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Calculus, MET2 2025 VCAA 1

Let  \(g: R \rightarrow R\)  be defined by  \(g(x)=4 x^3-3 x^4\).

  1. Find the coordinates of both stationary points of \(g\).   (2 marks)

  2. Sketch the graph of \(y=g(x)\) on the axes below, labelling the stationary points and axial intercepts with their coordinates.   (2 marks)

   

  1. Complete the following gradient table with appropriate values of \(x\) and \(g^{\prime}(x)\) to show that \(g\) has a stationary point of inflection.   (2 marks)

\begin{array}{|c|c|c|c|}
\hline \rule{0pt}{2.5ex}x  \rule[-1ex]{0pt}{0pt}& \quad \quad \quad \quad& \quad \quad \quad \quad & \quad \quad \quad \quad\\
\hline\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& & & \\
\hline
\end{array}

  1. Find the average value of \(g\) between  \(x=0\)  and  \(x=2\).   (2 marks)
  2. Let \(h\) be the result after applying a sequence of transformations to \(g\), such that \(h\) has a stationary point of inflection at  \((1,0)\) and a local maximum at \((-1,1)\).
  3. Write down a possible sequence of three transformations to map from \(g\) to \(h\).   (3 marks)

  4. Let  \(X \sim \operatorname{Bi}(4, p)\) be a binomial random variable.
  5. Show that  \(\operatorname{Pr}(X \geq 3)=g(p)\) for all \(p \in[0,1]\).   (2 marks)

Show Answers Only

a.    \(\text{SP’s at} \ (0,0) \ \text{and}\ (1,1)\).
 

b.   

c.

\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}&  \quad -1 \quad & \quad \quad 0 \quad \quad & \quad \quad \frac{1}{2} \quad \quad\\
\hline
\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& 24 & 0 & \frac{3}{2} \\
\hline
\end{array}

 
d.    \(I=-\dfrac{8}{5}\)
 

e.    \(\text{Possible sequences include:}\)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 1}\rule[-1ex]{0pt}{0pt}&\text{Sequence 2} \\
\hline
\rule{0pt}{2.5ex}\text {1. Reflect in the } y \text {-axis.}\rule[-1ex]{0pt}{0pt}&\text{1. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}& \text{2. Translate 1 unit left.} \\
\hline
\rule{0pt}{2.5ex}\text {3. Translate 1 unit right.}\quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.}  \\
\hline
\end{array}

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 3}\rule[-1ex]{0pt}{0pt}&\text{Sequence 4} \\
\hline
\rule{0pt}{2.5ex}\text{1. Reflect in the \(y\)-axis.} \rule[-1ex]{0pt}{0pt}&\text{1. Translate 0.5 units left.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Translate 0.5 units right.}\rule[-1ex]{0pt}{0pt}&\text{2. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{3. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.} \\
\hline
\end{array}

 
f.    \(X \sim \operatorname{Bi}(4, p)\)

\(\operatorname{Pr}(X \geqslant 3)\) \(=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)\)
  \(=4 p^3(1-p)+p^4\)
  \(=4 p^3-4 p^4+p^4\)
  \(=4 p^3-3 p^4\)
  \(=g(p)\)
Show Worked Solution

a.    \(g(x)=4 x^3-3 x^4\)

\(g^{\prime}(x)=12 x^2-12 x^3=12 x^2(1-x)\)

\(\text{Solve} \ \ g^{\prime}(x)=0:\)

\(x=0,1\)

\(\text{SP’s at} \ (0,0) \ \text{and}\ (1,1)\).
 

b.    \(\text{Find} \ x \text {-intercepts:}\)

\(\text{Solve} \ \ g(x)=0 \ \ \Rightarrow\ \ x=0, \dfrac{4}{3}\)
 

c.

\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}&  \quad -1 \quad & \quad \quad 0 \quad \quad & \quad \quad \frac{1}{2} \quad \quad\\
\hline
\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& 24 & 0 & \frac{3}{2} \\
\hline
\end{array}

 
d.    \(\text{Solve integral (by CAS):}\)

\(I=\dfrac{1}{2-0} \displaystyle \int_0^2 g(x)\ d x=-\dfrac{8}{5}\)
 

e.    \(\text{Possible sequences include:}\)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 1}\rule[-1ex]{0pt}{0pt}&\text{Sequence 2} \\
\hline
\rule{0pt}{2.5ex}\text {1. Reflect in the } y \text {-axis.}\rule[-1ex]{0pt}{0pt}&\text{1. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}& \text{2. Translate 1 unit left.} \\
\hline
\rule{0pt}{2.5ex}\text {3. Translate 1 unit right.}\quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.}  \\
\hline
\end{array}

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 3}\rule[-1ex]{0pt}{0pt}&\text{Sequence 4} \\
\hline
\rule{0pt}{2.5ex}\text{1. Reflect in the \(y\)-axis.} \rule[-1ex]{0pt}{0pt}&\text{1. Translate 0.5 units left.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Translate 0.5 units right.}\rule[-1ex]{0pt}{0pt}&\text{2. Dilate by factor 2 from the \(y\)-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{3. Dilate by factor 2 from the \(y\)-axis.} \quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the \(y\)-axis.} \\
\hline
\end{array}

 
f.    \(X \sim \operatorname{Bi}(4, p)\)

\(\operatorname{Pr}(X \geqslant 3)\) \(=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)\)
  \(=4 p^3(1-p)+p^4\)
  \(=4 p^3-4 p^4+p^4\)
  \(=4 p^3-3 p^4\)
  \(=g(p)\)
♦ Mean mark (e) 40%.
Mean mark (f) 51%

ENGINEERING, TE 2025 HSC 7 MC

Which of the pictorial drawings is an example of isometric drawing?
 

 

Show Answers Only

\(A\)

Show Worked Solution
  • In an isometric drawing, all three axes are drawn at 120° to each other, with vertical lines remaining vertical.
  • Drawing A shows all receding axes at 30° to the horizontal — the defining feature of isometric projection.
  • Drawings B and C show one face drawn true shape with receding lines at an oblique angle — characteristic of oblique projection.
  • Drawing D shows receding lines at an angle inconsistent with isometric construction.

\(\Rightarrow A\)

Functions, EXT1 F1 2025 MET2 5 MC

Which of the following sets represents a function that has an inverse function?

  1. \(\{(1,3),(2,0),(2,1)\}\)
  2. \(\{(-1,3),(2,2),(3,1)\}\)
  3. \(\{(-1,3),(0,1),(1,3)\}\)
  4. \(\{(1,0),(2,3),(1,3)\}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Option A and D are not functions (one x-value has two y-values).}\)

\(\text{Consider option B:}\)

\(\{(-1,3),(2,2),(3,1)\} \ \ \text{is a} \  1:1 \ \text {function.}\)

\(\therefore \ \text {Inverse exists}\)

\(\text{Note that option C is not one-to-one because two different inputs (-1 and 1) map to}\)

\(\text{the same output (3), so its inverse would not be a function.}\)

\(\Rightarrow B\)

Trigonometry, 2ADV T3 2025 MET2 1 MC

A function that has a range of \([6,12]\) is

  1. \(f(x)=6+3 \cos (9 x)\)
  2. \(f(x)=6+6 \cos (3 x)\)
  3. \(f(x)=9-3 \cos (6 x)\)
  4. \(f(x)=9-6 \cos (3 x)\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{By trial and error,}\)

\(\text{Consider option C:}\ \  f(x)=9-3 \cos (6 x)\)

\(\text{Since}\ \ -1 \leqslant \cos (6 x) \leqslant 1\)

\(\text{Range is} \ \ [-3+9,3+9]=[6,12]\)

\(\Rightarrow C\)

Statistics, SPEC2 2025 VCAA 6

The volume of water, \(V\) mL, consumed by a student during a school day may be assumed to be normally distributed with a mean of 1000 mL and a standard deviation of 80 mL .

    1. Write down the mean and standard deviation of the sampling distribution for the average volume of water consumed by randomly selected samples of 25 students.
    2. Give your answers in millilitres.   (1 mark)

    3. What is the probability, correct to four decimal places, that the average volume of water consumed by a random sample of 25 students on a particular school day is more than 970 mL?   (1 mark)

The canteen at a particular school stocks two brands of water in bottles, Wasser and Apa.

The manufacturer of Wasser bottled water knows that the volume of water dispensed into bottles may be assumed to be normally distributed with a standard deviation of 5 mL. Engineers at the company take a random sample of 30 bottles and measure the volume of water in each bottle. The sample mean is found to be 750 mL.

