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CHEMISTRY, M2 EQ-Bank 3 MC

The following equation represents the reaction of calcium disilicide \(\ce{(CaSi2)}\) with antimony trichloride \(\ce{(SbCl3)}\) to produce calcium chloride \(\ce{(CaCl2)}\), silicon \(\ce{(Si)}\), and antimony \(\ce{(Sb)}\):

\(\ce{a CaSi2 + b SbCl3 -> c Si + d Sb + e CaCl2}\)

What are the stoichiometric values for \( a, b, c, d\) and \( e\) in the balanced equation?

  1. \(2, 3, 6, 3, 2\)
  2. \(3, 2, 2, 6, 3\)
  3. \(3, 3, 6, 2, 3\)
  4. \(3, 2, 6, 2, 3\)
Show Answers Only

\(D\)

Show Worked Solution

→ Balancing the chlorine atoms: 

\(\ce{a CaSi2 + 2SbCl3 -> c Si + d Sb +  3CaCl2}\)

→ Since \(\ce{Si}\) and \(\ce{Sb}\) are by themselves on the right hand side, they can be easily balanced last.

→ Next, balance the \(\ce{Ca}\) atoms on the left hand side:

\(\ce{3CaSi2 + 2SbCl3 -> c Si + d Sb +  3CaCl2}\)

→ The values for \(c\) and \(d\) can be added to balance the remaining atoms.

\(\ce{3CaSi2 + 2SbCl3 -> 6 Si + 2 Sb +  3CaCl2}\)

\(\Rightarrow D\)

Calculus, EXT1 C2 2024 HSC 13b

  1. Show that  \(\cos ^4 x+\sin ^4 x=\dfrac{1+\cos ^2 2 x}{2}\).   (2 marks)

  2. Hence, or otherwise, evaluate  \(\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x\).  (3 marks)

Show Answers Only

i.     \(\text{LHS}\) \(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
    \(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
    \(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
    \(=\dfrac{1+\cos ^2(2 x)}{2}\)

  
ii.   \(\dfrac{3 \pi}{16}\)

Show Worked Solution

i.     \(\text{LHS}\) \(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
    \(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
    \(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
    \(=\dfrac{1+\cos ^2(2 x)}{2}\)

  

ii.     \(\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x\)
    \(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\cos ^2(2 x) d x\)
    \(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\dfrac{1}{2}(1+\cos (4 x)) d x\)
    \(=\dfrac{1}{2}\left[\dfrac{3}{2}x +\dfrac{1}{8} \sin (4 x)\right]_0^{\frac{\pi}{4}}\)
    \(=\dfrac{1}{2}\left[\dfrac{3}{2} \times \dfrac{\pi}{4}+\dfrac{1}{8} \sin \pi-0\right]\)
    \(=\dfrac{3 \pi}{16}\)

Calculus, EXT1 C3 2024 HSC 13a

In an experiment, the population of insects, \(P(t)\), was modelled by the logistic differential equation

\(\dfrac{d P}{d t}=P(2000-P)\)

where \(t\) is the time in days after the beginning of the experiment.

The diagram shows a direction field for this differential equation, with the point \(S\) representing the initial population.
 

  1. Explain why the graph of the solution that passes through the point \(S\) cannot also pass through the point \(T\).   (1 mark)

  2. Clearly sketch the graph of the solution that passes through the point \(S\).   (1 mark)

  3. Find the predicted value of the population, \(P(t)\), at which the rate of growth of the population is largest.   (2 marks)

Show Answers Only

 i.    \(\text{Any solution that passes through}\ S\ \text{will follow the}\)

\(\text{slope field (not crossing any lines) and approach a horizontal}\)

\(\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).\)

ii.    
       

iii.  \(P= 1000\)

Show Worked Solution

 i.    \(\text{Any solution that passes through}\ S\ \text{will follow the}\)

\(\text{slope field (not crossing any lines) and approach a horizontal}\)

\(\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).\)

ii.    
       

iii.  \(\dfrac{d P}{d t}=P(2000-P)\)

\(\text {Find \(P\) where  \(\dfrac{d P}{d t}\)  is a maximum.}\)

\(\text{Consider the graph}\ \ y=P(2000-P): \)

\(\Rightarrow \ \text {Graph is a concave down quadratic cutting at}\ \ P=0\ \ \text{and}\ \ P=2000\)

\(\Rightarrow \ \text{Max value of}\ \ P(2000-P)\ \ \Big(\text{i.e.}\ \dfrac{dP}{dt}\Big)\ \ \text{occurs at}\ \ P=1000\ \text{(axis).}\)

♦ Mean mark (iii) 51%.

Functions, EXT1 F1 2024 HSC 12e

The diagram shows the graph of  \(y=\dfrac{1}{\abs{x-5}}\).
 

For what values of \(x\) is  \(\dfrac{x}{6} \geq\dfrac{1}{\abs{x-5}}\) ?   (3 marks)

Show Answers Only

\(x \in[2,3] \cup[6, \infty)\)

Show Worked Solution

\(\dfrac{x}{6} \geqslant \dfrac{1}{|x-5|}\)

\(x|x-5| \geqslant 6\)

\(\text{Case  1:}\)

\(x(x-5) \geqslant 6\)

\(x^2-5 x-6 \geqslant 0\)

\((x-6)(x+1) \geqslant 0\)

\(x \leqslant-1\ \ \text{or}\ \ x \geqslant 6\)

\(\text {By inspection of graph} \ \Rightarrow \ x \leqslant -1\ \text{is not a solution}\)

\(\Rightarrow x \geqslant 6\)

Mean mark 55%.

\(\text {Case 2: }\)

\(-x(x-5) \geqslant 6\)

\(-x^2+5 x-6 \geqslant 0\)

\(x^2-5 x+6 \leqslant 0\)

\((x-3)(x-2) \leqslant 0\)

\(\Rightarrow 2 \leqslant x \leqslant 3\)

\(\therefore x \in[2,3] \cup[6, \infty)\)

Statistics, EXT1 S1 2024 HSC 12c

A charity employs a worker to collect donations. There is a 0.31 chance that when the charity worker talks to someone a donation is made to the charity.

Each day the charity worker must talk to exactly 100 people.

Use a standard normal distribution table to approximate the probability that, on a particular day, at least 35% of the people talked to made a donation.   (3 marks)

Show Answers Only

\(P(\hat{p} \geqslant 0.35) =0.1922\)

Show Worked Solution

\(E(\hat{p})=p=0.31, n=100\)

\(\sigma^2=\dfrac{p(1-p)}{n}=\dfrac{0.31 \times 0.69}{100}=0.002139\)

\(\hat{p} \sim N\left(\mu, \sigma^2\right) \sim N(0.31,0.002139)\)

  \(P(\hat{p} \geqslant 0.35)\) \(=P\left(Z \geqslant \dfrac{x-\mu}{\sigma}\right)\)
    \(=P\left(Z \geqslant \dfrac{0.35-0.31}{\sqrt{0.002139}}\right)\)
    \(=P(Z \geqslant 0.87)\)
    \(=1-P(Z<0.87)\)
    \(=1-0.8078 \text { (from table)}\) 
    \(=0.1922\)

CHEMISTRY, M2 EQ-Bank 2

In an experiment, calcium carbonate \(\ce{(CaCO3)}\) is heated strongly to produce calcium oxide \(\ce{(CaO)}\) and carbon dioxide according to the reaction below:

\(\ce{CaCO3(s) -> CaO(s) + CO2(g)}\)

A student starts with 50.0 g of calcium carbonate. After heating, they collect 28.0 g of calcium oxide.

  1. Using the law of conservation of mass, calculate the mass of carbon dioxide gas produced in this reaction.   (2 marks)
  1. Explain how the law of conservation of mass applies to this reaction.   (2 marks)
Show Answers Only

a.    \(22.0\ \text{g}\)

b.    The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

Show Worked Solution

a.    Law of conservation of mass:

→ The total mass of reactants must equal the total mass of products.

