5}{100}=5600+980=$6580$

Financial Maths, STD2 EQ-Bank 9 MC

A company calculates holiday pay for employees entitled to four weeks annual leave with 17.5% leave loading on four weeks pay.

A spreadsheet is used to calculate the total holiday pay.

Which formula has been used in cell B9?

$=\text{B4}^*\text{B5}+\text{B4}^*\text{B5}^*\text{B6}$
$=\text{B4}^*\text{B5}+\text{B4}^*\text{B5}^*\text{B6}/100$
$=\text{B4}^*\text{B5}^*\text{B6}/100$
$=\text{B4}+\text{B5}+\text{B6}$

Show Answers Only

$B$

Show Worked Solution

$\text{Total holiday pay}=\text{weekly wage}\times 4 +\text{weekly wage}\times 4\ \times 17.5\%$

$\text{Formula: }\text{weekly wage}\times 4 +\text{weekly wage}\times 4\ \times \dfrac{17.5}{100}$

$\text{Using cell references: }=\text{B4}^*\text{B5}+\text{B4}^*\text{B5}^*\text{B6}/100$

$\text{Check: }1450\times 4+1450\times 4\times\dfrac{17.5}{100}=5800+1015=6815$

$\Rightarrow B$

Algebra, STD2 EQ-Bank 29

The formula below is used to estimate the number of hours you must wait before your blood alcohol content (BAC) will return to zero after consuming alcohol.

$\text{Number of hours}\ =\dfrac{\text{BAC}}{0.015}$

The spreadsheet below has been created by Ben so his 21st birthday attendees can monitor their alcohol consumption if they intend to drive, given their BAC reading.

By using appropriate grid references, write down a formula that could appear in cell B5. (2 marks)

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Ben's friend Ryan's reading reflects that he will have to wait 11 hours for his BAC to return to zero. Using the formula, calculate the value Ryan entered into cell B3. (2 marks)

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

Show Answers Only

a. $=\text{B3}/0.015$

b. $0.165$

Show Worked Solution

a. $=\text{B3}/0.015$

b.	$11$	$=\dfrac{\text{BAC}}{0.015}$
	$\text{BAC}$	$=11\times 0.015=0.165$

$\text{Ryan entered 0.165 into cell B3.}$

Algebra, STD2 EQ-Bank 28

Fried's formula for determining the medicine dosage for children aged 1 - 2 years is:

$\text{Dosage}=\dfrac{\text{Age of infant (months)}\ \times \ \text{adult dose}}{150}$

The spreadsheet below is used as a calculator for determining an infant's medicine dosage according to Fried's formula.

Amber, a 12 month old child, is being discharged from hospital with two medications. Medicine A has an adult dosage of 325 milligrams and she is to take 26 milligrams. She must also take 9.6 milligrams of Medicine B but the equivalent adult dosage has been left off the spreadsheet.

By using appropriate grid references, write down a formula that could appear in cell B10. (2 marks)

--- 1 WORK AREA LINES (style=lined) ---
Calculate the equivalent adult dosage for Medicine B (cell B7) using the information in the spreadsheet. (2 marks)

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

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a. $=\text{B5}^*\text{B6}/150$

b. $120 \ \text{milligrams}$

Show Worked Solution

a. $=\text{B5}^*\text{B6}/150$

b. $\text{Let} \ A= \text{Adult dose}$

$\text{Using given formula:}$

$9.6$	$=\dfrac{12 \times A}{150}$
$A$	$=\dfrac{9.6 \times 150}{12}=120 \ \text{milligrams}$

Algebra, STD2 EQ-Bank 12 MC

Young's formula for determining the medicine dosage for children aged 1 - 12 years is:

$\text{Dosage} = \dfrac{\text{Age of child (years)}\ \times\ \text{Adult dose}}{\text{Age of child (years) } +\ 12}$

The spreadsheet below is used as a calculator for determining an child's medicine dosage according to Young's formula.

Young's formula for calculating an 8 year old child's dosage has been used in cell B9. Using appropriate cell references, the correct formula to input into cell B9 is:

$=\text{B5}^*\text{B6}/\text{B5}+12$
$=(\text{B5}^*\text{B6})/\text{B5}+12$
$=\text{B5}^*\text{B6}/(\text{B6}+12)$
$=(\text{B5}^*\text{B6})/(\text{B5}+12)$

Show Answers Only

$D$

Show Worked Solution

$=(\text{B5}^*\text{B6})/(\text{B5}+12)$

$\Rightarrow D$

Algebra, STD2 EQ-Bank 26

Fried's formula for determining the medicine dosage for children aged 1 - 2 years is:

$\text{Dosage}=\dfrac{\text{Age of infant (months)}\ \times \ \text{adult dose}}{150}$

The spreadsheet below is used as a calculator for determining an infant's medicine dosage according to Fried's formula.

By using appropriate grid references, write down a formula that could appear in cell B9. (2 marks)

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Another infant requiring the same medicine has been recommended a dosage of 2 millilitres. What is the age of the infant? (2 marks)

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a. $=\text{B5}^*\text{B6}/150$

b. $15 \ \text{months}$

Show Worked Solution

a. $=\text{B5}^*\text{B6}/150$

b. $\text{Let} \ n= \text{age of infant}$

$\text{Using given formula:}$

$2$	$=\dfrac{n \times 20}{150}$
$n$	$=\dfrac{2 \times 150}{20}=15 \ \text{months}$

Financial Maths, STD2 EQ-Bank 32

Nigel's weekly wages are calculated using the partially completed spreadsheet below.