  1. Find a 95% confidence interval for the mean volume of water dispensed into each Wasser bottle.
  2. Give your values in millilitres, correct to one decimal place.   (1 mark)

  3. The engineers decide to take 300 random samples, each containing 30 bottles, and calculate the respective 95% confidence intervals. All samples are independent.
  4. In how many of these confidence intervals would the engineers expect the value of the true mean volume dispensed to be included?   (1 mark)

  5. What is the minimum size of the sample required to ensure that the difference between the sample mean and the mean volume dispensed is no more than 1 mL at the 95% confidence level?   (1 mark)

The volume of water dispensed into Apa water bottles may be assumed to be normally distributed with a mean of 750 mL and a standard deviation of 5 mL. After a service, a random sample of 50 bottles gave a sample mean of 748 mL. The company now claims that the mean volume of water dispensed is less than the stated mean of 750 mL.

A one-tailed statistical test at the 1% level of significance is proposed.

  1. Write down the null and alternative hypotheses that will be used in testing the company's claim.   (1 mark)

    1. Determine the \(p\) value for this test.
    2. Give your answer correct to four decimal places.   (1 mark)

    3. Is the company's claim correct?
    4. Explain your conclusion in terms of the \(p\) value.   (1 mark)

  3. At the 1% level of significance for a sample size of 50 bottles, find the critical value of the sample mean, below which a sample mean value would support the conclusion that the mean volume of water dispensed is now less than 750 mL.
  4. Give your answer correct to three decimal places.   (1 mark)

  5. Assume that, after the service, the true mean volume of water in the Apa bottles was found to be 747.5 mL and that the population standard deviation, \(\sigma\), is 5 mL.
  6. At the 1% level of significance, for a sample size of 50 , find the probability that the company will conclude that the service has not reduced the mean volume of water in an Apa bottle.
  7. Give your answer correct to three decimal places.   (1 mark)

Show Answers Only

a.i.  \(\mu=1000,\ \ \sigma=16\)

a.ii.  \(P(V>970)=0.9696\)

b.    \(\text{95% CI}=(748.2,751.8)\) 

c.    \(95 \% \times 300=285\)

d.    \(n=97\) 

e.    \(H_0: \ \mu=750, \ H_1: \ \mu<750\)

f.i.  \(p=0.0023\) 

f.ii.   \(\text{Since \(0.0023<0.01\) (significance level), claim is correct.}\)

g.    \(\overline{X}=748.355 \ \text{mL}\)

h.    \(p=0.113\)

Show Worked Solution
a.i.   \(\mu\) \(=1000\)
  \(\sigma\) \(=\dfrac{80}{\sqrt{16}}=16\)

 

a.ii.  \(P(V>970)=0.9696\)
 

b.    \(\overline{X} \sim N\left(750, \dfrac{5}{\sqrt{30}}\right)\)

\(\text{95% CI}\) \(=\left(750-1.96 \times \dfrac{5}{\sqrt{30}}, 750+1.96\times\dfrac{5}{\sqrt{30}}\right)\)
  \(=(748.2,751.8)\)

 

c.    \(95 \% \times 300=285\)
 

d.    \(\text {Solve for} \ n:\)

\(1\) \(\geqslant 1.96 \times \dfrac{5}{\sqrt{n}}\)
\(n\) \(\geqslant 96.04\)
\(n\) \(=97\)

 

e.    \(H_0: \ \mu=750\)

\(H_1: \ \mu<750\)
 

f.i.    \(p\) \(=\operatorname{Pr}\left(z<\dfrac{748-750}{\frac{5}{\sqrt{50}}}\right)\)
    \(=\operatorname{Pr}\left(z<-\dfrac{2 \sqrt{50}}{5}\right)\)
    \(=0.0023\)

 

f.ii.   \(\text{Since \(0.0023<0.01\) (significance level), claim is correct.}\)
 

g.    \(\text{Solve for} \ c:\ \ \operatorname{Pr}(\overline{X}<c)=0.01\)

\(c=748.355 \ \text{mL}\)
 

h.    \(p=0.113\)

Vectors, SPEC2 2025 VCAA 5

Consider three planes defined by the equations  \(\Pi_1: \ 2 x+9 z=8, \Pi_2: \ 3 x+6 y+5 z=7\)  and  \(\Pi_3: \ x+9 y-3 z=7\).

  1. Find the point of intersection of the three planes.   (1 mark)

    1. Find a vector that gives the direction of the line of intersection of the planes \(\Pi_2\) and \(\Pi_3\).   (2 marks)

    2. Find a set of parametric equations that give the coordinates of the points that lie on this line of intersection.   (1 mark)

  3. Find the shortest distance from the point \((1,1,2)\) to the plane \(\Pi_3\).   (2 marks)

  4. Consider a family of planes, \(\Psi\), with equation  \(6 x+27 z=m\), where \(m \in N\).
    1. Show that the plane \(\Pi_1\) is parallel to each member of \(\Psi\).   (1 mark)

    2. Find all values of \(m\) for which the shortest distance between plane \(\Pi_1\) and the plane of the form  \(6 x+27 z=m\)  is  \(\dfrac{23}{3 \sqrt{85}}\).   (3 marks)

Show Answers Only

a.    \(\text{Intersection at}\ (-5,2,2)\)
 

b.i.  \(\underset{\sim}{d}=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)\)
 

b.ii. \(x=-5-63 \lambda\)

\(y=2 + 14 \lambda\)

\(z=2+21 \lambda\)
 

c.    \(\dfrac{3}{\sqrt{91}}\)
 

d.i.  \(\Psi: \ 6 x+27 z=m\)

\(\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}\)

\(\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}\)

\(\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}\)

d.ii.  \(m=1,47\)

Show Worked Solution

a.    \(\Pi_1: \ 2 x+9 z=8\)

\(\Pi_2: \ 3 x+6 y+5 z=7\)

\(\Pi_3: \ x+9 y-3 z=7\)

\(\text {Solve simultaneously (by CAS):}\)

\(\text{Intersection at}\ (-5,2,2)\)
 

b.i.  \(\underset{\sim}{d} \ \text{is} \perp \text{to normals of} \ \Pi_2 \ \text{and}\  \Pi_3.\)

\(\underset{\sim}{d}=n_2 \times n_3=\left|\begin{array}{ccc}i & j & k \\ 3 & 6 & 5 \\ 1 & 9 & -3\end{array}\right|=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)\)
 

b.ii. \((-5,2,2)\ \text{lies on both planes (from part a.)}\)

\(x=-5-63 \lambda\)

\(y=2 + 14 \lambda\)

\(z=2+21 \lambda\)

♦ Mean mark (b.ii) 51%.

c.    \(\text{Find shortest distance from (1, 1, 2) to \(\Pi_3\).}\)

\(\Pi_3: \ x+9 y-3 z=7 \ \ \Rightarrow \ \ (7,0,0)\ \text{lies on plane.}\)

\(\text {Spanning vector to} \ (1,1,2) \ \text{is} \ (-6 \underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})\)

\(\text{Distance}=\dfrac{1}{\sqrt{91}} \times \abs{\left(\begin{array}{c}-6 \\ 1 \\ 2\end{array}\right)\left(\begin{array}{c}1 \\ 9 \\ -3\end{array}\right)}=\dfrac{3}{\sqrt{91}}\)
 

d.i.  \(\Psi: \ 6 x+27 z=m\)

\(\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}\)

\(\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}\)

\(\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}\)
  

d.ii. \(\text{Point on}\ \Pi_1: (4,0,0)\)

\(\text{Point on} \ \Psi:\left(\dfrac{m}{6}, 0,0\right)\)

\(\text{Spanning vector}=\left(4-\dfrac{m}{6}\right) \underset{\sim}{i}\)

\(\text{Distance}=\abs{\left(4-\dfrac{m}{6}\right) \underset{\sim}{i} \cdot \dfrac{(2 \underset{\sim}{i}+9\underset{\sim}{k})}{\sqrt{85}}}=\dfrac{23}{3 \sqrt{85}}\)

\(\text{Solve}\ \ \abs{8-\dfrac{m}{3}}=\dfrac{23}{3} \ \ \text{for} \ m:\)

\(m=1,47\)

♦ Mean mark (d.ii) 45%.

Vectors, SPEC2 2025 VCAA 4

The path of a moving particle with position vector

\(\underset{\sim}{r}(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

is shown below for time \(t \geq 0\).

All lengths are in metres and time is measured in seconds.
 