\(m\ce{(CO2)}\) \(=m\ce{(CaCO3)}-m\ce{(CaO)}\)  
  \(= 50.0-28.0\)  
  \(=22.0\ \text{g}\)  

 

b.    Application of law to reaction:

→ The mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

CHEMISTRY, M2 EQ-Bank 1

Balance the following chemical equations:

  1. \(\ce{C2H6(l) + O2(g) -> H2O(l) + CO2(g)}\)   (1 mark)
  2. \(\ce{H3PO4(aq) + CuCO3(aq) -> Cu3(PO4)2(aq) + CO2(g) + H2O(l)}\)   (1 mark)
Show Answers Only

a.    \(\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}\)

b.    \(\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}\)

Show Worked Solution

a.    \(\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}\)

b.    \(\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}\)

CHEMISTRY, M2 EQ-Bank 10 MC

When measuring very low concentrations of pollutants in water, which unit is most suitable?

  1. Molarity (mol L\(^{-1}\))
  2. Parts per million (ppm)
  3. Percentage by mass
  4. Volume percent (% v/v)
Show Answers Only

\(B\)

Show Worked Solution

→ Parts per million (ppm) is a unit used to express very low concentrations. One ppm means that there is 1 part of solute per 1,000,000 parts of solution. This unit is ideal for detecting small amounts of pollutants in large volumes of water.

→ Molarity is a common unit for concentration but is better suited for more moderate to high concentrations in laboratory and industrial contexts, rather than for trace pollutants in water.

→ Percentage by mass and volume percent (% v/v) are typically used for relatively larger concentrations and are not as effective for expressing concentrations as low as parts per million.

\(\Rightarrow B\)

CHEMISTRY, M2 EQ-Bank 11

A student is investigating the concentration of copper ions in a water sample collected from a local river. They use an instrument to determine that the sample contains copper ions at a concentration level of 1.75 ppm.

  1. Calculate the mass of copper ions in a 2 L sample of water.   (2 marks)
  1. Explain why parts per million is a suitable unit for measuring low concentrations of ions in environmental samples like river water.   (2 marks)
Show Answers Only

a.    3.5 mg

b.    Benefits of using ppm as a concentration measurement.

→ Parts per million (ppm) is suitable for measuring low concentrations of ions because it represents the mass of solute per million parts of solution, making it ideal for detecting trace levels of substances.

→ This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.

Show Worked Solution

a.    \(1\ \text{ppm} = 1\ \text{mg/L}\)

\(\Rightarrow \ce{[Cu^{2+}]} = 1.75\ \text{mg/L}\)

\(m\ce{(Cu^{2+})} = 1.75 \times 2 = 3.5\ \text{mg}\)

 

b.    Benefits of using ppm as a concentration measurement.

→ Parts per million (ppm) is suitable for measuring low concentrations of ions because it represents the mass of solute per million parts of solution, making it ideal for detecting trace levels of substances.

→ This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.

Statistics, EXT1 S1 2024 HSC 8 MC

A local council is proposing to ban dog-walking on the beach. It is known that the proportion of households that have a dog is \(\dfrac{7}{12}\).

The local council wishes to poll \(n\) households about this proposal.

Let \(\hat{p}\) be the random variable representing the proportion of households polled that have a dog.

What is the smallest sample size, \(n\), for which the standard deviation of \(\hat{p}\) is less than 0.06?

  1. \(67\)
  2. \(68\)
  3. \(94\)
  4. \(95\)
Show Answers Only

\(B\)

Show Worked Solution

\(E(\hat{p}) = p = \dfrac{7}{12} \)

\(\sigma^{2} = \dfrac{p(1-p)}{n}\)

\(\text{Find}\ n\ \text{such that}\ \ \sigma \lt 0.06 :\)

\(\Bigg( \dfrac{6}{100}\Bigg)^{2} \) \( \gt \dfrac{ \frac{7}{12} \times \frac{5}{12}}{n} \)  
\(n\) \( \gt \dfrac{35}{144} \times \dfrac{100^2}{6^2}\)   
  \( \gt 67.5\)  

 
\(\Rightarrow B\)

CHEMISTRY, M2 EQ-Bank 9 MC

Which of the following statements best describes a primary standard solution?

  1. It is a solution that must be prepared from a substance with a low molar mass to increase accuracy.
  2. It is a solution prepared using a substance with a precisely known and stable concentration, suitable for use in standardising other solutions.
  3. It is a solution prepared from any chemical, as long as the concentration is measured with a volumetric flask.
  4. It is a solution that can vary in concentration over time and requires frequent standardisation.
Show Answers Only

\(B\)

Show Worked Solution

→ Primary standards are used to prepare primary solutions with known and accurate concentrations.

→ A primary standard is a substance that is pure, stable, has a high molar mass (to minimise weighing errors), and is not affected by the atmosphere (such as moisture absorption or \(\ce{CO2}\) interaction).

\(\Rightarrow B\)

Trigonometry, EXT1 T1 2024 HSC 4 MC

What are the domain and range of the function  \(y = 2 \cos^{-1}(2x) + 2 \sin^{-1}(2x)\)?

  1. Domain: \([-0.5, 0.5]\) and Range: \(\{\pi\}\)
  2. Domain: \([-0.5, 0.5]\) and Range: \([-\pi, 3 \pi ]\)
  3. Domain: \([-2, 2]\) and Range: \(\{\pi\}\)
  4. Domain: \([-2, 2]\) and Range: \([-\pi, 3\pi]\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Domain:}\ \ -1 \leqslant 2x \leqslant 1 \ \ \Rightarrow\ \ -\dfrac{1}{2} \leqslant x \leqslant \dfrac{1}{2} \)

\(\text{Range:}\ \ 2\Big(\cos^{-1}(2x)+ \sin^{-1}(2x)\Big) = 2 \times \dfrac{\pi}{2} = \pi\)

\(\Rightarrow A\)

Combinatorics, EXT1 A1 2024 HSC 3 MC

Students from 4 different schools come together to form a choir.

What is the minimum size of the choir to know that there must be at least 20 students in the choir from one of the schools?

  1. \(76\)
  2. \(77\)
  3. \(80\)
  4. \(81\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Pigeonholes}\ (k) = 4\)

\(\text{Let pigeons}\ (n) = x\)

\(\dfrac{n}{k} = \dfrac{x}{4} \gt 19\ \ \Rightarrow \  \gt 76\)

\(x_{min} = 77\)

\(\Rightarrow B\)

Statistics, 2ADV S3 2024 HSC 25

A function \(f(x)\) is defined as

\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
1-\dfrac{x}{h}, & \text { for}\ \ 0 \leq x \leq h, \\
0, & \text { for}\ \  x \gt h \end{array}\right.\)

where \(h\) is a constant.

  1. Find the value of \(h\) such that \(f(x)\) is a probability density function.   (2 marks)

  2. By first finding a formula for the cumulative distribution function, sketch its graph.   (2 marks)

  3. Find the value of the median of the probability density function \(f(x)\) . Give your answer correct to 3 decimal places.   (2 marks)

Show Answers Only

a.   \(h=2\)

b.   

c.   \(\text{Median}\ =0.586\)

Show Worked Solution

a.   \(f(x)\ \text{is a PDF if:}\)

\(\displaystyle \int_{0}^{h} 1-\dfrac{x}{h}\,dx\) \(=1\)  
\(\Big[x-\dfrac{x^2}{2h} \Big]_0^{h}\) \(=1\)  
\(h-\dfrac{h}{2}\) \(=1\)  
\(h\) \(=2\)  

  
b.   \( \displaystyle \int 1-\dfrac{x}{2}\,dx = x-\dfrac{x^2}{4} + c\)

\(\text{At}\ \ x=0, \ \ F(0)=0,\ \ c=0 \)

\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
x-\dfrac{x^2}{4}, & \text { for}\ \ 0 \leq x \leq 2, \\
1, & \text { for}\ \  x \gt 2 \end{array}\right.\)
 

♦♦ Mean mark (b) 27%.

c.   \(\text{Let}\ \ m=\ \text{median of}\ f(x) \)

\(\displaystyle \int_0^{m} 1-\dfrac{x}{2}\,dx\)  \(=0.5\)  
\(\Big[ x-\dfrac{x^2}{4} \Big]_0^m\) \(=0.5\)  
\(m-\dfrac{m^2}{4}\) \(=0.5\)  
\(4m-m^2\) \(=2\)  
\(m^2-4m+2\) \(=0\)  
\((m-2)^2\) \(=2\)  
\(m\) \(=2 \pm \sqrt{2}\)  

 

\(\therefore \ \text{Median}\) \(=2-\sqrt{2}\ \ (x \in [0,2]) \)   
  \(=0.586\ \text{(3 d.p.)}\)  
♦♦ Mean mark (c) 38%.