Calculate the total wages Nigel earned on Wednesday. (2 marks)

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Determine the time Nigel finished work on Saturday, given he earned $196.35 on the day. (2 marks)

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

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a. $\text{Total wages}=119.00+35.70=\$ 154.70$

b. $14:30$

Show Worked Solution

a. $\text{Wednesday wages:}$

$\text{Regular hours:} \ 5 \times 23.80=\$ 119.00$

$\text{Time-and-a-half:} \ 1 \times 1.5 \times 23.80=\$35.70$

$\text{Total wages}=119.00+35.70=\$ 154.70$

b. $\text{Let \(h=$ total hours worked:}\)

$\text{Since Saturday wages are time-and-a-half rate:}$

$h \times 1.5 \times 23.80$	$=196.35$
$h$	$=\dfrac{196.35}{1.5 \times 23.80}=5.5\ \text{hours}$

$\text{Nigel’s shift started at 09:00 and lasted 5.5 hours.}$

$\therefore\ \text{Nigel finished work at 14:30.}$

Financial Maths, STD2 EQ-Bank 33

Trust Us Realty has three salespeople, Ralph, Ritchie, and Fonzi.

Their June monthly wages include a base wage and commission earned, which is modelled in the spreadsheet below.

Write down the formula that was used in cell C9, using appropriate grid references. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Calculate Fonzi's total pay for the month of June. (2 marks)

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Ralph's total pay for June is $5850. Determine Ralph's total sales for the month. (2 marks)

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

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a. $=0.01^* \text{B9}$

b. $\$ 20\,200$

c. $\$ 385\,000$

Show Worked Solution

a. $=0.01^* \text{B9}$

b. $\text{Fonzi’s Commission:}$

$\text{Sales}\ \$0-\$500\,000=0.01 \times 500\,000=\$ 5000$

$\text{Sales over} \ \$500\, 000=(2\,150\,000-500\,000) \times 0.008=\$13\,200$

$\text{Total June wages}=5000+13\,200+2000=\$ 20\,200$

c. $\text{Ralph’s sales commission}\ =5850-2000=\$3850$

$\text {Since Ralph earned} \ \$3850 \ \text{in commission:}$

$\text{Sales} \times 0.01$	$=3850$
$\text{Sales}$	$=\dfrac{3850}{0.01}=\$ 385\,000$

Algebra, STD2 EQ-Bank 31

Clark's formula for determining the medicine dosage for children is:

$\text{Dosage}=\dfrac{\text{weight in kilograms}\ \times \ \text{adult dosage}}{70}$

The spreadsheet below is used as a calculator for determining a child's medicine dosage according to Clark's formula.

By using appropriate grid references, write down a formula that could appear in cell E5. (2 marks)

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Another child requiring the same medicine has been recommended a dosage of 62.5 milligrams. How much does the child weigh? (2 marks)

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a. $=\text{B6}^*\text{B5}/70$

b. $17.5 \ \text{kilograms}$

Show Worked Solution

a. $=\text{B6}^*\text{B5}/70$

b. $\text{Let} \ w= \text{weight of the child.}$

$\text{Using given formula:}$

$62.5$	$=\dfrac{w \times 250}{70}$
$w$	$=\dfrac{62.5 \times 70}{250}=17.5 \ \text{kilograms}$

Algebra, STD2 EQ-Bank 23

Sharon drinks three glasses of chardonnay over a 180-minute period, each glass containing 1.6 standard drinks.

Sharon weighs 78 kilograms, and her blood alcohol content (BAC) at the end of this period can be calculated using the following formula:

$\text{BAC}_{\text {female }}=\dfrac{10 N-7.5 H }{5.5 M}$

where $N$	= number of standard drinks consumed
$H$	= the number of hours drinking
$M$	= the person's mass in kilograms

The spreadsheet below can be used to calculate Sharon's $\text{BAC}$.

What value should be input into cell B5. (1 mark)

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Write down the formula that has been used in cell E4, using appropriate grid references. (2 marks)

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a. $\text{Cell B5 value}=3$

b. $=\left(10^* \text{B4}-7.5^* \text{B5}\right) /(5.5^* \text{B6})$

Show Worked Solution

a. $\text{180 minutes}\ =\ \text{3 hours}$

$\therefore \ \text{Cell B5 value}=3$

b. $=\left(10^* \text{B4}-7.5^* \text{B5}\right) /(5.5^* \text{B6})$

HMS, HAG 2025 HSC 32b

To what extent do THREE factors that create health inequities affect ONE population group in Australia? ( 12 marks)

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Population group: The aged (65+ years)

Judgement Statement

Three factors—access to services, socioeconomic disadvantage and social isolation— create significant health inequities for aged Australians, with compounding effects that substantially reduce health outcomes and quality of life.

Access to Services and Transport

Limited access to healthcare services creates major health inequities for older Australians.
Rural and remote aged populations face substantial barriers reaching specialist care, diagnostic services and preventative health programs.
Transport difficulties compound these issues as declining mobility reduces medical appointment attendance.
This results in delayed diagnosis of cardiovascular disease, cancer and dementia related illnesses, increasing mortality rates among the elderly.
Fewer preventative visits mean chronic conditions progress undetected whilst inadequate allied health access reduces rehabilitation opportunities.
Geographic isolation intensifies these barriers, creating cycles of declining health status compared to urban counterparts.