  1. Write down the coordinates of the particle's starting point.   (1 mark)

  2. On the graph above, draw an arrow from the point \((9,0)\) to indicate the direction of motion of the particle.   (1 mark)

  3. Find the value of \(t\) for which the particle will first return to its starting point.   (1 mark)

  4. Show that the speed of the particle, in m s\(^{-1}\), at time \(t\) can be expressed as  \(\sqrt{125-100 \cos \left(\dfrac{3 t}{2}\right)}\).   (3 marks)

  5. What is the maximum speed of the particle in m s\(^{-1}\)?   (1 mark)

  6. On the graph above, trace the path of the particle for  \(t \in[0, \pi]\).   (1 mark)

  7. Find the length of the path traced in part f, giving your answer in metres, correct to one decimal place.   (2 marks)

Show Answers Only

a.    \(\text{Initial coordinates:}\ (1,0)\)
 

b.    
     

c.   \(t=4 \pi \ \text{seconds}\)

d.    \(r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}\)

\(\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

\(\text {Speed}=\abs{\dot{r}(t)}:\)

\(\text{Speed}^2\) \(=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2\)
  \(=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)\)
       \(\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)\)
  \(=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)\)
  \(=125-100 \cos \left(\dfrac{5 t}{2}-t\right)\)
\(\text{speed}\) \(=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}\)

 

e.    \(\text{Speed}_{\text {max}}=15 \ \text{ms}^{-1}\)
 

f.    \(\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).\)
 


 

g.    \(\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}\)

Show Worked Solution

a.    \(\text{At} \ \ t=0:\)

\(r(t)=(5 \cos 0-4 \cos 0)\underset{\sim}{i} + (5 \sin 0-4 \sin 0) {\underset{\sim}{j}}=\underset{\sim}{i}\)

\(\text{Initial coordinates:}\ (1,0)\)
 

b.    
     

c.   \(\text{Solve for}\ t \ \text{(by CAS):}\)

\(5 \cos (t)-4 \cos \left(\dfrac{5t}{2}\right)=1\)

\(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)=0\)

\(t=4 \pi \ \text{seconds}\)

Mean mark (c) 53%.
 

d.    \(r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}\)

\(\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

\(\text {Speed}=\abs{\dot{r}(t)}:\)

\(\text{Speed}^2\) \(=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2\)
  \(=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)\)
       \(\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)\)
  \(=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)\)
  \(=125-100 \cos \left(\dfrac{5 t}{2}-t\right)\)
\(\text{speed}\) \(=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}\)

 

e.    \(\text{Speed}_{\text {max }} \ \text {occurs when} \ \ \cos \left(\frac{3 t}{2}\right)=-1\)

\(\text{Speed}_{\text {max}}=\sqrt{125+100}=15 \ \text{ms}^{-1}\)
 

f.    \(\text{At} \ \ t=\pi, \quad r(\pi)=-5\underset{\sim}{i}-4\underset{\sim}{j}\)

\(\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).\)
 


 

g.    \(\text{Length of curve (by CAS):}\)

\(\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}\)

Functions, MET2 2025 VCAA 5 MC

Which of the following sets represents a function that has an inverse function?

  1. \(\{(1,3),(2,0),(2,1)\}\)
  2. \(\{(-1,3),(2,2),(3,1)\}\)
  3. \(\{(-1,3),(0,1),(1,3)\}\)
  4. \(\{(1,0),(2,3),(1,3)\}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Option A and D are not functions (x-values have two y-values).}\)

\(\text{Consider option B:}\)

\(\{(-1,3),(2,2),(3,1)\} \ \text{is a one-to-one function.}\)

\(\therefore \ \text {Inverse exists}\)

\(\text{Note that option C is not one-to-one because two different inputs (-1 and 1) map to}\)

\(\text{the same output (3), so its inverse would not be a function.}\)

\(\Rightarrow B\)

Graphs, MET2 2025 VCAA 1 MC

A function that has a range of \([6,12]\) is

  1. \(f: R \rightarrow R, \ f(x)=6+3 \cos (9 x)\)
  2. \(f: R \rightarrow R, \ f(x)=6+6 \cos (3 x)\)
  3. \(f: R \rightarrow R, \ f(x)=9-3 \cos (6 x)\)
  4. \(f: R \rightarrow R, \ f(x)=9-6 \cos (3 x)\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{By trial and error,}\)

\(\text{Consider option C:}\ \  f(x)=9-3 \cos (6 x)\)

\(\text{Since}\ \ -1 \leqslant \cos (6 x) \leqslant 1\)

\(\text{Range is} \ \ [-3+9,3+9]=[6,12]\)

\(\Rightarrow C\)

Complex Numbers, SPEC2 2025 VCAA 2

  1. Sketch \(\{z: z \bar{z}=4, z \in C\}\) on the Argand plane below.   (1 mark)
     

    1. Show that \(\{z:|z-2 i|=|z-\sqrt{3}-i|, z \in C\}\) may be expressed as  \(y=\sqrt{3} x\).   (2 marks)

    2. Sketch \(\{z:|z-2 i|=|z-\sqrt{3}-i|, z \in C\}\) on the Argand plane in part a.   (1 mark)

    1. Find the points of intersection of the curves defined in part a and in part b.i, expressing your answers in the form  \(a+i b\), where  \(a, b \in R\).   (2 marks)

    2. Label these points on the Argand plane in part a.   (1 mark)

Consider the points \(P\) and \(Q\) labelled on the Argand plane below.
 

  1. A ray originating at point \(P\) and passing through point \(Q\) has the equation  \(\operatorname{Arg}\left(z-z_0\right)=\theta\), where \(\theta\) is a radian measure.
  2. Write down the values of \(z_0\) and \(\theta\).   (1 mark)

  3. Find the area of the minor segment bounded by the chord connecting the points \(P\) and \(Q\) and the circle given by  \(|z|=3\).
  4. Give your answer in the form \(c \pi+d\), where \(c, d \in R\).   (2 marks)

Show Answers Only

a.    \(z \bar{z}=4\ \ \Rightarrow\ \ \text{Circle, centre (0, 0), radius = 2}\)
 

b.i  \(\abs{z-2 i}=\abs{x+(y-2) i}=x^2+(y-2)^2\)

\(\abs{z-\sqrt{3}-i}=\abs{x-\sqrt{3}+(y-1)^2}=(x-\sqrt{3})^2+(y-1)^2\)

\(\text{Equating expressions:}\)

\(x^2+(y-2)^2=(x-\sqrt{3})^2+(y-1)^2\)

\(x^2+y^2-4 y+4=x^2-2 \sqrt{3} x+3+y^2-2 y+1\)

\(-2 y\) \(=-2 \sqrt{3} x\)
\(y\) \(=\sqrt{3} x\)

 

b.ii    \(\text {See image in part (a).}\)

  \(\abs{z-2 i}=\abs{z-\sqrt{3}-i} \ \Rightarrow \ \text{see straight line in image.}\)
 

c.i   \(\text{Find intersection:}\)

\(x^2+y^2=4 \ \ \text{and} \ \ y=\sqrt{3} x\)

\(x^2+3 x^2=4 \ \ \Rightarrow \ \ x^2=1 \ \ \Rightarrow \ \ x= \pm 1\)

\(\text{Intersection co-ordinates:}\ \ (1, \sqrt{3}),(-1,-\sqrt{3})\)

\(z=1+\sqrt{3} i\)

\(z=-1-\sqrt{3} i\)
 

c.ii  \(\text{See image in part (a)}\)
 

d.    \(z_0=\dfrac{3}{2}+\dfrac{3 \sqrt{3}}{2} i\)

\(\theta=-\dfrac{\pi}{3}\)
 

e.    \(A\) \(=\dfrac{1}{2} \times 3^2 \times\left(\dfrac{\pi}{3}-\sin \dfrac{\pi}{3}\right)\)
    \(=\dfrac{3}{2} \pi-\dfrac{9 \sqrt{3}}{4}\)
Show Worked Solution

a.    \(z \bar{z}=4\ \ \Rightarrow\ \ \text{Circle, centre (0, 0), radius = 2}\)
 

b.i  \(\abs{z-2 i}=\abs{x+(y-2) i}=x^2+(y-2)^2\)

\(\abs{z-\sqrt{3}-i}=\abs{x-\sqrt{3}+(y-1)^2}=(x-\sqrt{3})^2+(y-1)^2\)

\(\text{Equating expressions:}\)

\(x^2+(y-2)^2=(x-\sqrt{3})^2+(y-1)^2\)

\(x^2+y^2-4 y+4=x^2-2 \sqrt{3} x+3+y^2-2 y+1\)

\(-2 y\) \(=-2 \sqrt{3} x\)
\(y\) \(=\sqrt{3} x\)

 

b.ii    \(\text {See image in part (a).}\)

  \(\abs{z-2 i}=\abs{z-\sqrt{3}-i} \ \Rightarrow \ \text{see straight line in image.}\)
 

c.i   \(\text{Find intersection:}\)

\(x^2+y^2=4 \ \ \text{and} \ \ y=\sqrt{3} x\)

\(x^2+3 x^2=4 \ \ \Rightarrow \ \ x^2=1 \ \ \Rightarrow \ \ x= \pm 1\)

\(\text{Intersection co-ordinates:}\ \ (1, \sqrt{3}),(-1,-\sqrt{3})\)

\(z=1+\sqrt{3} i\)

\(z=-1-\sqrt{3} i\)
 

c.ii  \(\text{See image in part (a)}\)
 

d.    \(z_0=\dfrac{3}{2}+\dfrac{3 \sqrt{3}}{2} i\)

\(\theta=-\dfrac{\pi}{3}\)

Mean mark (d) 51%.
e.    \(A\) \(=\dfrac{1}{2} \times 3^2 \times\left(\dfrac{\pi}{3}-\sin \dfrac{\pi}{3}\right)\)
    \(=\dfrac{3}{2} \pi-\dfrac{9 \sqrt{3}}{4}\)