Algebra, STD1 A3 2024 HSC 23

Carrie is organising a fundraiser.

The cost of hiring the venue and the band is $2500. The cost of providing meals is $50 per person.

  1. Complete the table of values to show the total cost of the fundraiser.   (1 mark)

\begin{array} {|l|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number of people} \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \  & 25 & 50 & 75 & 100 & 125 & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Cost} \rule[-1ex]{0pt}{0pt} & & 3750 & 5000 & 6250 & 7500 & 8750 & 10\,000 \\
\hline
\end{array}

  1. Carrie decides that tickets should be sold at $70 per person. The graph shows the expected revenue at this ticket price. Using the information in part (a), plot the line that shows the cost of the fundraiser.   (2 marks)

     

  1. How many tickets need to be sold for the fundraiser to break even?   (1 mark)

  2. Carrie sold 300 tickets. How much profit did the fundraiser make?   (3 marks)

Show Answers Only

a.    \(\text{Cost when 0 people }= $2500\)

b.


c.    \(100\ \text{tickets}\)

d.    \(\text{Profit }=$3500\)

Show Worked Solution

a.    \(\text{Cost when 0 people }= $2500\)

b.


c.    \(\text{Point of intersection}\ \ \Rightarrow\ \ \text{break-even}\)

\(\therefore\ \text{Break-even when 100 tickets sold.}\)
 

Mean mark (c) 53%.

d.    \(\text{Revenue}\ (R)=70n\ \ (n=\ \text{number of people)}\)

\(\text{Cost}\ (C)=2500 + \Big(\dfrac{1250}{25}\Big)n=2500+50n\)

\(\text{Find profit}\ (P)\ \text{when}\ \ n=300:\)

\(P\) \(=R-C\)
  \(=70\times 300-(2500+50\times300)\)
  \(=21\,000-17\,500\)
  \(=$3500\)
♦ Mean mark (d) 46%.

Measurement, STD1 M5 2024 HSC 29

A floor plan for a living area is shown. All measurements are in millimetres.
 

  1. What is the length and width of the cupboard, in metres?   (1 mark)

  2. The floor of the living area is to be tiled. Tiles will NOT be placed under the cupboard.
  3. Each tile is 0.2 m × 0.5 m. The tiles are supplied in boxes of 15 at a cost of $100 per box. Only full boxes can be purchased.
  4. What is the cost of the tiles for the living area?   (4 marks)
Show Answers Only

a.    \(4\ \text{m}\times 0.5\ \text{m}\)

b.    \($1100\)

Show Worked Solution

a.    \(\text{1 metre = 1000 mm}\)

\(\text{Cupboard}\ = 4\ \text{m}\times 0.5\ \text{m}\)

Mean mark (a) 53%.

b.    \(\text{Method 1}\)

\(\text{Total area to be tiled}\ =6\times 3-4\times 0.5=16\ \text{m}^2\)

\(\text{Area of each tile}\ =0.2\times0.5=0.1\ \text{m}^2\)

\(\text{Tiles needed}\ =\dfrac{16}{0.1}=160\ \text{tiles}\)

\(\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}\)

\(\therefore\ \text{Total cost}\ =100\times 11=$1100\)

   
\(\text{Method 2}\)

\(\text{Tiles to fit 6 m width}\ =\dfrac{6}{0.2}=30 \)

\(\text{Tiles to fit 2 m width}\ =\dfrac{2}{0.2}=10 \)
 

\(\text{Total tiles}\ =30\times 5\ \text{rows}+10\times 1\ \text{rows}\ =160\)

\(\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}\)

\(\therefore\ \text{Total cost}\ =100\times 11=$1100\)

♦♦ Mean mark (b) 35%.

CHEMISTRY, M2 EQ-Bank 9

Sulfuric acid and potassium carbonate undergo neutralisation according to the following equation:

\(\ce{H2SO4(aq) + K2CO3(aq) -> K2SO4(aq) + CO2(g) + H2O(l)}\)

  1. Calculate the mass of potassium carbonate required to prepare 200 mL of a 0.15 mol L\(^{-1}\) solution of potassium carbonate.   (2 marks)
  1. 30.00 mL of the solution in part (a.) was used to neutralise 20.00 mL of sulfuric acid. Calculate the concentration of the acid.   (2 marks)
Show Answers Only

a.    \(4.1\ \text{g}\)

b.    \(0.225\ \text{mol L}^{-1}\)

Show Worked Solution

a.    \(MM\ce{(K2CO3)} = 2(39.10) + 12.01 + 3(16.00) = 138.21\ \text{g mol}^{-1}\)

\(n\ce{(K2CO3)} = c \times V = 0.15 \times 0.2 = 0.03\ \text{mol}\)

\(\Rightarrow m\ce{(K2CO3)} = 138.21 \times 0.03 = 4.1\ \text{g}\)

 

b.    The number of moles of \(\ce{K2CO3}\) used to neutralise:

\(c \times V = 0.15 \times 0.03 = 0.0045\ \text{mol} = n\ce{(H2SO4)}\)

\(\ce{[H2SO4]} = \dfrac{0.0045}{0.02} = 0.225\ \text{mol L}^{-1}\)

CHEMISTRY, M2 EQ-Bank 7 MC

A solution is prepared by dissolving 2.00 g of calcium hydroxide \(\ce{(Ca(OH)2)}\) in enough water to produce a 500.0 mL solution. What is the concentration of hydroxide ions in the final solution?

  1. 0.108 mol L\(^{-1}\)
  2. 0.0540 mol L\(^{-1}\)
  3. 0.0270 mol L\(^{-1}\)
  4. 0.0162 mol L\(^{-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(n\ce{(Ca(OH)2)} = \dfrac{2.00}{40.08 + 2(1.008) + 2(16.00)} = 0.0270\ \text{mol}\)

\(\ce{Ca(OH)2(s) -> Ca^{2+}(aq) + 2OH-(aq)}\)

Using molar ratios:

\(n\ce{(OH^-)} = 0.0270 \times 2 = 0.0540\ \text{mol}\)

\(c\ce{(OH^-)} = \dfrac{n}{V} = \dfrac{0.0540}{0.5} = 0.108\ \text{mol L}^{-1}\)

\(\Rightarrow A\)

CHEMISTRY, M2 EQ-Bank 8

Calculate the concentration of sulfate ions present in a 750.0 mL aqueous solution containing 20.0 g of dissolved iron\(\text{(III)}\) sulfate.   (3 marks)

Show Answers Only

\(0.200\ \text{mol L}^{-1}\)

Show Worked Solution

\(\ce{Fe2(SO4)3(s) -> 2Fe^{2+}(aq) + 3SO4^{2-}(aq)}\)

\(n\ce{(Fe2(SO4)3)} = \dfrac{m}{MM} = \dfrac{20.00}{2(55.85) + 3(32.07) + 12(16.00)} = 0.0500\ \text{mol}\)

Molar ratio: \(\ce{Fe2(SO4)3:SO4^{2-}}\ =\ 1:3\)

\(\Rightarrow n\ce{(SO4^{2-})}= 3 \times 0.0500 = 0.1500\ \text{mol}\)

\(c\ce{(SO4^{2-})} = \dfrac{0.1500}{0.75} = 0.200\ \text{mol L}^{-1}\)

CHEMISTRY, M2 EQ-Bank 6 MC

A chemist is given 200.0 mL of a 1.2 M solution of sodium sulfate and is asked to dilute it to form 50.0 mL of a 0.30 M sodium sulfate solution.