Socioeconomic Factors

Financial constraints significantly affect aged Australians relying solely on age pensions.
Lower incomes restrict access to private healthcare, specialists and medications not fully covered by Medicare.
Gap payments and medication costs force delays in seeking care, resulting in untreated chronic conditions.
This leads to preventable hospitalisations and poorer disease management outcomes.
Socioeconomically disadvantaged older adults experience higher rates of inadequate nutrition and poor housing quality, further compromising health.

Social Isolation

Social isolation profoundly impacts mental and physical health for older adults living alone following bereavement or family relocation.
Reduced social connections correlate strongly with increased depression, anxiety and cognitive decline including dementia progression.
Isolation decreases motivation for self-care and physical activity whilst poor housing modifications increase fall risk and injury rates.
Limited support networks mean delayed help-seeking when unwell, worsening health outcomes and recovery times.

Reaffirmation

These three factors interact substantially to create significant health inequities for aged Australians.
The combined effect exceeds individual factors as limited access, financial barriers and isolation reinforce each other.
Aged populations experiencing multiple disadvantages demonstrate markedly worse health outcomes including higher chronic disease rates and premature mortality.

Show Worked Solution

Population group: The aged (65+ years)

Judgement Statement

Three factors—access to services, socioeconomic disadvantage and social isolation— create significant health inequities for aged Australians, with compounding effects that substantially reduce health outcomes and quality of life.

Access to Services and Transport

Limited access to healthcare services creates major health inequities for older Australians.
Rural and remote aged populations face substantial barriers reaching specialist care, diagnostic services and preventative health programs.
Transport difficulties compound these issues as declining mobility reduces medical appointment attendance.
This results in delayed diagnosis of cardiovascular disease, cancer and dementia related illnesses, increasing mortality rates among the elderly.
Fewer preventative visits mean chronic conditions progress undetected whilst inadequate allied health access reduces rehabilitation opportunities.
Geographic isolation intensifies these barriers, creating cycles of declining health status compared to urban counterparts.

Socioeconomic Factors

Financial constraints significantly affect aged Australians relying solely on age pensions.
Lower incomes restrict access to private healthcare, specialists and medications not fully covered by Medicare.
Gap payments and medication costs force delays in seeking care, resulting in untreated chronic conditions.
This leads to preventable hospitalisations and poorer disease management outcomes.
Socioeconomically disadvantaged older adults experience higher rates of inadequate nutrition and poor housing quality, further compromising health.

Social Isolation

Social isolation profoundly impacts mental and physical health for older adults living alone following bereavement or family relocation.
Reduced social connections correlate strongly with increased depression, anxiety and cognitive decline including dementia progression.
Isolation decreases motivation for self-care and physical activity whilst poor housing modifications increase fall risk and injury rates.
Limited support networks mean delayed help-seeking when unwell, worsening health outcomes and recovery times.

Reaffirmation

These three factors interact substantially to create significant health inequities for aged Australians.
The combined effect exceeds individual factors as limited access, financial barriers and isolation reinforce each other.
Aged populations experiencing multiple disadvantages demonstrate markedly worse health outcomes including higher chronic disease rates and premature mortality.

♦♦ Mean mark 45%.

HMS, TIP 2025 HSC 31b

Justify THREE elements a coach needs to consider when designing a training session for a sport. ( 12 marks)

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Sport: Netball

Position Statement:

Warm-up, skill instruction and conditioning are essential elements when designing effective training sessions.
These elements prepare athletes physically, develop technical competence and build sport-specific fitness for optimal performance.

Warm-Up Preparation:

Structured warm-ups prepare athletes physiologically and psychologically for training. Research confirms gradual cardiovascular activation and dynamic stretching increase muscle temperature whilst reducing injury risk.
For example, netball players performing 10 minutes jogging at 70% maximum heart rate followed by side-stepping and pivoting experience improved muscle elasticity. This justifies warm-up inclusion because prepared muscles respond effectively to training whilst preventing strains.
Athletes completing structured warm-ups demonstrate noticeably better performance quality during skill work, proving warm-up effectiveness in optimising session outcomes.

Skill Instruction and Practice:

Deliberate skill instruction develops technical proficiency essential for competitive success. Structured practice with immediate feedback accelerates skill acquisition compared to unguided repetition.
Netball coaches implementing progression models—basic chest pass drills advancing to contested passing under defensive pressure—enable athletes to master fundamentals whilst introducing pressure gradually. This justifies systematic instruction because controlled complexity builds confidence and competence simultaneously.
Athletes receiving structured instruction show significantly faster technique improvement than those practising without guidance, demonstrating critical importance for training efficiency.

Conditioning for Sport-Specific Fitness:

Sport-specific conditioning builds physiological capacities for competition demands. Targeted fitness development enhances performance whilst reducing fatigue-related errors.
Combining agility ladder drills with defensive shadowing in netball develops quick directional changes and sustained movement capacity players need. This justifies conditioning because it transfers directly to game performance.
Athletes following sport-specific conditioning reportedly maintain higher intensity for longer during competition, supporting conditioning as essential for advantage.

Show Worked Solution

Sport: Netball

Position Statement:

Warm-up, skill instruction and conditioning are essential elements when designing effective training sessions.
These elements prepare athletes physically, develop technical competence and build sport-specific fitness for optimal performance.