Calculus, SPEC2 2025 VCAA 1

  1. Sketch the graph of  \(y(x)=\dfrac{3 x}{x^3+x+2}\)  on the axes below.
  2. Label the asymptotes with their equations, and label the turning point and the point of inflection with their coordinates. Give the coordinates of the point of inflection correct to one decimal place.   (3 marks)

  1. The region bounded by the graph of  \(y=\dfrac{3 x}{x^3+x+2}\), the coordinate axes and the line  \(x=2\)  is rotated about the \(x\)-axis to form a solid of revolution.
    1. Write down a definite integral that, when evaluated, will give the volume of the solid of revolution.   (1 mark)

    2. Find the volume of the solid of revolution correct to two decimal places.   (1 mark)

  2. Find the equations of the vertical asymptotes of the curve given by  \(y=\dfrac{3 x}{x^3-5 x+2}\).   (1 mark)

  3. A family of curves is given by  \(y(x)=\dfrac{3 x}{x^3+a x+2}\), where  \(a \in R\).
    1. Consider the case where the graph has a stationary point \(P\).
    2. Find the \(y\)-coordinate of \(P\) in terms of \(a\).   (1 mark)

    3. For a given value of \(a\), the graph has no stationary points.
    4. Find the equations of the vertical asymptotes of the graph in this case.   (1 mark)

    5. For a given value of \(a\), the graph will have a point of inflection at  \(x=2\).
    6. Find the value of \(a\).   (2 marks)

Show Answers Only

a.    

b.i.    \(V=\displaystyle \int_0^2 \pi\left(\frac{3 x}{x^3+x+2}\right)^2 \, d x\)

b.ii.  \(V=2.29\ \text{u}^3\)

c.   \(x=-\sqrt{2}-1, \ x=\sqrt{2}-1, \ x=2\)

d.i.  \(\text{SP at} \ \left(1, \dfrac{3}{3+a}\right)\)
 

d.ii. \(\text{No SP’s exist when}\ \ 3+a=0 \ \ \Rightarrow \ \ a=-3\)

\(x=-2,1\)

d.iii.  \(a=\dfrac{24}{5}\)

Show Worked Solution

a.    

b.i.    \(V=\displaystyle \int_0^2 \pi\left(\frac{3 x}{x^3+x+2}\right)^2 \, d x\)
 

b.ii.  \(V=2.29\ \text{u}^3\)
 

c.   \(\text{Find vertical asymptotes.}\)

\(\text {Solve:} \ \ x^3-5 x+\dfrac{1}{2}=0\)

\(x=-\sqrt{2}-1, \ x=\sqrt{2}-1, \ x=2\)
 

d.i.  \(y=\dfrac{3 x}{x^3+a x+2}\)

\(\text{Find \(x\) when \(y^{\prime}=0 \ \ \Rightarrow \ \ x=1\)}\)

\(\text{SP at} \ \ \left(1, \dfrac{3}{3+a}\right)\)
 

d.ii. \(\text{No SP’s exist when}\ \ 3+a=0 \ \ \Rightarrow \ \ a=-3\)

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

\(\text{Vertical asymptotes occur when}\ \ x^3-3 x+2=0\)

\(\text{(By CAS):} \ \ x=-2,1\)

♦ Mean mark (d.ii) 46%.

d.iii.  \(\text{If POI exists at} \ \ x=2:\)

\(y^{\prime \prime}(2)=0 \ \Rightarrow \ \dfrac{-3(5 a-24)}{2(a+5)^3}=0\)

\(a=\dfrac{24}{5}\)

Calculus, EXT1 C3 2025 SPEC2 8 MC

Consider the direction field below.
 

The direction field best represents the differential equation

  1. \(\dfrac{d y}{d x}=x^2-y\)
  2. \(\dfrac{d y}{d x}=x-y^2\)
  3. \(\dfrac{d y}{d x}=y-x\)
  4. \(\dfrac{d y}{d x}=x-y\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{At \((-1,0)\), gradient is positive (eliminate B and D)}\).

\(\text{At \((1,0)\), gradient is positive (eliminate C)}\).

\(\Rightarrow A\)

Proof, EXT2 P1 2025 SPEC2 1 MC

A tiger is a type of cat.

Consider the following statement.

'If I have a tiger, then I have a cat.'

The contrapositive of this statement is

  1. if I do not have a tiger, then I do not have a cat.
  2. if I have a cat, then I have a tiger.
  3. if I do not have a cat, then I do not have a tiger.
  4. if I do not have a tiger, then I have a different type of cat.
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Statement: If} \ \ X \Rightarrow Y\)

\(\text{Contrapositive: If}\ \ \neg \ Y\ \Rightarrow \neg \ X\)

\(\text{If I don’t have a cat} \ \ \Rightarrow \ \ \text{I don’t have a tiger.}\)

\(\Rightarrow C\)

Probability, 2ADV EQ-Bank 12

A new test has been developed for determining whether or not people are carriers of the Gaussian virus.

Two hundred people are tested. A two-way table is being used to record the results.
 

  1.  What is the value of `A`?   (1 mark)

  2.  A person randomly selected from the tested group is a carrier of the virus.
  3. What is the probability that the test results would show this?   (1 mark) 
  4. If the person randomly selected has a negative result from their test, what is the probability they are not a carrier of the virus?   (2 marks)

Show Answers Only

a.    `98`

b.    `37/43`

c.    `49/55`

Show Worked Solution

a.    `A= 200-(74 + 12 + 16)= 98`
 

b.    `P` `= text(#Positive carriers)/text(Total carriers)`
  `= 74/86`
  `= 37/43`

 

c.    `text(Total number with negative results) = 110`

`text{Total non-carriers with negative results}\ = 98`

`P\text{(not a carrier)|negative result}\ = 98/110=49/55`

Probability, 2ADV EQ-Bank 14

Lie detector tests are not always accurate. A lie detector test was administered to 200 people.

The results were:

• 50 people lied. Of these, the test indicated that 40 had lied;
• 150 people did NOT lie. Of these, the test indicated that 20 had lied.

  1. Complete the table using the information above   (1 mark)
      
        

  2. For what percentage of the people tested was the test accurate?   (1 mark)

  3. What is the probability that the test indicated a lie for a person who did NOT lie?   (1 mark)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `text(85%)`

c.    `2/15`

Show Worked Solution

a.

b.  `text(Percentage of people with accurate readings)`

`= text(# Accurate readings)/text(Total readings) xx 100`

`= (40+130)/200`

`= 85 text(%)`
 

c.  `text{P(lie detected when NOT a lie)}`

`= 20/150`

`= 2/15`

Statistics, 2ADV EQ-Bank 4 MC

A random variable \(X\) is defined as the volume of fuel (in litres) dispensed by a bowser per use. Which statement best describes \(X\) ?

  1. \(X\) is discrete because the machine dispenses a fixed amount each time.
  2. \(X\) is discrete because volume can only take positive values.
  3. \(X\) is continuous because the machine is used many times per day.
  4. \(X\) is continuous because volume can take any value within an interval.
Show Answers Only

\(D\)

Show Worked Solution
  • Volume is a measurement variable that can take any value within a range, so \(X\) is a continuous random variable.

\(\Rightarrow D\)

Statistics, 2ADV EQ-Bank 3 MC

Which of the following is an example of a continuous random variable?

  1. The number of text messages sent by a student in a day
  2. The time taken for a swimmer to complete 100 m race
  3. The number of cars in a school car park at 9 am
  4. The number of siblings a student has
Show Answers Only

\(B\)

Show Worked Solution
  • Time is a measurement that can take any value within an interval and is therefore continuous.
  • Options A, B and D are all counts that can only take whole number values, making them discrete.

\(\Rightarrow B\)

Statistics, 2ADV EQ-Bank 2 MC

A hospital records the number of patients admitted to the emergency department each day. How would this data best be classified?

  1. Discrete, because the number of patients changes every day
  2. Discrete, because the count can only take whole number values
  3. Continuous, because the number of patients can be very large
  4. Continuous, because patient admission is an ongoing process
Show Answers Only

\(B\)

Show Worked Solution
  • The number of patients is a count and can only take whole number values, making it a discrete numerical variable.

\(\Rightarrow B\)

Statistics, 2ADV EQ-Bank 1 MC

A scientist measures the wingspan of butterflies in a study. How would this data best be classified?

  1. Discrete, because wingspan can be measured precisely
  2. Discrete, because each butterfly has a unique wingspan
  3. Continuous, because wingspan can take any value within a range
  4. Continuous, because there are many butterflies in the study
Show Answers Only

\(C\)

Show Worked Solution
  • Wingspan is a measurement that can take any value within a range, making it a continuous numerical variable.

\(\Rightarrow C\)

Complex Numbers, EXT2 N2 2025 SPEC1 8

Consider the function with rule  \(f(z)=z^4+6 z^2+25\), where \(z \in C\).