Which of the following options regarding the dilution is correct?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\ & \\
\rule[-1ex]{0pt}{0pt} \ & \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Volume of 1.2 M solution needed } & \textbf{Glassware to make} \\
\textbf{to dilute solution (mL)} \rule[-1ex]{0pt}{0pt}& \textbf{accurately known solution} \\
\hline
\rule{0pt}{2.5ex} 12.5 \rule[-1ex]{0pt}{0pt}& \text{beaker, measuring cylinder} \\
\hline
\rule{0pt}{2.5ex} 75 \rule[-1ex]{0pt}{0pt}& \text{beaker, measuring cylinder} \\
\hline
\rule{0pt}{2.5ex} 12.5 \rule[-1ex]{0pt}{0pt}& \text{volumetric flask, pipette} \\
\hline
\rule{0pt}{2.5ex} 75 \rule[-1ex]{0pt}{0pt}& \text{volumetric flask, pipette} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

→ Using the dilution formula, \(c_1V_1=c_2V_2\)

\(V_1 = \dfrac{c_2V_2}{c_1} = \dfrac{0.3 \times 50}{1.2} = 12.5\ \text{mL}\)

→ To prepare an accurately diluted solution, a volumetric flask and a pipette should be used, as these provide precise measurements. 

\(\Rightarrow C\)

CHEMISTRY, M2 EQ-Bank 7

  1. 4.56 g of potassium chloride \(\ce{KCl}\) is dissolved in 250 mL of water. What is the concentration of this solution?   (2 marks)
  1. How many grams of calcium chloride \(\ce{CaCl2}\) will be needed to make 1.50 L of a 0.250 M solution?   (2 marks)
Show Answers Only

a.    \(0.24\ \text{mol L}^{-1}\)

b.    \(41.6\ \text{g}\)

Show Worked Solution

a.    \(MM\ce{(KCl)}= 39.10 + 35.45 = 74.55\ \text{g mol}^{-1}\)

\(n\ce{(KCl)}= \dfrac{m}{MM} = \dfrac{4.56}{74.55} = 0.061\ \text{mol}\)

\(c\ce{(KCl)}=\dfrac{n}{V}= \dfrac{0.061}{0.25} = 0.24\ \text{mol L}^{-1}\)

 

b.    \(n\ce{(CaCl2)} = c \times V = 0.25 \times 1.5 = 0.375\ \text{mol}\)

\(m\ce{(CaCl2)} = MM \times n = (40.08 + 2(35.45)) \times 0.375 = 41.6\ \text{g}\)

Algebra, STD1 A1 2024 HSC 19

The length \((L)\) of a shark, in centimetres, can be modelled by the formula

\(L = 7.3a + 38\),

where \(a\) is the age of the shark, in years.

At what age should the shark reach a length of 156 cm, correct to the nearest year?   (2 marks)

Show Answers Only

\(16\ \text{years}\)

Show Worked Solution
\(L\) \(=7.3a+38\)
\(156\) \(=7.3a+38\)
\(7.3a\) \(=156-38\)
\(a\) \(=\dfrac{118}{7.3}\)
  \(=16.16\dots\)
  \(=16\ \text{years (nearest year)}\)

Measurement, STD1 M1 2024 HSC 9 MC

An equilateral triangle and an isosceles triangle are shown.

The triangles have the same perimeter.
 

What is the value of \(x\)?

  1. \(8\)
  2. \(9\)
  3. \(11\)
  4. \(12\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Perimeter triangle 1 }=3\times 11=33\ \text{m}\)

\(\text{Perimeter triangle 2 }=2\times x+9=(2x+9)\ \text{m}\)

\(\therefore\ 2x+9\) \(=33\)
\(2x\) \(=24\)
\(x\) \(=12\)

  
\(\Rightarrow D\)

Measurement, STD1 M4 2024 HSC 3 MC

The travel graph shows the distance of a runner from a town.
 

Between what times was the runner travelling at their greatest speed?

  1. 10 am and 11 am
  2. 11 am and 11:30 am
  3. 11:30 am and 1 pm
  4. 1 pm and 2:30 pm
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Greatest speed is when the graph is steepest.}\)

\(\therefore\ \text{Greatest speed is 10 km/hour and occurs between 10 am and 11 am.}\)

\(\Rightarrow A\)

CHEMISTRY, M2 EQ-Bank 6

Iron \(\text{(III)}\) hydroxide can be precipitated from the reaction of iron \(\text{(III)}\) nitrate solution with sodium hydroxide solution.

\(\ce{Fe(NO3)3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaNO3(aq)}\)

  1. Calculate the mass of precipitate formed when 25.0 mL of 0.150 mol L\(^{-1}\) iron \(\text{(III)}\) nitrate solution is added to 40.0 mL of 0.250 mol L\(^{-1}\) sodium hydroxide solution.   (3 marks)
  1. Calculate the concentration of nitrate ions in the final solution.   (2 marks)
Show Answers Only

a.    \(0.356\ \text{g}\)

b.    \(0.173\ \text{mol L}^{-1}\)

Show Worked Solution

a.    \(n\ce{(Fe(NO3)3)} = c \times V = 0.150 \times 0.025 = 0.00375\ \text{mol}\)

\(n\ce{(NaOH)} = 0.25 \times 0.040 = 0.01\ \text{mol}\)

→ The molar ratio of  \(\ce{Fe(NO3)3:NaOH} = 1:3\)

\(3 \times 0.00375 = 0.01125\ \text{mol}\) of \(\ce{NaOH}\) is required to react with 0.00375 mol of \(\ce{Fe(NO3)3}\).

→ Due to there only being 0.01 mol of \(\ce{NaOH}\) present, \(\ce{NaOH}\) will be the limiting reagent. 

\(n\ce{(Fe(OH)3)}\) formed \(=n\ce{(NaOH)} \times \dfrac{1}{3} = 0.01 \times \dfrac{1}{3} = 0.00333\ \text{mol}\)

\(m\ce{(Fe(OH)3)}= n \times MM = 0.00333 \times (55.85 + 3(1.008) + 3(16.00))= 0.356\ \text{g}\)

 

b.    Volume of final solution \(=0.025 + 0.040 = 0.065\ \text{L}\)

\(n\ce{(NO3^-)}= 3 \times n(Fe^{3+}) = 3 \times 0.00375 = 0.01125\ \text{mol}\)

\(c\ce{(NO3^-)} = \dfrac{n}{V} = \dfrac{0.01125}{0.065} = 0.173\ \text{mol L}^{-1}\)

CHEMISTRY, M2 EQ-Bank 3 MC

What volume of water needs to be added to dilute 15.0 mL of a 0.200 mol L\(^{-1}\) solution to a 0.0500 mol L\(^{-1}\) solution?

  1. 45.0 mL
  2. 60.0 mL
  3. 75.0 mL
  4. 90.0 mL
Show Answers Only

\(A\)

Show Worked Solution

→ Use the dilution formula  \(c_1V_1=c_2V_2\), where:

    • \(c_1 = 0.200\ \text{mol L}^{-1}\) (initial concentration)
    • \(V_1 = 15\ \text{mL}\) (initial volume)
    • \(c_2 = 0.0500\ \text{mol L}^{-1}\) (final concentration)

\(V_2 = \dfrac{c_1V_1}{c_2} = \dfrac{0.2 \times 15}{0.05} = 60\ \text{mL}\)

→ Volume of water to be added \(=60-15=45\ \text{mL}\)

\(\Rightarrow A\)

CHEMISTRY, M2 EQ-Bank 5

The compound potassium nitrate has the formula \(\ce{KNO3}\).