Warm-Up Preparation:

Structured warm-ups prepare athletes physiologically and psychologically for training. Research confirms gradual cardiovascular activation and dynamic stretching increase muscle temperature whilst reducing injury risk.
For example, netball players performing 10 minutes jogging at 70% maximum heart rate followed by side-stepping and pivoting experience improved muscle elasticity. This justifies warm-up inclusion because prepared muscles respond effectively to training whilst preventing strains.
Athletes completing structured warm-ups demonstrate noticeably better performance quality during skill work, proving warm-up effectiveness in optimising session outcomes.

Skill Instruction and Practice:

Deliberate skill instruction develops technical proficiency essential for competitive success. Structured practice with immediate feedback accelerates skill acquisition compared to unguided repetition.
Netball coaches implementing progression models—basic chest pass drills advancing to contested passing under defensive pressure—enable athletes to master fundamentals whilst introducing pressure gradually. This justifies systematic instruction because controlled complexity builds confidence and competence simultaneously.
Athletes receiving structured instruction show significantly faster technique improvement than those practising without guidance, demonstrating critical importance for training efficiency.

Conditioning for Sport-Specific Fitness:

Sport-specific conditioning builds physiological capacities for competition demands. Targeted fitness development enhances performance whilst reducing fatigue-related errors.
Combining agility ladder drills with defensive shadowing in netball develops quick directional changes and sustained movement capacity players need. This justifies conditioning because it transfers directly to game performance.
Athletes following sport-specific conditioning reportedly maintain higher intensity for longer during competition, supporting conditioning as essential for advantage.

♦♦ Mean mark 37%.

HMS, TIP 2025 HSC 30b

Justify the use of heat and cold and progressive mobilisation as rehabilitation procedures for a shoulder dislocation. ( 12 marks)

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Position Statement:

Heat and cold therapy combined with progressive mobilisation provides optimal rehabilitation for shoulder dislocations.
These procedures effectively manage inflammation whilst restoring range of motion and facilitating safe return to function.

Cold Therapy in Acute Phase:

Immediate cold application reduces inflammatory response and controls swelling after dislocation. Evidence confirms cold constricts blood vessels, limiting fluid accumulation around damaged tissue.
Research shows applying ice for 15-20 minutes every two hours during first 48-72 hours significantly decreases pain and tissue damage. This demonstrates cold therapy’s effectiveness in protecting injured structures during the acute inflammatory phase.
Cold application enables earlier mobilisation by controlling pain and swelling. Athletes experience reduced discomfort, allowing gentle movement exercises to begin sooner whilst preventing excessive inflammation that delays healing. This justifies cold as essential for initial injury management.

Heat Application in Later Stages:

After initial inflammation subsides (typically 72 hours post-injury), heat therapy increases blood flow to promote tissue repair. Studies indicate heat dilates vessels, delivering oxygen and nutrients essential for healing damaged ligaments and capsule tissue.
Heat application before mobilisation exercises improves muscle elasticity and joint flexibility. Evidence shows warming tissues reduces stiffness, allowing greater range of movement during rehabilitation exercises without causing re-injury or excessive discomfort.

Progressive Mobilisation Throughout Recovery:

Progressive mobilisation systematically restores shoulder function through graduated exercises matched to healing stages. This approach begins with passive pendulum movements, advancing to active-assisted exercises, then resistance training as tissue strength develops.
Evidence demonstrates controlled movement prevents joint stiffness, maintains neuromuscular patterns and reduces muscle atrophy that prolonged immobilisation causes.
Athletes following progressive protocols can achieve full range of motion faster than those using rest alone.

Reinforcement:

Some argue rest alone suffices for recovery. However, research consistently demonstrates controlled movement combined with appropriate thermal therapy optimises healing timeframes whilst minimising complications.
This evidence-based protocol remains valid because it addresses both tissue healing requirements and functional restoration needs.
The combined approach facilitates complete recovery whilst significantly reducing re-dislocation risk during return to sport.

Show Worked Solution

Position Statement:

Heat and cold therapy combined with progressive mobilisation provides optimal rehabilitation for shoulder dislocations.
These procedures effectively manage inflammation whilst restoring range of motion and facilitating safe return to function.

Cold Therapy in Acute Phase:

Immediate cold application reduces inflammatory response and controls swelling after dislocation. Evidence confirms cold constricts blood vessels, limiting fluid accumulation around damaged tissue.
Research shows applying ice for 15-20 minutes every two hours during first 48-72 hours significantly decreases pain and tissue damage. This demonstrates cold therapy’s effectiveness in protecting injured structures during the acute inflammatory phase.
Cold application enables earlier mobilisation by controlling pain and swelling. Athletes experience reduced discomfort, allowing gentle movement exercises to begin sooner whilst preventing excessive inflammation that delays healing. This justifies cold as essential for initial injury management.

Heat Application in Later Stages:

After initial inflammation subsides (typically 72 hours post-injury), heat therapy increases blood flow to promote tissue repair. Studies indicate heat dilates vessels, delivering oxygen and nutrients essential for healing damaged ligaments and capsule tissue.
Heat application before mobilisation exercises improves muscle elasticity and joint flexibility. Evidence shows warming tissues reduces stiffness, allowing greater range of movement during rehabilitation exercises without causing re-injury or excessive discomfort.

Progressive Mobilisation Throughout Recovery:

Progressive mobilisation systematically restores shoulder function through graduated exercises matched to healing stages. This approach begins with passive pendulum movements, advancing to active-assisted exercises, then resistance training as tissue strength develops.
Evidence demonstrates controlled movement prevents joint stiffness, maintains neuromuscular patterns and reduces muscle atrophy that prolonged immobilisation causes.
Athletes following progressive protocols can achieve full range of motion faster than those using rest alone.