  1. Consider  \(z_1=1+2 i\).
  2. Plot and label \(z_1\) and \(\overline{z}_1\) on the Argand plane below.   (1 mark)  

     
  3. Given that  \(1+2 i\)  is a solution of  \(f(z)=0\), find a quadratic factor of \(f(z)\).   (2 marks)  

  4. Hence, find all remaining solutions of  \(f(z)=0\).   (2 marks)

Show Answers Only

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(z^2-2 z+5\)

c.    \(z=-1+2 i, z=-1-2 i\)

Show Worked Solution

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)

\(\Rightarrow 1-2 i \ \ \text{is also a solution (conjugate factor theorem).}\)

\(\text {Express as a quadratic factor:}\)

\((z-(1+2 i))(z-(1-2 i))\) \(=((z-1)-2 i)((z-1)+2 i)\)
  \(=(z-1)^2-4 i^2\)
  \(=z^2-2 z+5\)

 

c.    \(f(z)=z^4+6 z^2+25, \quad z \in C\)

\(\text{Since \(\ z^2-2 z+5\ \) is a factor:}\)

\(f(z)\) \(=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)\)
  \(=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)\)
  \(=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b\)
  \(=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b\)
Mean mark (c) 53%.

\(\text{Equating co-efficients:}\)

\(a-2=0 \ \Rightarrow \ a=2\)

\(5 b=25 \ \Rightarrow \ b=5\)

\(f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)\)
 

\(\text{Solve:} \ \ z^2+2 z+5=0\)

\(z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i\)

\(\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i\)

Calculus, SPEC2 2025 VCAA 8 MC

Consider the direction field below.
 

The direction field best represents the differential equation

  1. \(\dfrac{d y}{d x}=x^2-y\)
  2. \(\dfrac{d y}{d x}=x-y^2\)
  3. \(\dfrac{d y}{d x}=y-x\)
  4. \(\dfrac{d y}{d x}=x-y\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{At \((-1,0)\), gradient is positive (eliminate B and D)}\).

\(\text{At \((1,0)\), gradient is positive (eliminate C)}\).

\(\Rightarrow A\)

Calculus, SPEC2 2025 VCAA 7 MC

Using the substitution  \(u=\cos (\theta), \dfrac{1}{2} \displaystyle \int_0^{\tfrac{\pi}{2}} \dfrac{\sin (2 \theta)}{1+\cos (\theta)} \, d \theta\)  can be expressed as

  1. \(\displaystyle\int_0^{\tfrac{\pi}{2}} u \sqrt{\frac{1-u}{1+u}}\ d u\)
  2. \(\displaystyle\int_0^1\left(1+\frac{1}{1+u}\right) d u\)
  3. \(\displaystyle\int_0^{\tfrac{\pi}{2}}\left(1-\frac{1}{1+u}\right) d u\)
  4. \(\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) d u\)
Show Answers Only

\(D\)

Show Worked Solution

\(u=\cos (\theta) \ \Rightarrow \ du=-\sin (\theta)\ d \theta\)

\(\text{When} \ \ \theta=\dfrac{\pi}{2}, u=0\)

\(\text{When}\ \  \theta=0, u=1\)

\(I\) \(=\displaystyle\frac{1}{2} \int_0^{\tfrac{\pi}{2}} \frac{\sin (2 \theta)}{1+\cos \theta} \, d \theta\)
  \(=\displaystyle\int_0^{\tfrac{\pi}{2}} \frac{\sin (\theta) \cos (\theta)}{1+\cos (\theta)} \, d \theta\)
  \(=\displaystyle \int_1^0-\frac{u}{1+u} \, d u\)
  \(=\displaystyle\int_0^1 \frac{1+u-1}{1+u} \, d u\)
  \(=\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) \, d u\)

 

\(\Rightarrow D\)

Statistics, STD2 S1 2010 HSC 26b*

A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.

The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.
 

2010 26b

  1. Find the value of  Χ  in the table.   (1 mark)

  2. On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data.   (1 mark)
  3. Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer.   (1 mark)
Show Answers Only

a.    `15`

b. & c.
               

`text(Median)\ ~~155`

Show Worked Solution

a.    `X= 25-10= 15`

b. & c.
               

`text(Median)\ ~~155`

Complex Numbers, SPEC1 2025 VCAA 8

Consider the function with rule  \(f(z)=z^4+6 z^2+25\), where \(z \in C\).

  1. Consider  \(z_1=1+2 i\).
  2. Plot and label \(z_1\) and \(\overline{z}_1\) on the Argand plane below.   (1 mark)  

     
  3. Given that  \(1+2 i\)  is a solution of  \(f(z)=0\), find a quadratic factor of \(f(z)\).   (2 marks)  

  4. Hence, find all remaining solutions of  \(f(z)=0\).   (2 marks)

Show Answers Only

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(z^2-2 z+5\)

c.    \(z=-1+2 i, z=-1-2 i\)

Show Worked Solution

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)

\(\Rightarrow 1-2 i \ \ \text{is also a solution.}\)

\(\text {Express as a quadratic factor:}\)

\((z-(1+2 i))(z-(1-2 i))\) \(=((z-1)-2 i)((z-1)+2 i)\)
  \(=(z-1)^2-4 i^2\)
  \(=z^2-2 z+5\)

 

c.    \(f(z)=z^4+6 z^2+25, \quad z \in C\)

\(\text{Since \(\ z^2-2 z+5\ \) is a factor:}\)

\(f(z)\) \(=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)\)
  \(=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)\)
  \(=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b\)
  \(=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b\)
Mean mark (c) 53%.

\(\text{Equating co-efficients:}\)

\(a-2=0 \ \Rightarrow \ a=2\)

\(5 b=25 \ \Rightarrow \ b=5\)

\(f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)\)
 

\(\text{Solve:} \ \ z^2+2 z+5=0\)

\(z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i\)

\(\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i\)

Trigonometry, 2ADV T3 2025 MET1 3

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

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

  2. Solve  \(f(x)=0\)  for \(x\).   (3 marks)

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

Show Answers Only

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

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

c.   

   

Show Worked Solution

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

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

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

 
\(\text{Base angle}=\dfrac{\pi}{3}\)

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

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

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

c.   

   

♦ Mean mark (c) 49%.

Calculus, MET1 2025 VCAA 7

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

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

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

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

     

   

  1. 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)

    2. Write down the values of \(x\) for which  \(f(x) g(x) \geq 0\).   (1 mark)

Show Answers Only

a.    \(f(5)=5^3-5^2-16 \times 5-20 =0\)

b.    \(f(x)=(x+2)^2(x-5)\)

c.   
       

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.    \(\text{By long division:}\)
 
           

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

c.   
       
 

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.
     

\(\text{By inspection of graph:}\)

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

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

Probability, MET1 2025 VCAA 4

The probability distribution for the discrete random variable \(X\) is given in the table below, where \(k\) is a positive real number.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad \quad 0 \quad \quad & \quad \quad 1 \quad \quad & \quad \quad 2 \quad \quad & \quad \quad 3 \quad \quad \\
\hline
\rule{0pt}{2.5ex}\operatorname{Pr}(X=x) \rule[-1ex]{0pt}{0pt}& \dfrac{4}{k} &\dfrac{2 k}{75} &\dfrac{k}{75} & \dfrac{2}{k} \\
\hline
\end{array}

  1. Show that  \(k=10\)  or  \(k=15\).   (2 marks)

  2. Let  \(k=15\).
    1. Find \(\operatorname{Pr}(X>1)\).   (1 mark)

    2. Find \(E (X)\).   (1 mark)

Show Answers Only

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

  
\(\therefore k=10 \ \ \text{or 15}\)

b.i.   \(\operatorname{Pr}(X>1)=\dfrac{1}{3}\) 

b.ii.  \(E(X)=\dfrac{90}{75}\)

Show Worked Solution

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

 

\(\therefore k=10 \ \ \text{or 15}\)
 

b.i.   \(\operatorname{Pr}(X>1)\) \(=\operatorname{Pr}(X=2)+\operatorname{Pr}(X=3)\)
    \(=\dfrac{15}{75}+\dfrac{2}{15}\)
    \(=\dfrac{1}{3}\)

 

b.ii.   \(E(X)\) \(=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}\)
    \(=\dfrac{30}{75}+\dfrac{30}{75}+\dfrac{30}{75}\)
    \(=\dfrac{90}{75}\)

BIOLOGY, M8 EQ-Bank 5 MC

Which coordination system detects changes in blood glucose and sends signals to effector organs to restore homeostasis?

  1. Circulatory
  2. Digestive
  3. Hormonal
  4. Neural
Show Answers Only

`D`

Show Worked Solution
  • D is correct: Neural pathways detect changes and send signals to effectors to restore homeostasis.

Other Options:

  • A is incorrect: Circulatory system transports substances; does not detect or signal.
  • B is incorrect: Digestive system processes nutrients; not a coordination system.
  • C is incorrect: Hormonal is a valid coordination system but uses chemical messengers via blood, not neural signals.