  1. A student makes a solution of this compound by dissolving 40.0 g in 250.0 mL of distilled water. Calculate the concentration of this solution in mol L\(^{-1}\).   (2 marks)
  1. The student now requires 500.0 mL of a 5.0% (w/v) solution. What volume of the solution in part (a) is required to make this?   (3 marks)
Show Answers Only

a.    \(1.58\ \text{mol L}^{-1}\)

b.    \(0.156\ \text{L}\)

Show Worked Solution

a.    \(n\ce{(KNO3)}= \dfrac{m}{MM} = \dfrac{40.0}{39.10 + 14.01 + 3(16.00)} =0.396\ \text{mol}\)

Concentration of solution \(=\dfrac{n}{V} = \dfrac{0.396}{0.25} = 1.58\ \text{mol L}^{-1}\)

 

b.    A 5% solution requires 5 grams of solute in 100 mL of solution.

Therefore a 500 mL solution requires 25 grams of solute.

\(n\ce{(KNO3)} = \dfrac{25}{101.11} = 0.247\ \text{mol}\)

\(c\ce{(KNO3)} = \dfrac{n}{V} = \dfrac{0.247}{0.5} = 0.494\ \text{mol L}^{-1}\)

\(c_1V_1\) \(=c_2V_2\)  
\(1.58 \times V_1\) \(=0.494 \times 0.5\)  
\(V_1\) \(=\dfrac{0.247}{1.58}\)  
\(V_1\) \(=0.156\ \text{L}\)  

Calculus, 2ADV C3 2024 HSC 19

Sketch the curve  \(y=x^4-2 x^3+2\)  by first finding all stationary points, checking their nature, and finding the points of inflection.   (5 marks)

Show Answers Only

Show Worked Solution
\(y\) \( = x^4-2 x^3+2\)  
\(y^{\prime}\) \( = 4 x^3-6 x^2 =2 x^2(2 x-3)\)  
\(y^{\prime\prime}\) \( = 12 x^2-12 x =12 x(x-1)\)  

 
\(\text{SP’s when}\ \ y^{′}=0:\)

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

\(\text{At}\ \ x=0, \ y^{\prime \prime} =0\)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -1 & \ \ \ \ 0\ \ \ \  & 1 \\
\hline
\rule{0pt}{2.5ex} y^{′} \rule[-1ex]{0pt}{0pt} & -10 & 0 & -5 \\
\hline
\end{array}

\(\Rightarrow\ \text{Horizontal POI at}\ (0,2)\)
 

\(\text{At}\ \ x=\dfrac{3}{2}:\)

\(\ y^{\prime \prime} =12 \times \dfrac{3}{2}\left(\dfrac{3}{2}-1\right)=9 \gt 0, \ \ y=\Bigg(\dfrac{3}{2}\Bigg)^{4}-2\Bigg(\dfrac{3}{2}\Bigg)^{3}+2=\dfrac{5}{16}\)

\(\Rightarrow \text{MIN at}\ \left(\dfrac{3}{2}, \dfrac{5}{16}\right)\)
 

\(\text{POI when}\ \ y^{″}=0:\)

\(12 x(x-1)=0 \ \Rightarrow\ \  x=1\ \ \text{or}\ \ x=0\ \text{(see above)}\)

\(\text{Test concavity change at}\ \ x=1:\)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \dfrac{1}{2} & \ \ \ \ 1\ \ \ \  &  \dfrac{3}{2} \\
\hline
\rule{0pt}{2.5ex} y^{″} \rule[-1ex]{0pt}{0pt} & \ \ -3\ \  & 0 & \ \ \ 9\ \ \  \\
\hline
\end{array}

\(\Rightarrow\ \text{POI at}\ (1,1)\)
 

Financial Maths, STD1 F1 2024 HSC 27

Zazu works a 38-hour week and is paid at an hourly rate of $45. Any overtime hours worked are paid at time-and-a-half.

In a particular week, Zazu worked the regular 38 hours and some overtime hours. In that week Zazu earned $2790.

How many hours of overtime did Zazu work in that week?   (3 marks)

Show Answers Only

\(\text{16 hours of overtime}\)

Show Worked Solution

\(\text {Let X = overtime hours}\)

  \(\text{Total pay}\) \(=(38 \times 45)+\Big(X \times \dfrac{3}{2}\times 45\Big) \)
  \(2790\) \(=1710+\dfrac{135}{2}X\)
  \(\dfrac{135}{2}X\) \(=1080\)
  \(X\) \(=\dfrac{1080 \times 2}{135}\)
    \(=16 \text{ hours of overtime}\)

Financial Maths, 2ADV M1 2024 HSC 24

Jack intends to deposit $80 into a savings account on the first day of each month for 24 months. The interest rate during this time is 6% per annum, compounded monthly.

  1. Calculate how much money Jack will have in his account at the end of the 24 months.   (3 marks)
  2. A table of future value interest factors could have been used to calculate how much money Jack would have in his account at the end of the 24 months.
  3. Part of the table is shown.
     

  1. What is the value of \(A\) in the table? Give your answer correct to 3 decimal places.   (1 mark)
Show Answers Only

a.   \(A_{24}=\$2044.73\)

b.   \(A=25.559\)

Show Worked Solution

a.   \(r=\dfrac{6\%}{12}=0.5\%\ \text{per month}\)

\(\text{Let}\ \ A_n=\ \text{Amount after}\ n\ \text{months}\)

\(A_1\) \(=80(1.005)\)  
\(A_2\) \(=[80(1.005)+80](1.005)\)  
  \(=80(1.005)^2+80(1.005) \)  
\(\vdots\)    
\(A_n\) \(=80(1.005)^n+80(1.005)^{n-1}+ … + 80(1.005) \)  
  \(=80\underbrace{(1.005+1.005^2+ … + 1.005^n)}_{\text{GP where}\ \ a=1.005, r=1.005} \)  
  \(=80\Bigg(\dfrac{1.005(1.005^{n}-1)}{1.005-1}\Bigg) \)  

 
\(\text{After 24 months:}\)

\(A_{24}\) \(=80\Bigg(\dfrac{1.005(1.005^{24}-1)}{1.005-1}\Bigg) \)  
  \(=$2044.73 \ \text{(nearest cent)}\)  
♦♦ Mean mark (a) 38%.
b.    \(FV\) \(= 80 \times A\)
  \(2044.73\) \(=80 \times A\)
  \(A\) \(=\dfrac{2044.73}{80}\)
    \(=25.5591…\)
    \(=25.559\ \text{(3 d.p.)}\)
Mean mark (b) 51%.

Trigonometry, 2ADV T1 2024 HSC 20

A vertical tower \(T C\) is 40 metres high. The point \(A\) is due east of the base of the tower \(C\). The angle of elevation to the top \(T\) of the tower from \(A\) is 35°. A second point \(B\) is on a different bearing from the tower as shown. The angle of elevation to the top of the tower from \(B\) is 30°. The points \(A\) and \(B\) are 100 metres apart.
 

  1. Show that distance \(A C\) is 57.13 metres, correct to 2 decimal places.  (1 mark)
  1. Find the bearing of \(B\) from \(C\) to the nearest degree.  (3 marks)
Show Answers Only

\(194^{\circ}\)

Show Worked Solution

a.   \(\text{In}\ \Delta TCA:\)

\(\tan 35°\) \( =\dfrac{40}{AC}\)  
\(AC\) \( =\dfrac{40}{\tan 35°}\)  
  \(=57.125…\)  
  \(=57.13\ \text{m (2 d.p.)}\)  

 
b.   \(\text{In}\ \Delta TCB:\)

\(\tan 30°\) \( =\dfrac{40}{BC}\)  
\(BC\) \( =\dfrac{40}{\tan 30°}\)  
  \(=69.28\ \text{m}\)  

  
\( \text{Find} \ \angle BCA \ \text{using cosine rule:}\)

\(\cos \angle B CA\) \( = \dfrac{57.13^2+69.28^2-100^2}{2 \times 57.13 \times 69.28}\)  
  \(= -0.2446 \)…  
\(\angle BCA\) \( = 104.2° \)  

\(\therefore\ \text{Bearing of}\ B\ \text{from}\ C= 90+104=194^{\circ} \text{(nearest degree)} \)

Statistics, 2ADV S2 2024 HSC 16

Flowers were planted in two gardens (Garden A and Garden B).

On a particular day, 25 flowers were randomly selected from each garden and their heights measured in millimetres.

The data are represented in parallel box-plots.
 