Reinforcement:

Some argue rest alone suffices for recovery. However, research consistently demonstrates controlled movement combined with appropriate thermal therapy optimises healing timeframes whilst minimising complications.
This evidence-based protocol remains valid because it addresses both tissue healing requirements and functional restoration needs.
The combined approach facilitates complete recovery whilst significantly reducing re-dislocation risk during return to sport.

♦♦ Mean mark 52%.

HMS, HIC 2025 HSC 28aii

Explain how becoming involved in community service can assist young people in attaining better health. ( 5 marks)

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Community service creates opportunities for social connection with like-minded peers. This occurs when young people work alongside others towards shared goals in volunteer settings.
For example, a young person volunteering at environmental clean-up events develops teamwork and communication abilities. As a result, their social health improves through new friendships and increased confidence in group interactions.
Volunteering enhances mental health by providing sense of purpose and direction. This happens because contributing meaningfully to community needs generates feelings of value and belonging.
For instance, a young person supporting elderly residents through hospital visits experiences improved emotional wellbeing. The reason for this is that helping others creates perspective on personal challenges whilst building resilience through meaningful relationships.
Community service leads to increased physical activity in many volunteer roles. Consequently, young people gain health benefits from active engagement rather than sedentary screen time.

Show Worked Solution

Community service creates opportunities for social connection with like-minded peers. This occurs when young people work alongside others towards shared goals in volunteer settings.
For example, a young person volunteering at environmental clean-up events develops teamwork and communication abilities. As a result, their social health improves through new friendships and increased confidence in group interactions.
Volunteering enhances mental health by providing sense of purpose and direction. This happens because contributing meaningfully to community needs generates feelings of value and belonging.
For instance, a young person supporting elderly residents through hospital visits experiences improved emotional wellbeing. The reason for this is that helping others creates perspective on personal challenges whilst building resilience through meaningful relationships.
Community service leads to increased physical activity in many volunteer roles. Consequently, young people gain health benefits from active engagement rather than sedentary screen time.

♦♦ Mean mark 51%.

HMS, HIC 2025 HSC 27

Explain the responsibilities of individuals, communities and governments in creating supportive environments to promote health. Support your answer with examples. ( 8 marks)

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Individual Responsibilities:

Individuals must adopt health-promoting behaviours that create safe environments for themselves and others. This occurs when people apply health literacy to make informed decisions.
For example, a parent choosing active transport to school reduces vehicle emissions and models healthy physical activity patterns. This leads to improved air quality and encourages children to adopt active lifestyles.

Community Responsibilities:

Communities must advocate for and support their members through accessible programs and resources. This enables individuals to access culturally appropriate health services.
For instance, neighbourhood walking groups organised by local councils provide social connection and physical activity opportunities. Consequently, participants experience improved mental and physical health through regular engagement and peer support networks.

Government Responsibilities:

Governments must develop and enforce policies that facilitate health-promoting environments. This works through legislation creating safe public spaces and restricting harmful exposures.
For example, mandatory bicycle helmet laws and dedicated cycling infrastructure protect cyclists from injury whilst encouraging active transport adoption. As a result, communities experience reduced traffic congestion, improved air quality and increased population physical activity levels.

Collective Impact:

These responsibilities work together to create comprehensive supportive environments. The significance is that sustained health improvements require coordinated action across all three levels rather than isolated individual efforts.

Show Worked Solution

Individual Responsibilities:

Individuals must adopt health-promoting behaviours that create safe environments for themselves and others. This occurs when people apply health literacy to make informed decisions.
For example, a parent choosing active transport to school reduces vehicle emissions and models healthy physical activity patterns. This leads to improved air quality and encourages children to adopt active lifestyles.

Community Responsibilities:

Communities must advocate for and support their members through accessible programs and resources. This enables individuals to access culturally appropriate health services.
For instance, neighbourhood walking groups organised by local councils provide social connection and physical activity opportunities. Consequently, participants experience improved mental and physical health through regular engagement and peer support networks.

Government Responsibilities:

Governments must develop and enforce policies that facilitate health-promoting environments. This works through legislation creating safe public spaces and restricting harmful exposures.
For example, mandatory bicycle helmet laws and dedicated cycling infrastructure protect cyclists from injury whilst encouraging active transport adoption. As a result, communities experience reduced traffic congestion, improved air quality and increased population physical activity levels.

Collective Impact:

These responsibilities work together to create comprehensive supportive environments. The significance is that sustained health improvements require coordinated action across all three levels rather than isolated individual efforts.

HMS, TIP 2025 HSC 26

Analyse the relationship between training thresholds and TWO physiological adaptations. In your answer, provide examples of both aerobic and resistance training. (8 marks)

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Overview Statement:

Training thresholds represent critical intensity levels that trigger specific physiological adaptations.
Understanding how aerobic and resistance thresholds connect to metabolic and muscular changes can open up pathways to enhanced athletic performance.

Aerobic Threshold and Cardiovascular Adaptations:

The aerobic training threshold occurs at approximately 70% of maximum heart rate. Training at this intensity influences fuel utilisation and cardiovascular function.
This causes the body to shift from primarily using fat to using carbohydrates for energy. For example, a marathon runner training at 70% max heart rate stimulates this metabolic adaptation.
The threshold also triggers increased stroke volume through enhanced left ventricle filling capacity. This relationship results in improved cardiac output and oxygen delivery to working muscles.
Consequently, athletes sustain effort over extended periods with greater efficiency. This shows that aerobic threshold training enables both metabolic and cardiovascular improvements for endurance performance.