Graphs, MET1 2025 VCAA 3

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

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

  2. Solve  \(f(x)=0\)  for \(x\).   (3 marks)

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

Show Answers Only

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

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

c.   

   

Show Worked Solution

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

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

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

 
\(\text{Base angle}=\dfrac{\pi}{3}\)

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

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

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

c.   

   

♦ Mean mark (c) 49%.

Statistics, STD2 S1 EQ-Bank 23

A boxplot for the sale prices of a sample of 203 homes is shown.
 

  1. Calculate the range of the sale price data in the boxplot.   (1 mark)

  2. Calculate the upper fence for any outliers within the sale price data of the boxplot.   (2 marks)

Show Answers Only

a.    \(\text{Range}=900\,000\)

b.    \(1\,350\,000\)

Show Worked Solution

a.    \(\text{Range}=1\,300\,000-400\,000=900\,000\)
 

b.    \(IQR=900\,000-600\,000=300\,000\)

\(Q_3=900\,000\)

\(\text{Upper Fence}\) \(=Q_3+1.5 \times IQR\)
  \(=900\,000+1.5 \times 300\,000\)
  \(=1\,350\,000\)

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.
 

  1. Calculate the sum of the degrees of all the vertices in this graph.   (1 mark)
  2. Euler's formula, \(v+f=e+2\), holds for this graph.
  3. Complete the formula by writing the appropriate numbers in the boxes below.   (1 mark)

  1. Frances is at the gym. She would like to visit each of the other locations once and end at her home.
  2. What is the mathematical term used to describe this route?   (1 mark)
  3. Using edges from the original graph, construct a spanning tree below.   (1 mark)

     


Show Answers Only

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

Data Analysis, GEN2 2025 VCAA 2

A boxplot for the sale prices of a sample of 203 homes is shown.
 

  1. Calculate the range of the sale price data in the boxplot.   (1 mark)

  2. Calculate the upper fence for the sale price data in the boxplot.   (1 mark)

Show Answers Only

a.    \(\text{Range}=900\,000\)

b.    \(1\,350\,000\)

Show Worked Solution

a.    \(\text{Range}=1\,300\,000-400\,000=900\,000\)
 

b.    \(IQR=900\,000-600\,000=300\,000\)

\(Q_3=900\,000\)

\(\text{Upper Fence}\) \(=Q_3+1.5 \times IQR\)
  \(=900\,000+1.5 \times 300\,000\)
  \(=1\,350\,000\)

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}

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

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

    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)

    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)

  4. Table 3, below, shows the percentage of houses and apartments with prices in the given ranges. Some information is missing.
  5. Use the data from Table 1 to complete Table 3.
  6. Table 3 
Show Answers Only

a.    \(\text{Median}=920\,000\)

b.    \(\text{Variable is nominal}\)

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

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

\(\text{spread of house sale prices.}\)

d.

\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%.

d.

\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}

Networks, GEN1 2025 VCAA 34 MC

Consider the following graph.
 

The number of bridges in this graph is

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

\(C\)

Show Worked Solution

\(\text{Bridge – an edge, if removed, will disconnect the network.}\)

\(\text{The top 3 edges satisfy this test.}\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 29 MC

The female population of an animal species has been divided into four age groups (1, 2, 3 and 4), with age group 1 being the youngest.

The age groups, birth rates and survival rates are presented in the life-cycle diagram below.
 

 
Which one of the following is a Leslie matrix that corresponds with this life-cycle diagram?
 

  1. \(\begin{bmatrix}0 & 2.1 & 4.6 & 1.8 \\ 0.9 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.2 & 0\end{bmatrix}\)
     
  2. \(\begin{bmatrix}0 & 0.9 & 0.7 & 0.2 \\ 2.1 & 0 & 0 & 0 \\ 0 & 4.6 & 0 & 0 \\ 0 & 0 & 1.8 & 0\end{bmatrix}\)
     
  3. \(\begin{bmatrix}2.1 & 4.6 & 1.8 & 0 \\ 0.9 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.2 & 0\end{bmatrix}\)
     
  4. \(\begin{bmatrix}2.1 & 4.6 & 1.8 & 0.8 \\ 0 & 0.9 & 0 & 0 \\ 0 & 0 & 0.7 & 0 \\ 0 & 0 & 0 & 0.2\end{bmatrix}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Age group 1 has no birthrate \(\Rightarrow  \ e_{11}=0\) (eliminate C, D)}\)

\(\text{Only option A has birth/death rates in the correct place.}\)

\(\Rightarrow A\)

Matrices, GEN1 2025 VCAA 27 MC

Consider the matrix \(E\) where

\begin{align*}
E=\begin{bmatrix}
m & -9 \\
4 & n
\end{bmatrix}
\end{align*}

For the inverse of \(E\) to exist, the values of \(m\) and \(n\), respectively, cannot be

  1. \(3\) and \(12\)
  2. \(12\) and \(3\)
  3. \(3\) and \(-12\)
  4. \(-3\) and \(-12\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Inverse exists}\ \ \Rightarrow\ \ \operatorname{det}\,E \neq 0\)

\(\operatorname{det}\,E=m \times n-(-9 \times 4)=mn+36\)

\(\text {Only option C gives} \ \ mn=-36\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 26 MC

Consider the matrices \(A, B\) and \(C\) where

\begin{align*}
A=\begin{bmatrix}
2 & 0 \\
1 & 6
\end{bmatrix}, B=\begin{bmatrix}
3 \\
5
\end{bmatrix} \ \text{and}\ \  C=AB.
\end{align*}

The calculation that correctly determines element \(c_{21}\) is

  1. \(2 \times 3+0 \times 5\)
  2. \(2 \times 5+3 \times 0\)
  3. \(1 \times 3+6 \times 5\)
  4. \(1 \times 5+6 \times 3\)
Show Answers Only

\(C\)

Show Worked Solution

\(AB=\begin{bmatrix}2 & 0 \\ 1 & 6\end{bmatrix}\begin{bmatrix}3 \\ 5\end{bmatrix}=\begin{bmatrix} 6 \\ 33\end{bmatrix}\)

\(c_{12}=1 \times 3+6 \times 5=33\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 25 MC

Consider the matrix \(G\) where

\begin{align*}
G=\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}

Which one of the following correctly describes matrix \(G\)?

  1. a binary matrix
  2. a permutation matrix
  3. an identity matrix
  4. a diagonal matrix
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Binary matrix – made up of 0’s and 1’s only.}\)

\(\text{Permutation matrix – single 1 in each row and column (not B).}\)

\(\text{Identity matrix – main diagonal of 1’s (not C).}\)

\(\text{Diagonal matrix – all entries not on main diagonal =0 (not D).}\)

\(\Rightarrow A\)

Data Analysis, GEN1 2025 VCAA 15 MC

The number of drinks sold daily by a juice bar, drinks, over a 10-day period is shown in the table below.
 

The four-mean smoothed number of drinks, with centring, sold on day 8 is closest to

  1. 134.25
  2. 140.0
  3. 142.75
  4. 145.75
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Calculation 1:} \ \ \dfrac{78+187+106+166}{4}=134.25\)

\(\text{Calculation 2:} \ \ \dfrac{187+106+166+124}{4}=145.75\)

\(\text{Centring:} \ \ \dfrac{134.25+145.75}{2}=140\)

\(\Rightarrow B\)

Data Analysis, GEN1 2025 VCAA 13-14 MC

The following time series graph shows the margin, in points, for a team winning all of its games in the first 18 weeks of a season. The team plays one game per week.
 

Part 1

The time series graph is smoothed using five-median smoothing.

The smoothed value for the margin in week 8, in points, is

  1. 32
  2. 35
  3. 38
  4. 41

 
Part 2

Which one of the following options best describes the qualitative features of the time series graph above?

  1. decreasing trend only
  2. irregular fluctuations only
  3. structural change only
  4. decreasing trend with irregular fluctuations
Show Answers Only

\(\text{Part 1:}\ C\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\)

\(\text{Five values with week 8 in the middle:} \ 50,38,32,35,41\)

\(\text{Ranking values (above):} \ 32,35,38,41,50\)

\(\text{Median}=38\)

\(\Rightarrow C\)
 

\(\text{Part 2}\)

\(\text{Qualitative features of graph:}\)

\(\text{Decreasing trend (eliminate B and C)}\)

\(\text{Irregular fluctuations}\)

\(\Rightarrow D\)

Data Analysis, GEN1 2025 VCAA 11-12 MC

The table below shows the life expectancy in years, life, and the number of doctors per 1000 people, doctors, for a sample of 10 countries in 2024. A scatterplot displaying the data is also shown.
 

 

Part 1

A logarithmic (base 10) transformation was applied to the variable life.