Compare the two datasets by examining the skewness of the distributions, and the measures of central tendency and spread.   (3 marks)

Show Answers Only

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely clustered} \)
     \(\text{around the median than A.}\)

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is less than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread of data points of B.}\)

\(\text{Overall, flowers from Garden B will tend to be higher than Garden A flowers as well as exhibiting}\)
\(\text{a larger range of heights.}\)

Show Worked Solution

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely clustered}\)
    \(\text{around the median than A.}\)
 

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is less than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread of data points of B.}\)
 

\(\text{Overall, flowers from Garden B will tend to be higher than Garden A flowers as well as exhibiting}\)
\(\text{a larger range of heights.}\)

L&E, 2ADV E1 2024 HSC 13

The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .

Population \(W\) is modelled by the equation  \(y=A \times(1.055)^x\).

Population \(K\) is modelled by the equation  \(y=B \times(0.97)^x\).
 

Complete the table using the information provided.   (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

Show Answers Only

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

\(\text{Calculations:}\)

\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)

\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)

\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)

Financial Maths, STD2 F4 2024 HSC 25

Alex and Jun each invest $1800 for 5 years.

    • Alex's investment earns simple interest at a rate of 7.5% per annum.
    • Jun's investment earns interest at a rate of 6.0% per annum, compounding quarterly.

By calculating the interest earned over the 5 years, determine who will have the greater amount.   (3 marks)

Show Answers Only

\(\text {Alex’s investment:}\)

\(\text{Interest}=Prn=1800 \times 0.075 \times 5=\$ 675\)
 

\(\text {Jun’s investment:}\)

\(r=\dfrac{6.0\%}{4}=1.5 \% \text { per quarter}\)

\(\text {Compounding periods }=5 \times 4=20\)

\(F V=P V(1+r)^n=1800(1+0.015)^{20}=\$ 2424.34\)

\(\text{Total interest}=F V-P V=2424.34-1800=\$ 624.34\)
 

\(\text {Alex’s interest }>\text { Jun’s interest.}\)

\(\Rightarrow \text{ Alex will have a greater amount (since original investment the same)}\)

Show Worked Solution

\(\text {Alex’s investment:}\)

\(\text{Interest}=Prn=1800 \times 0.075 \times 5=\$ 675\)
 

\(\text {Jun’s investment:}\)

\(r=\dfrac{6.0\%}{4}=1.5 \% \text { per quarter}\)

\(\text {Compounding periods }=5 \times 4=20\)

\(F V=P V(1+r)^n=1800(1+0.015)^{20}=\$ 2424.34\)

\(\text{Total interest}=F V-P V=2424.34-1800=\$ 624.34\)
 

\(\text {Alex’s interest }>\text { Jun’s interest.}\)

\(\Rightarrow \text{ Alex will have a greater amount (since original investment the same)}\)

Algebra, STD2 A4 2024 HSC 22

The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .

Population \(W\) is modelled by the equation  \(y=A \times(1.055)^x\).

Population \(K\) is modelled by the equation  \(y=B \times(0.97)^x\).
 

Complete the table using the information provided.   (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

Show Answers Only

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

\(\text{Calculations:}\)

\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)

\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)

\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)

Mean mark 51%.

Financial Maths, STD2 F4 2024 HSC 21

William has a reducing balance loan on which he owes $5590. He makes monthly repayments of $110.

The loan company charges interest at 24% per annum, compounded monthly.

The spreadsheet shows some of the information for the next two months of the loan.

  1. Complete the entries in the spreadsheet to show the balance owing on the loan at the end of two months.   (2 marks)
     


  1. Explain why the loan will never be repaid if William continues to make repayments of $110 per month.   (1 mark)

Show Answers Only

a.
       

b.   \(\text{The interest charged exceeds the amount that is repaid.}\)

\(\text{Interest charges will gradually increase with the monthly repayment staying the same \(\Rightarrow\) loan will never be repaid.}\)

Show Worked Solution

a.
       

\(\text{Calculations:}\)

\(\text {Cell E2 }= 5590+111.80-110=\$ 5591.80\)

\(\text{Cell B3 }=\text{ Cell E2}\)

\(\text{Cell C3}=5591.80 \times 0.02 = \$111.84 \ \text{(monthly r/i = 2%)}\)

\(\text{Cell E3}=5591.80+111.84-110 = \$5593.64\)
 

b.   \(\text{The interest charged exceeds the amount that is repaid.}\)

\(\text{Interest charges will gradually increase with the monthly repayment staying the same \(\Rightarrow\) loan will never be repaid.}\)

Financial Maths, STD2 F5 2024 HSC 20

The table shows the future value for an annuity of $1 for varying interest rates and time periods.
 

  1. Ken invests $200 at the start of each year for eight years, at an interest rate of 5% per annum.
  2. Calculate the future value of Ken's investment.   (1 mark)

  3. Shay is planning to take a holiday in three years. She needs $4500 for this holiday and will make regular six-monthly payments into an account that earns interest at the rate of 4% per annum, compounded 6 monthly.
  4. What is the minimum amount Shay needs to pay into this account every 6 months? Give your answer to the nearest $10. Support your answer with calculations.  (2 marks)

Show Answers Only

a.    \(F V=\$ 2005.32\)

b.    \(\$700\)

Show Worked Solution

a.    \(\text {8 annual periods at 5% p.a.} \Rightarrow \text { Factor}=10.0266\)

\(F V=200 \times 10.0266=\$ 2005.32\)
 

b.    \(r=\dfrac{4 \%}{2}=2 \%\ \text{per 6 months}\)

\(\text {Compounding periods}=3 \times 2=6\)

\(\Rightarrow \text {Factor }=6.4343\)

\(4500\) \(=\ \text{Annuity} \times 6.4343\)  
\(\text{Annuity}\) \(=\dfrac{4500}{6.4343}\)  
  \(=699.38\)  
  \(=\$700\ \text{(nearest \$10)}\)  
♦ Mean mark (b) 51%.

Networks, STD2 N2 2024 HSC 18

The diagram shows a network with weighted edges.
 

  1. Draw a minimum spanning tree for this network and determine its weight.   (2 marks)
     


  1. Is it possible to find another spanning tree with the same weight? Give a reason for your answer.   (1 mark)

Show Answers Only

a.
         

 
b.    \(\text{Yes.}\)

\(\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC\  \text{to create}\)

\(\text{a second MST (with equivalent weight = 24)}\)

Show Worked Solution

a.
         

b.    \(\text{Yes.}\)

\(\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC\  \text{to create}\)

\(\text{a second MST (with equivalent weight = 24)}\)

♦ Mean mark (b) 45%.

Statistics, STD2 S4 2024 HSC 19

A teacher was exploring the relationship between students' marks for an assignment and their marks for a test. The data for five different students are shown on the graph.

The least-squares regression line is also shown.
 

  1. What is the equation of the least-squares regression line for this dataset?   (2 marks)

  2. Another student, whose marks are not on the graph, scored 5 for the assignment and 12 on the test.
  3. Did this student do better or worse on the test than the regression line predicts? Provide a reason for your answer.   (1 mark)

Show Answers Only

a.    \(\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)\)

\(y \,\text {-intercept}=6\)

\(m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4\)

\(\text{LSRL}\ \ \Rightarrow \ y=1.4x+6\)
 

b.    \(\text {The student’s mark \((5,12)\) sits below the LSRL.}\)

 \(\text {Therefore the student did worse than expected.}\)

Show Worked Solution

a.    \(\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)\)

\(y\,\text {-intercept}=6\)

\(m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4\)

\(\text{LSRL}\ \ \Rightarrow \ y=1.4x+6\)

♦ Mean mark (a) 46%.

b.    \(\text {The student’s mark \((5,12)\) sits below the LSRL.}\)

 \(\text {Therefore the student did worse than expected.}\)

Measurement, STD2 M7 2024 HSC 17

The cost of electricity is 30.13 cents per kWh .