Resistance Threshold and Muscular Adaptations:

Resistance training thresholds involve working at 70-85% of one-rep maximum with 6-12 repetitions. This intensity creates sufficient mechanical stress to stimulate muscle hypertrophy.
For instance, a weightlifter performing squats at 80% of 1RM bmicroscopic muscle fibre damage. This initiates increased protein synthesis and muscle repair processes.
The threshold works through progressive overload that leads to enlarged muscle fibres with increased actin and myosin filaments. As a result, greater force production capacity develops.
The significance is that athletes gain strength and power output essential for explosive movements.

Implications and Synthesis:

These thresholds work together as intensity markers that determine adaptation type. Aerobic thresholds influence metabolic and cardiovascular systems whilst resistance thresholds affect muscular structure.
Therefore, coaches must apply appropriate threshold intensities to achieve specific performance goals. This reveals that training success depends on understanding the precise relationship between intensity levels and resulting physiological changes.

Show Worked Solution

Overview Statement:

Training thresholds represent critical intensity levels that trigger specific physiological adaptations.
Understanding how aerobic and resistance thresholds connect to metabolic and muscular changes can open up pathways to enhanced athletic performance.

Aerobic Threshold and Cardiovascular Adaptations:

The aerobic training threshold occurs at approximately 70% of maximum heart rate. Training at this intensity influences fuel utilisation and cardiovascular function.
This causes the body to shift from primarily using fat to using carbohydrates for energy. For example, a marathon runner training at 70% max heart rate stimulates this metabolic adaptation.
The threshold also triggers increased stroke volume through enhanced left ventricle filling capacity. This relationship results in improved cardiac output and oxygen delivery to working muscles.
Consequently, athletes sustain effort over extended periods with greater efficiency. This shows that aerobic threshold training enables both metabolic and cardiovascular improvements for endurance performance.

Resistance Threshold and Muscular Adaptations:

Resistance training thresholds involve working at 70-85% of one-rep maximum with 6-12 repetitions. This intensity creates sufficient mechanical stress to stimulate muscle hypertrophy.
For instance, a weightlifter performing squats at 80% of 1RM bmicroscopic muscle fibre damage. This initiates increased protein synthesis and muscle repair processes.
The threshold works through progressive overload that leads to enlarged muscle fibres with increased actin and myosin filaments. As a result, greater force production capacity develops.
The significance is that athletes gain strength and power output essential for explosive movements.

Implications and Synthesis:

These thresholds work together as intensity markers that determine adaptation type. Aerobic thresholds influence metabolic and cardiovascular systems whilst resistance thresholds affect muscular structure.
Therefore, coaches must apply appropriate threshold intensities to achieve specific performance goals. This reveals that training success depends on understanding the precise relationship between intensity levels and resulting physiological changes.

♦♦ Mean mark 36%.

Probability, 2ADV EQ-Bank 24

In Year 11 there are 80 students. The students may choose to study Spanish (S), Japanese (J) and Mandarin (M).

The Venn diagram shows their choices.

Two of the students are selected at random.

What is the probability that both students study only Spanish? (2 marks)

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What is the probability that at least one of the students studies two languages. (2 marks)

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a. $\dfrac{93}{632}$

b. $\dfrac{81}{158} $

Show Worked Solution

a. $\text{Students only studying Spanish = 31}$

$P(\text{both study only Spanish}\ =\dfrac{31}{80} \times \dfrac{30}{79} = \dfrac{93}{632}$

b. $\text{1st student chosen:}$

$P(2L) =\dfrac{6+4+14}{80} = \dfrac{24}{80}\ \ \Rightarrow\ \ P(\overline{2L})=\dfrac{56}{80} $

$\text{2nd student chosen:}$

$P(\overline{2L})=\dfrac{55}{79} $

$P(\text{at least one studies two languages})$

$= 1- P(\text{both don’t study two languages)}$

$=1-\dfrac{56}{80} \times \dfrac{55}{79} $

$=\dfrac{81}{158} $

Functions, 2ADV EQ-Bank 28

The cost of hiring an open space for a music festival is $120 000. The cost will be shared equally by the people attending the festival, so that `C` (in dollars) is the cost per person when `n` people attend the festival.

Complete the table below and draw the graph showing the relationship between `n` and `C`. (2 marks)
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & & & & 60 & 48\ & 40 \ \\
\hline
\end{array}
What equation represents the relationship between `n` and `C`? (1 mark)

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Give ONE limitation of this equation in relation to this context. (1 mark)

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\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}

b. `C = (120\ 000)/n`

c. `text(Limitations can include:)`

`•\ n\ text(must be a whole number)`

`•\ C > 0`

Show Worked Solution

b. `C = (120\ 000)/n`

c. `text(Limitations can include:)`

`•\ n\ text(must be a whole number)`

`•\ C > 0`

HMS, TIP 2025 HSC 18 MC

An elite athlete is training to enhance their muscular strength.