With \(\log _{10}(\textit{life})\) as the response variable, the equation of the least squares line fitted to the transformed data is closest to

  1. \(\log _{10}(life)=1.57+0.0123 \times doctors\) 
  2. \(\log _{10}(life)=1.63+0.0326 \times doctors\)
  3. \(\log _{10}(life)=1.79+0.0383 \times doctors\)
  4. \(\log _{10}(life)=1.85+0.0403 \times doctors\)

 
Part 2

A squared transformation was applied to the variable \(doctors\).

The equation of the least squares line fitted to this transformed data is of the form  \(\textit{life}=a+b \times(\textit{doctors})^2\).

Using this equation, the predicted \(life\), in years, for a country with two \(doctors\) per 1000 people is closest to

  1. 73.6
  2. 74.0
  3. 74.5
  4. 74.9
Show Answers Only

\(\text{Part 1:}\ C\)

\(\text{Part 2:}\ C\)

Show Worked Solution

\(\text{Transform data in table (use for parts 1 and 2):}\)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{doctors} \ \ \rule[-1ex]{0pt}{0pt} & \log_{10}(\textit{life}) & \textit{doctors}^2 & \quad \textit{life} \quad \\
\hline
\rule{0pt}{2.5ex} 0.21 \rule[-1ex]{0pt}{0pt} & 1.810 &0.044 &64.6\\
\hline
\rule{0pt}{2.5ex} 0.42 \rule[-1ex]{0pt}{0pt} & 1.809 &0.176 &64.4\\
\hline
\rule{0pt}{2.5ex} 0.59 \rule[-1ex]{0pt}{0pt} & 1.811 &0.348 &64.7\\
\hline
\rule{0pt}{2.5ex} 0.79 \rule[-1ex]{0pt}{0pt} & 1.814 &0.624 &65.1\\
\hline
\rule{0pt}{2.5ex} 1.12 \rule[-1ex]{0pt}{0pt} & 1.819 &1.254 &65.9\\
\hline
\rule{0pt}{2.5ex} 1.42 \rule[-1ex]{0pt}{0pt} & 1.824 &2.016 &66.7\\
\hline
\rule{0pt}{2.5ex} 1.72 \rule[-1ex]{0pt}{0pt} & 1.836 &2.958 &68.6\\
\hline
\rule{0pt}{2.5ex} 1.77 \rule[-1ex]{0pt}{0pt} & 1.851 &3.133 &70.9\\
\hline
\rule{0pt}{2.5ex} 1.94 \rule[-1ex]{0pt}{0pt} & 1.868 &3.764 &73.8\\
\hline
\rule{0pt}{2.5ex}2.05 \rule[-1ex]{0pt}{0pt} & 1.898 &4.203 &79.1\\
\hline
\end{array}

 
\(\text{Part 1}\)

\(\text{Calculate LSRL (by CAS):}\)

\(\log _{10}(\textit{life})=1.7879 \ldots+0.03829 \ldots \times \textit{doctors}\)

\(\Rightarrow C\)
 

\(\text{Part 2}\)

\(\text{Calculate LSRL (by CAS):}\)

\(\textit{life}=63.1165+2.8419 \times \textit{doctors}^2\)
 

\(\text{Find}\ \textit{life}\ \text{when} \ \textit{doctors}=2:\)

\(\textit{life}=63.1165+2.8419 \times 2^2=74.48\ldots \ \text{years}\)

\(\Rightarrow C\)

Data Analysis, GEN1 2025 VCAA 7-8 MC

The data in the table below shows the preferred car colour for a sample of female and male car buyers.
 

Part 1

From this table, the percentage of female car buyers whose preferred car colour is silver is closest to

  1. 33%
  2. 40%
  3. 42%
  4. 57%

 
Part 2

Which one of the following is the most appropriate way to graphically display the data shown in the table above?

  1. a histogram
  2. a back-to-back stem plot
  3. parallel boxplots
  4. a segmented bar chart
Show Answers Only

\(\text{Part 1:}\ B\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\ \)

\(\text{Total females}\ =28+42+35=105\)

\(\text{Females who prefer silver}\ = 42\)

\(\text{Percentage}\ = \dfrac{42}{105}=0.40\)

\(\Rightarrow B\)
 

\(\text{Part 2}\)

\(\text{Most appropriate graphical display is a segmented bar chart.}\)

\(\text{Each bar = a gender with colours represented by segments in each bar.}\)

\(\Rightarrow D\)

Data Analysis, GEN1 2025 VCAA 6 MC

The following information relating to life expectancy comes from a sample of nations in the Oceania region.

  • The first quartile is 74.9 years.
  • The third quartile is 78.5 years.
  • The lowest five values recorded are 68.5, 68.6, 69.0, 70.1 and 74.8 years.

How many outliers would be displayed at the lower end of a boxplot showing this sample of Oceania data?

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

\(C\)

Show Worked Solution

\(IQR=78.5-74.9=3.6\)

\(\text{Lower fence}\ = 74.9-1.5 \times 3.6=69.5\)

\(\text{68.5, 68.6 and 69.0 are all less than the lower fence.}\)

\(\Rightarrow C\)

Data Analysis, GEN1 2025 VCAA 5 MC

The heights of females aged between 16 and 18 years within a population are normally distributed.

Analysis of the heights of this group of females showed that:

  • 2.5% of the heights were greater than 178.9 cm
  • 16% of the heights were less than 157.6 cm.

Using the 68-95-99.7% rule, the mean and standard deviation of the heights of these females are respectively

  1. 150.5 and 21.3
  2. 154.9 and 7.1
  3. 164.7 and 7.1
  4. 171.8 and 14.2
Show Answers Only

\(C\)

Show Worked Solution
\(\bar{x}+2z\) \(=178.9\ \ …\ (1)\)  
\(\bar{x}-z\) \(=157.6\ …\ (2)\)  

 
\(\text{Subtract}\ (1)-(2):\)

\(3z=21.3\ \ \Rightarrow\ \ z=7.1\)

\(\text{Substitute}\ \ z=7.1\ \ \text{into (2):}\)

\(\bar{x}-7.1=157.6\ \ \Rightarrow\ \ \bar{x}=164.7\)

\(\Rightarrow C\)

Data Analysis, GEN1 2025 VCAA 3 MC

The boxplots below show the life expectancy, in years, for a sample of countries from two different continents (Sample H and Sample T).
 

Which one of the following statements is correct?

  1. The interquartile range for Sample T is greater than the interquartile range for Sample H.
  2. The median for Sample T is more than 10 years greater than the median for Sample H.
  3. The third quartile for Sample H is greater than the first quartile for Sample T.
  4. Life expectancy in all Sample T countries exceeds the median life expectancy in Sample H.
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Options A to C are all demonstrably incorrect.}\)

\(\text{Consider option D:}\)

\(\text{Sample T’s low is higher than Sample H’s median.}\ \checkmark \)

\(\Rightarrow D\)

Data Analysis, GEN1 2025 VCAA 1-2 MC

At a fruit shop, customers can buy avocados in bags.

The bag size ranges from one to six avocados per bag.

The histogram below shows the number of customers who bought each bag size on a particular day.
 

Part 1

The median bag size bought by customers on the day was

  1. 2
  2. 3
  3. 3.5
  4. 4


Part 2

The total number of avocados sold in bags was

  1. 39
  2. 134
  3. 136
  4. 138
Show Answers Only

\(\text{Part 1:}\ D\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\)

\(\text{Sum of columns}\ = 4+11+3+9+5+7=39\)

\(\text{Median = 20th value which is in the 4th column (bag size 4).}\)

\(\Rightarrow D\)
 

\(\text{Part 2}\)

\(\text{Total number of avocados}\)

\(=1 \times 4 + 2 \times 11+3 \times 3+4 \times 9+5 \times 5+6 \times 7 =138\)

\(\Rightarrow D\)

Calculus, 2ADV C1 EQ-Bank 12

A block of ice is melting. The mass \(M\) kilograms of the ice block remaining at time \(t\) hours after it begins to melt is given by  \(M(t)=50(12-3t)^2, 0 \leqslant t \leqslant 4\).

  1. Find the rate of change of the ice block's mass at any time \(t\).   (1 mark)

  2. How long does it take for the ice block to completely melt?   (1 mark)

  3. At what time is the ice melting at a rate of 2100 kilograms per hour?   (2 marks)

Show Answers Only

a.    \(\dfrac{dM}{dt}=-300(12-3t)\)

b.    \(4\ \text{hours}\)

c.    \(t=\dfrac{5}{3}\ \text{hours}\)

Show Worked Solution

a.    \(M(t)=50(12-3t)^2\)

\(\dfrac{dM}{dt}=50 \times 2 \times (-3) \times(12-3t)=-300(12-3t)\)
 

b.    \(\text{Find}\ t\ \text{when}\ \ M(t)=0:\)

\(50(12-3t)^2=0 \ \Rightarrow \ t=4\)

\(\text{Ice block is completely melted at} \ \ t=4 \ \ \text {hours}\)
 

c.    \(\text{Find}\ t \ \text{when}\ \ \dfrac{d M}{d t}=-2100:\)

\(-300(12-3t)\) \(=-2100\)
\(12-3t\) \(=7\)
\(-3t\) \(=-5\)
\(t\) \(=\dfrac{5}{3}\ \text{hours}\)

Calculus, 2ADV C1 EQ-Bank 11

An oil slick on the surface of water forms a circular shape. The radius \(r\) metres of the oil slick is increasing according to the formula  \(r(t)=3 \sqrt{t}\), where \(t\) is the time in minutes after the oil begins to spread,  \(t \geqslant 0\).