Calculate the cost of using a 650 W air conditioner for 6 hours.   (2 marks)

Show Answers Only

\(\text{Cost} =\$ 1.18\)

Show Worked Solution

\(\text{Usage}=6 \times 650=3900\, \text{W}=3.9\, \text{kW}\)

\(\text{Cost} =3.9 \times 30.13=117.507 \,\text{c}=\$ 1.18 \, \text {(nearest cent)}\)

CHEMISTRY, M2 EQ-Bank 2 MC

What volume of 0.50 M hydrochloric acid  is required to neutralise 25.0 mL of 0.40 M sodium hydroxide?

\(\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}\)

  1. \(15\ \text{mL}\)
  2. \(20\ \text{mL}\)
  3. \(25\ \text{mL}\)
  4. \(30\ \text{mL}\)
Show Answers Only

\(B\)

Show Worked Solution

→ Calculate the number of moles of \(\ce{NaOH}\):

\(\ce{n(NaOH)=c \times V} = 0.4 \times 0.025 = 0.0100\ \text{mol}\)

→ Since the ratio of hydrochloric acid to sodium hydroxide is \(1:1\), the number of moles of hydrochloric acid required is also 0.0100 mol.

→ Calculate the volume of hydrochloric acid required:

\(\ce{V(HCl)= \dfrac{\text{n}}{\text{c}}}= \dfrac{0.0100}{0.5}=0.02\ \text{L}\ =20\ \text{mL}\)

\(\Rightarrow B\)

CHEMISTRY, M2 EQ-Bank 1

A solution is prepared by dissolving 5.00 g of sodium carbonate \(\ce{(Na2CO3)}\) in enough water to make 250.0 mL of solution.

Calculate the concentration of the sodium carbonate solution in mol L\(^{-1}\).   (2 marks)

Show Answers Only

\(0.189\ \text{mol L}^{-1}\)

Show Worked Solution

\(\ce{MM(Na2CO3)}= 105.99\ \text{g mol}^{-1}\)

\(\ce{n(Na2CO3)}= \dfrac{5}{105.99} = 0.0472\ \text{mol}\)
 

Concentration of \(\ce{Na2CO3}\):

\(c= \dfrac{\text{n}}{\text{V}} = \dfrac{0.0472}{0.250}=0.189\ \text{mol L}^{-1}\)

Networks, STD2 N3 2024 HSC 39

A project involving nine activities is shown in the network diagram.

The duration of each activity is not yet known.
 

The following table gives the earliest start time (EST) and latest start time (LST) for three of the activities. All times are in hours.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Activity} \rule[-1ex]{0pt}{0pt} & EST & LST \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ 0\ \ \ \ \ \  & \ \ \ \ \ \ 2\ \ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 0 & 1 \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 12 & 12 \\
\hline
\end{array}

  1. What is the critical path?   (1 mark)
  2. The minimum time required for this project to be completed is 19 hours.
  3. What is the duration of activity \(I\)?   (1 mark)

  4. The duration of activity \(C\) is 3 hours.
  5. What is the maximum amount of time that could occur between the start of activity \(F\) and the end of activity \(H\)?   (1 mark)

Show Answers Only

a.   \(\text{Critical Path:}\ BEGI\)

b.   \(\text{Duration of}\ I =7\ \text{hours}\)

c.  \(\text{Max tim}\ =9\ \text{hours}\)

Show Worked Solution

a.   \(\text{Activity}\ A\ \text{and}\ C: \ LST \gt EST\)

\(\Rightarrow\ \text{Activity}\ A\ \text{and}\ C\ \text{not on critical path.}\)

\(\text{Critical Path:}\ BEGI\)
 

♦ Mean mark (a) 43%.

b.   \(\text{Duration of}\ I = 19-12=7\ \text{hours}\)
 

c.   \(\text{Activity}\ I\ \text{must start at 12 hours (on critical path).}\)

\(\text{Given duration activity}\ C = 3:\)

\(\text{Max time from start of}\ F\ \text{to end of}\ H = 12-3=9\ \text{hours}\)

♦♦♦ Mean mark (c) 10%.

CHEMISTRY, M2 EQ-Bank 11

Iron forms a compound that contains iron (36.8%), sulfur (31.6%), and oxygen (31.6%). The compound boils at 120°C. For one mole of this compound, the density of its vapor at 150°C and 250 kPa is 42.0 g/L.

  1. Determine the empirical formula of the compound.   (2 mark)
  1. Calculate the molar mass of the compound.   (2 marks)
Show Answers Only

a.    \(\ce{Fe2S3O6}\)

b.    \(590.94\ \text{g mol}^{-1}\).

Show Worked Solution

a.   Divide each compound’s percentage by their molar masses:

\(\ce{Fe}: \dfrac{36.8%}{55.85} = 0.65 \ \Rightarrow \ \dfrac{0.659}{0.659}=1\)

\(\ce{S}: \dfrac{31.6%}{32.07} = 0.985 \ \Rightarrow \ \dfrac{0.985}{0.659}=\dfrac{3}{2}\)

\(\ce{O}: \dfrac{31.6%}{16.00} = 1.975 \ \Rightarrow \ \dfrac{1.975}{0.659}=3\)

→ Due to the fraction, each number must be doubled so there are only whole numbers.

→ Thus the empirical formula for the compound is \(\ce{Fe2S3O6}\).

 

b.    Using the Ideal Gas Law to find the volume of the vapour:

\(V=\dfrac{nRT}{P}=\dfrac{1 \times 8.314 \times (150+273)}{250}=14.07\ \text{L}\)

→ Molar mass of the vapour \(= 42 \times 14.07 = 590.94\ \text{g mol}^{-1}\).

CHEMISTRY, M2 EQ-Bank 9 MC

A hot air balloon contains 65 L of air at ground level, where the conditions are 98 kPa and 27°C.

What will be the new volume of the balloon when it rises to an altitude where the conditions are 10 kPa and –10°C?

  1. 400 L
  2. 484 L
  3. 558 L
  4. 610 L
Show Answers Only

\(C\)

Show Worked Solution

→ Using the Combined Gas Law: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)

→ \(T_1 = 300\ \text{K}\)  and  \(T_2 = 263\ \text{K}\)

→ \(V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{98 \times 65 \times 263}{300 \times 10}=558\ \text{L}\)

\(\Rightarrow C\)

Measurement, STD2 M6 2024 HSC 36

The diagram shows two vertical flagpoles, \(BE\) and \(CD\), set on sloping ground.
 

  1. What is the height of the flagpole \(BE\), correct to 1 decimal place?   (2 marks)
  2. What is the height of the flagpole \(CD\), correct to 1 decimal place?   (2 marks)
Show Answers Only

a.   \(BE=25.4\ \text{m}\)

b.   \(CD=19.7\ \text{m}\)

Show Worked Solution

a.   \(\text{Using the sine rule:}\)

\(\dfrac{BE}{\sin 27^{\circ}}\) \(=\dfrac{53.8}{\sin 106^{\circ}}\)  
\(\therefore BE\) \(=\dfrac{53.8 \times \sin 27^{\circ}}{\sin 106^{\circ}}\)  
  \(=25.408…\)  
  \(=25.4\ \text{m (1 d.p.)}\)  

 

b.   \(CD=EB-XB\)

\(\text{Consider}\ \Delta XBC:\)

\(\angle XBC=180-106=74^{\circ}, \ XC=ED=20\)

\(\tan 74^{\circ}\) \(=\dfrac{20}{XB}\)  
\(XB\) \(=\dfrac{20}{\tan 74^{\circ}}\)  
  \(=5.73\ \text{m}\)  

 
\(CD=25.4-5.73=19.7\ \text{m (1 d.p.)}\)

♦♦ Mean mark (b) 35%.

Statistics, 2ADV S3 2024 HSC 23

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable is less than \(z\).

\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} z \rule[-1ex]{0pt}{0pt} & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 \\
\hline
\rule{0pt}{2.5ex} \textit{Probability} \rule[-1ex]{0pt}{0pt} & 0.7257 & 0.7580 & 0.7881 & 0.8159 & 0.8413 & 0.8643 & 0.8849 & 0.9032 & 0.9192 \\
\hline
\end{array}

The probability values given in the table for different values of \(z\) are represented by the shaded area in the following diagram.
 