Which of the following approaches best demonstrates progressive overload for increased strength? (RM = repetition maximum)

Adding exercises while performing four sets of eight repetitions at 60% of one RM
Alternating between 80% and 90% of one RM weekly for three sets of 12 repetitions
Starting with weights at 50% of one RM and gradually increasing the total repetitions each week repetitions each week
Adjusting the resistance from 80% to 90% of one RM, performing three to five sets of four to six repetitions

Show Answers Only

$D$

Show Worked Solution

D is correct: Progressive resistance increase (80% to 90% 1RM) with low reps (4-6) targets absolute strength development

Other Options:

A is incorrect: 60% 1RM with 8 reps develops muscular endurance, not maximal strength; adding exercises doesn’t increase load
B is incorrect: Alternating loads weekly lacks progressive overload; 12 reps targets endurance rather than strength
C is incorrect: Starting at 50% 1RM and increasing repetitions develops endurance, not strength; requires higher loads

♦♦ Mean mark 43%.

HMS, TIP 2025 HSC 14 MC

Which row in the table describes both a valid and reliable test for measuring the speed of an athlete?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\textbf{}\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex} \textbf{A.}\ \rule[-1ex]{0pt}{0pt}&\\
\textbf{}\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex} \textbf{B.} \ \rule[-1ex]{0pt}{0pt}&\\
\textbf{} \ \rule[-1ex]{0pt}{0pt}&\\
\rule{0pt}{2.5ex} \textbf{C.} \ \rule[-1ex]{0pt}{0pt}& \\
\textbf{} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex} \rule{0pt}{2.5ex} \textbf{D.} \ \rule[-1ex]{0pt}{0pt}& \\
\textbf{}\ \rule[-1ex]{0pt}{0pt}& \\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex} \ \ \ \ \ Athlete \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \ Test \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Result \\
\hline
\rule{0pt}{2.5ex}\text{100 m sprinter} & \text{Reaction time} & \text{There are changes in the athlete’s sprint} \\
\text{} &\text{to starter } &\text{times.} \\
\hline
\rule{0pt}{2.5ex}\text{100 m sprinter} &\text{Reaction time} & \text{Results are consistent across multiple} \\
\text{} & \text{to starter} & \text{training sessions.} \\
\hline
\rule{0pt}{2.5ex} \text{Midfielder in } & \text{40 m sprint } & \text{There are changes in the athlete’s sprint} \\
\text{soccer } &\text{choice} & \text{times.} \\
\hline
\rule{0pt}{2.5ex}\text{Midfielder in } & \text{40 m sprint } & \text{Results are consistent across multiple} \\
\text{soccer } &\text{choice} & \text{training sessions. } \\
\hline
\end{array}
\end{align*}

Show Answers Only

$D$

Show Worked Solution

D is correct: 40m sprint test is valid (measures speed); consistent results demonstrate reliability

Other Options:

A is incorrect: Reaction time measures response speed, not running speed (lacks validity)
B is incorrect: Reaction time test is not valid for measuring running speed
C is incorrect: Changes in sprint times indicate inconsistent results (lacks reliability)

♦♦ Mean mark 50%.

HMS, BM 2025 HSC 13 MC

Which of the following best demonstrates how the characteristics of the learner can influence their progression through to the associative stage of skill acquisition?

A swimmer relies on their heredity traits and confidence to improve their freestyle technique.
A gymnast performs a routine, relying on additional practice and feedback from their coach.
A basketball player learns the technique of shooting by relying on demonstrations but gives up easily.
A tennis player is struggling to return serves, due to limited confidence and inconsistent attention.

Show Answers Only

$A$

Show Worked Solution

A is correct: Heredity traits and confidence are learner characteristics enabling progression from cognitive to associative stage

Other Options:

B is incorrect: Practice and feedback are teaching methods, not learner characteristics influencing progression
C is incorrect: Giving up easily indicates failure to progress beyond cognitive stage, not advancement
D is incorrect: Struggling with limited confidence shows barriers preventing progression, not successful advancement

♦♦♦ Mean mark 36%.

HMS, HAG 2025 HSC 12 MC

Which of the following lists only non-institutional health facilities or services?

Dentists, nursing homes and public hospitals
Dentists, general practitioners and pharmaceutical services
General practitioners, physiotherapists and public hospitals
Nursing homes, pharmaceutical services and physiotherapists

Show Answers Only

$B$

Show Worked Solution

B is correct: Dentists, GPs and pharmacies are all non-institutional services operating outside residential/hospital facilities

Other Options:

A is incorrect: Nursing homes and public hospitals are institutional facilities requiring overnight/residential care
C is incorrect: Public hospitals are institutional facilities providing inpatient care and accommodation
D is incorrect: Nursing homes are institutional facilities providing residential aged care with overnight stays

♦♦ Mean mark 45%.

HMS, HAG 2025 HSC 9 MC

Which of the following is a circulatory disease which causes the blood vessels to narrow, resulting in blockages that reduce the delivery of oxygen to the limbs, kidneys and stomach?

Angina
Coronary heart disease
Cerebrovascular disease
Peripheral vascular disease

Show Answers Only

$D$

Show Worked Solution

D is correct: Peripheral vascular disease affects blood vessels supplying limbs, kidneys and stomach with narrowed arteries

Other Options:

A is incorrect: Angina is chest pain symptom from reduced heart blood flow, not systemic vessel narrowing
B is incorrect: Coronary heart disease affects heart arteries specifically, not peripheral vessels to limbs/organs
C is incorrect: Cerebrovascular disease affects brain blood vessels, not limbs, kidneys or stomach circulation

♦♦ Mean mark 40%.

HMS, BM 2025 HSC 6 MC

A golfer is practising hitting the ball.