  1. Find the rate at which the radius is increasing at any time \(t\).   (1 mark)

  2. At what rate is the area of the oil slick increasing when \(t=16\) minutes?   (1 mark)

Show Answers Only

a.    \(\dfrac{d r}{d t}=\dfrac{3}{2 \sqrt{t}}\)

b.    \(\dfrac{d r}{d t}=\dfrac{3}{8} \ \text{metres/min}\)

Show Worked Solution

a.    \(r(t)=3 \sqrt{t}\)

\(\dfrac{d r}{d t}=\dfrac{1}{2} \times 3 \times t^{-\tfrac{1}{2}}=\dfrac{3}{2 \sqrt{t}}\)
 

b.    \(\text{Find} \ \dfrac{d r}{d t} \ \text{when} \ \ t=16:\)

\(\dfrac{d r}{d t}=\dfrac{3}{2 \times \sqrt{16}}=\dfrac{3}{8} \ \text{metres/min}\)

Calculus, 2ADV C1 EQ-Bank 13

A cylindrical water tank is being filled. The volume \(V\) litres of water in the tank at time \(t\) minutes after filling begins is given by  \(V(t)=500 \sqrt{(2 t+1)}, t \geqslant 0\).

  1. At what rate is water entering the tank at any time \(t\) ?   (1 mark)

  2. Find the rate at which the tank is being filled when \(t=12\) minutes.   (1 mark)

  3. At what time is the water flowing into the tank at a rate of 125 litres per minute?   (2 marks)

Show Answers Only

a.    \(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{2t+1}} \)

b.    \(100\ \text{L/min} \)

c.    \(\dfrac{15}{2}\ \text{mins}\)

Show Worked Solution

a.    \(V(t)=500(2t+1)^{\frac{1}{2}}\)

\(\dfrac{dV}{dt}=\dfrac{1}{2} \times 2 \times 500(2t+1)^{-\frac{1}{2}} = \dfrac{500}{\sqrt{2t+1}} \)
 

b.    \(\text{When}\ \ t=12:\)

\(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{25}} = 100\ \text{L/min} \)
 

c.    \(\text{Find}\ t\ \text{when}\ \dfrac{dV}{dt}=125:\)

\(125\) \(=\dfrac{500}{\sqrt{2t+1}}\)  
\(\sqrt{2t+1}\) \(=4\)  
\(2t+1\) \(=16\)  
\(t\) \(=\dfrac{15}{2}\ \text{mins}\)  

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.
  

 

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

  2. 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)

  3. 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)

Show Answers Only

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.    \(\text{Base wage}=650\)

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

\(\text{New commission rate}=4.2\%\)

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

Financial Maths, STD2 EQ-Bank 21

Liam works in a factory assembling electronic components and is paid on a piecework basis. A spreadsheet is used to calculate his weekly earnings.

Weekly earnings = Number of units completed \(\times\) Rate per unit

A spreadsheet showing Liam's earnings for one week is shown.
  

  1. Write down the formula used in cell B8, using appropriate grid references.   (1 mark)

  2. In the following week, Liam earned $2036.25. The rate per unit remained at $3.75. Calculate the number of units Liam completed that week.   (2 marks)

Show Answers Only

a.   \(=\text{B4}^*\text{B5}\)

b.    \(\text{Number of units completed}=543\)

Show Worked Solution

a.   \(\text{Weekly earnings} = \text{Number of units completed} \times \text{Rate per unit}\)

\(\text{Formula:}\ =\text{B4}^*\text{B5}\)
 

b.   \(\text{Weekly earnings} = \text{Number of units completed} \times \text{Rate per unit}\)

\(\text{Let the Weekly earnings}=E\)

\(2036.25\) \(=E\times 3.75 \)
\(E\) \(=\dfrac{2036.25}{3.75}=543\)

Financial Maths, STD2 EQ-Bank 26

Maria works as a freelance writer and earns income through royalties. A spreadsheet is used to calculate her monthly royalty earnings.

Royalties = Number of books sold \( \times \) Royalty rate per book

A spreadsheet showing Maria's royalty earnings is shown.
  

  1. Write down the formula used in cell B8, using appropriate grid references.   (1 mark)

  2. In April, Maria's total royalty earnings were $8960. Her royalty rate remained at $2.85 per book. How many books did Maria sell in April?   (2 marks)

Show Answers Only

a.   \(=\text{B4}^*\text{B5}\)

b.    \(\text{Number of books}=3144\)

Show Worked Solution

a.   \(\text{Total royalty earnings} = \text{Number of books sold} \times \text{Royalty rate per book}\)

\(\text{Formula:}\ =\text{B4}^*\text{B5}\)
 

b.   \(\text{Total royalty earnings} = \text{Number of books sold} \times \text{Royalty rate per book}\)

\(\text{Let the Number of books sold}=N\)

\(8960.40\) \(=N\times 2.85 \)
\(N\) \(=\dfrac{8960.40}{2.85}=3144\)

Measurement, STD2 EQ-Bank 23

2UG-2005-25b

Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle.   (1 mark)

Show Answers Only

`ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2:`

`a^2 + b^2= 5^2 + 12^2= 169= 13^2= c^2`

Show Worked Solution

`ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2:`

`a^2 + b^2= 5^2 + 12^2= 169= 13^2= c^2`

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.

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

  2. Write down the formula used in cell B9 to calculate the weekly wage, using appropriate grid references.   (2 marks)
  3. Verify the amount of Chen's total holiday pay using calculations.   (2 marks)
Show Answers Only

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

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

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. 

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

  2. 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)

Show Answers Only

a.    \(=\text{B3}/0.015\)

b.    \(0.165\)

Show Worked Solution

a.    \(=\text{B3}/0.015\)

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

  
\(\text{Ryan entered 0.165 into cell B3.}\)

Financial Maths, STD1 EQ-Bank 24

A company uses the spreadsheet below to calculate the fortnightly pay, after tax, of its employees.
 

  1. Write down the formula that was used in cell D5, using appropriate grid references.   (1 mark)
  2. Hence, calculate Kim's fortnightly pay.   (2 marks)
Show Answers Only

a.  \(=\text{B6}-\text{C6}\)

b.    \(\$3140.85\)

Show Worked Solution

a.    \(\text{Formula for Kim’s Salary after tax:}\)

\(=\text{B5}-\text{C5}\)
 

b.    \(\text{Kim’s salary after tax}\ = \$81\,662\)

\(\text{Kim’s fortnightly pay}\ = \dfrac{81\,662}{26}=\$3140.85\)

Financial Maths, STD2 EQ-Bank 17

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

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

A company's spreadsheet was created that calculates its employees' after tax fortnightly pay, based on the table and excluding the Medicare levy.

  1. Write down the formula that was used in cell D6, using appropriate grid references.   (1 mark)
  2. Write down the formula that was used in cell C5, using appropriate grid references.   (1 mark)
  3. Hence, calculate Greg's fortnightly pay.   (2 marks)
Show Answers Only

a.  \(=\text{B6}-\text{C6}\)

b.  \(=4288+(\text{B5}-45000)^*0.30 \)

c.    \(\$3140.85\)

Show Worked Solution

a.    \(\text{Formula for Ian’s Salary after tax:}\)

\(=\text{B6}-\text{C6}\)
 

b.    \(\text{Formula for Greg’s estimated tax:}\)

\(=4288+(\text{B5}-45000)^*0.30 \)
 

c.    \(\text{Greg’s estimated tax}\ =4288+(103\,500-45\,000) \times 0.30=\$21\,838\)

\(\text{Greg’s salary after tax}\ = 103\,500-21\,838=\$81\,662\)

\(\text{Greg’s fornightly pay}\ = \dfrac{81\,662}{26}=\$3140.85\)

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 \(\text{BAC}\).
 

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

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

Show Answers Only

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

Algebra, STD2 EQ-Bank 25

A fitness app calculates daily calorie requirements using the formula below.

Daily calories = Basal metabolic rate + Calorie burn rate per hour \( \times \) Hours of activity

The spreadsheet below has been used to calculate Jamal's daily calorie requirements when he has had 6 hours of activity.
  

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

  2. Jamal increases his hours of activity to 8 hours per day, while his basal metabolic rate and calorie burn rate remain the same.

    What will be Jamal's new daily calorie requirement?   (1 mark)

Show Answers Only

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)

b.    \(2770\ \text{calories}\)

Show Worked Solution

a.   \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)

b.    \(\text{Daily calories}=1650+140\times 8=2770\)

\(\therefore\ \text{Jamal’s new daily calorie requirement is}\ 2770.\)