Which of the following best describes the nature of the skill?

Fine, discrete and self-paced
Gross, discrete and self-paced
Fine, serial and externally paced
Gross, serial and externally paced

Show Answers Only

$B$

Show Worked Solution

B is correct: A golf swing uses large muscle groups, has clear beginning/end, controlled by performer’s timing.

Other Options:

A is incorrect: Fine motor skills involve small muscles; golf requires large muscle coordination.
C is incorrect: Serial skills link multiple actions; golf swing is single distinct movement.
D is incorrect: Externally paced means environment controls timing; golfer controls swing initiation.

♦♦ Mean mark 44%.

Functions, 2ADV EQ-Bank 28

Given $p$ and $q$ are rational numbers, and $p, q \neq 0$, show

$px^2-(p+q) x+q=0$

has rational roots. (3 marks)

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$\text{Proof (See Worked Solution)}$

Show Worked Solution

$\Delta$	$=b^2-4 a c$
	$=[-(p+q)]^2-4 \times p \times q$
	$=p^2+2 p q+q^2-4 p q$
	$=p^2-2 p q+q^2$
	$=(p-q)^2$

$\text{Roots of equation using quadratic formula:}$

$x$	$=\dfrac{(p+q) \pm \sqrt{(p-q)^2}}{2 p}$
	$=\dfrac{p+q+(p-q)}{2 p} \ \ \text{or} \ \ \dfrac{p+q-(p-q)}{2 p}$
	$=1 \ \ \text{or} \ \ \dfrac{q}{p}$.

$\text{Since \(p, q$ are rational, all roots are rational.}\)

Financial Maths, STD2 EQ-Bank 29

A used car is for sale at $19 500. Priya purchases it using a finance package with a 15% deposit and weekly repayments of $143.27 for 3 years.

What is the interest Priya will pay? (3 marks)

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$$5775.12$

Show Worked Solution

$\text{Deposit}=\dfrac{15}{100}\times 19\,500=$2925$

$\text{Weeks in 3 years}=3\times 52=156$

$\text{Total repayments}=156\times 143.27=$22\,350.12$

$\text{Total cost}=2925+22\,350.12=$25\,275.12$

$\therefore\ \text{Interest paid}=25\,275.12-19\,500=$5775.12$

Financial Maths, STD2 EQ-Bank 30

A smart TV is for sale at $2850. Liam purchases it using a finance package with a 20% deposit and monthly repayments of $87.63 for 3 years.

What is the interest Liam will pay? (3 marks)

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$$874.68$

Show Worked Solution

$\text{Deposit}=\dfrac{20}{100}\times 2850=$570$

$\text{Months in 3 years}=3\times 12=36$

$\text{Total repayments}=36\times 87.63=$3154.68$

$\text{Total cost}=570+3154.68=$3724.68$

$\therefore\ \text{Interest paid}=3724.68-2850=$874.68$

\(\operatorname{Pr}(X>k)\)	\(=\displaystyle\int_k^{\frac{4}{3}} \frac{3}{8}(4-3 x) d x\)
	\(=\displaystyle-\frac{3}{8} \int_k^{\frac{4}{3}}(3 x-4) d x\)
	\(=-\dfrac{3}{8} \cdot \dfrac{1}{2} \cdot \dfrac{1}{3}\left[(3 x-4)^2\right]_k^{\frac{4}{3}}\)
	\(=-\dfrac{1}{16}\left[(4-4)^2-(3 k-4)^2\right]\)
	\(=\dfrac{(3 k-4)^2}{16}\)

\(\dfrac{(3 k-4)^2}{16}\)	\(=\dfrac{9}{16}\)
\((3 k-4)^2\)	\(=9\)
\(3 k-4\)	\(= \pm 3\)

	\(\displaystyle\int_0^{\frac{4}{3}} h(x)\)	\(=\displaystyle \int_0^{\frac{4}{3}}(m f(x)+n) d x\)
		\(=m \underbrace{\displaystyle \int_0^{\frac{4}{3}} f(x) d x}_{=1\ \ \text{since p.d.f. }}+n \displaystyle\int_0^{\frac{4}{3}} 1\ d x\)
		\(=m+n[x]_0^{\frac{4}{3}}\)
		\(=m+\dfrac{4}{3} n\)

b.	\(2 \cos (2 x)+1\)	\(=0\)
	\(\cos (2 x)\)	\(=-\dfrac{1}{2}\)

\(x^2+2 x-3\)	\(=-3\)
\(x(x+2)\)	\(=0\)
\(x=0 \ \ \text{or} \ -2\)

\(y\)	\(=\left(\dfrac{w+1}{2}-1\right)\left(\dfrac{w+1}{2}-w\right)\)
	\(=\left(\dfrac{w-1}{2}\right)\left(\dfrac{1-w}{2}\right)\)
	\(=-\dfrac{1}{4}(w-1)^2\)

\(\dfrac{w^2}{(x-1)^2}\)	\(=(x-w)^2\)
\((x-w)^2(x-1)^2\)	\(=w^2\)
\([(x-w)(x-1)]^2-w^2\)	\(=0\)

\((x-w)(x-1)-w\)	\(=0\)
\(x^2-x-w x+w-w\)	\(=0\)
\(x^2-(w+1) x\)	\(=0\)
\(x[x-(w+1)]\)	\(=0\)

\((x-w)(x-1)+w\)	\(=0\)
\(x^2-(w+1) x+2 w\)	\(=0\)