The scores in a university examination with a large number of candidates are normally distributed with mean 58 and standard deviation 15.

  1. By calculating a \(z\)-score, find the percentage of scores that are between 58 and 70.   (2 marks)
  2. Explain why the percentage of scores between 46 and 70 is twice your answer to part (a).   (1 mark)
  3. By using the values in the table above, find an approximate minimum score that a candidate would need to be placed in the top 10% of the candidates.   (2 marks)
Show Answers Only

a.   \(28.81%\)

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)

c.   \(\text{Approx minimum score = 78%}\)

Show Worked Solution

a.   \(z\text{-score (58)}\ =\dfrac{x-\mu}{\sigma} = \dfrac{58-58}{15}=0\)

\(z\text{-score (70)}\ = \dfrac{70-58}{15}=0.8\)

\(\text{Using table:}\)

\(\text{% between 58–70}\ =0.7881-0.5=0.2881=28.81%\)
 

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)
 

c.   \(z\text{-score 1.3 has a table value 0.9032}\)

\(1-0.9032=0.0968\ \Rightarrow\ \text{i.e. 9.68% of students score higher.}\)

\(\text{Find}\ x\ \text{for a}\ z\text{-score of 1.3:}\)

\(1.3\) \(=\dfrac{x-58}{15}\)  
\(x\) \(=1.3 \times 15 +58\)   
  \(=77.5\)  

 
\(\therefore\ \text{Approx minimum score = 78%}\)

Statistics, STD2 S4 2024 HSC 30

A researcher is studying anacondas (a type of snake).

A dataset recording the age (in years) and length (in cm) of female and male anacondas is displayed on the graph.

Anacondas reach maturity at about 4 years of age.
 

Write THREE observations about anacondas that may be made from the scatterplot. (Note: No calculations are required.)   (3 marks)

Show Answers Only

\(\text{Answers could include three of the following:}\)

\(→\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(→\ \text{Female anacondas grow more quickly than male anacondas from birth until}\)

\(\text{maturity which can be seen by the steeper gradient of the LOBF for each dataset}\)

\(\text{over this period.}\)

\(→\ \text{Female anacondas continue to grow to at least 10 years of age, well past their}\)

\(\text{age of maturity at 4 years of age.}\)

\(→\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their}\)

\(\text{age of maturity at 4 years old.}\)

Show Worked Solution

\(\text{Answers could include three of the following:}\)

\(→\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(→\ \text{Female anacondas grow more quickly than male anacondas from birth until}\)

\(\text{maturity which can be seen by the steeper gradient of the LOBF for each dataset}\)

\(\text{over this period.}\)

\(→\ \text{Female anacondas continue to grow to at least 10 years of age, well past their}\)

\(\text{age of maturity at 4 years of age.}\)

\(→\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their}\)

\(\text{age of maturity at 4 years old.}\)

Financial Maths, STD2 F4 2024 HSC 29

The graph shows the decreasing value of an asset.

For the first 4 years, the value of the asset depreciated by $1500 per year, using a straight-line method of depreciation.

After the end of the 4th year, the method of depreciation changed to the declining-balance method at the rate of 35% per annum.

What is the total depreciation at the end of 10 years?   (4 marks)

Show Answers Only

\(\text{Total depreciation}\ =$46\,681.57\)

Show Worked Solution

\(\text{Depreciation after 4 years}\ = 4 \times 1500 = $6000\)

\(\text{Value after 4 years}\ = 50\,000-6000=44\,000\)

\(\text{Declining balance used for the next 6 years:}\)

\(V_0=$44\,000, r=0.35, n=6\)

\(S\) \(=V_0(1-r)^n\)  
  \(=44\,000(1-0.35)^6\)  
  \(=$3318.43\)  

 
\(\therefore\ \text{Total depreciation}\ =50\,000-3318.43=$46\,681.57\)

Probability, STD2 S2 2024 HSC 12 MC

A survey of 370 people was conducted to investigate the association between watching Anime and the age of the person.

The two-way table shows the responses collected.

Approximately what percentage of the over 30-year-olds watch Anime?

  1. 9%
  2. 18%
  3. 22%
  4. 42%
Show Answers Only

\(C\)

Show Worked Solution

\(\text {Total over } 30=157\)

\(\text {Over 30s who watch anime = 34}\)

\(\text {% over 30s who watch anime}\ =\dfrac{34}{157}=21.7\%\)

\(\Rightarrow C\)

Financial Maths, STD2 F4 2024 HSC 27

A couple borrows $30 000 to be repaid in equal monthly repayments of $280 over 10 years.

After following this repayment plan for 5 years, they decide to decrease their monthly repayment to $250. As a result, it will take them an additional two years to pay off their loan.

How much more will they repay in total by making this change?   (3 marks)

Show Answers Only

\(\text{Extra repaid}\ =$4200\)

Show Worked Solution

\(\text{Original plan:}\)

\(\text{Total repayments}\ =10 \times 12=120\)

\(\text{Amount repaid}\ =120 \times 280=$33\ 600\)
 

\(\text{New plan:}\)

\(\text{Repayments at \$280}\ =5 \times 12=60\)

\(\text{Repayments at \$250}\ =7 \times 12=84\)

\(\text{Amount repaid}\ = 60 \times 280 + 84 \times 250 =$37\ 800\)

\(\text{Extra repaid}\ =37\ 800-33\ 600=$4200\)

Algebra, STD2 A1 2024 HSC 11 MC

A train left Richmond at 6:42 am and arrived at Central Station at 8:04 am. The distance travelled by the train from Richmond to Central was 61 km.

What was the average speed of this train, correct to the nearest km/h ?

  1. 38
  2. 45
  3. 50
  4. 74
Show Answers Only

\(B\)

Show Worked Solution

\(\text {Time of travel:  8:04 less 6:42 = 82 minutes.}\)

  \(\text{Speed (avg)}\) \(=\dfrac{\text{distance}}{\text{time}}\)
    \(=\dfrac{61}{82}\ \ \text{km/min}\)
    \(=\dfrac{61}{82} \times 60\ \ \text{km/hr}\)
    \(=44.6\ \ \text{km/hr}\)

 
\(\Rightarrow B\)

Algebra, STD2 A4 2024 HSC 9 MC

The time taken to paint a school varies inversely with the number of painters completing the task.

It takes 6 painters a total of 20 days to paint a school.

How many days would it take 15 painters to paint the same school?

  1. 4.5
  2. 8
  3. 15.5
  4. 50
Show Answers Only

\(B\)

Show Worked Solution

\(T \propto \dfrac{1}{N} \ \Rightarrow \ \ T=\dfrac{k}{N}\)

\(\text {Find}\ k\ \text{given}\ \  T=20\ \ \text {when}\ \ N=6 \text {:}\)

\(20=\dfrac{k}{6} \ \Rightarrow\ \ k=120\)

\(\therefore\ T=\dfrac{120}{N}\)

\(\text {Find}\ T \ \text {if}\ \ N=15:\)

\(T=\dfrac{120}{15}=8 \text { days }\)

\(\Rightarrow B\)

Trigonometry, 2ADV T3 2024 HSC 28

Anna is sitting in a carriage of a Ferris wheel which is revolving. The height, \(A(t)\), in metres above the ground of the top of her carriage is given by

\(A(t)=c-k\,\cos\Big( \dfrac{\pi t}{24}\Big) \),

where \(t\) is the time in seconds after Anna's carriage first reaches the bottom of its revolution and \(c\) and \(k\) are constants.
 

The top of each carriage reaches a greatest height of 39 metres and a smallest height of 3 metres.

  1. Find the value of \(c\) and \(k\).   (2 marks)
  2. How many seconds does it take for one complete revolution of the Ferris wheel?   (1 mark)
  3. Billie is in another carriage. The height, \(B(t)\), in metres above the ground of the top of her carriage is given by

\(B(t)=c-k\,\cos\Big( \dfrac{\pi}{24}(t-6)\Big) \),

  1. where \(c\) and \(k\) are as found in part (a).
  2. During each revolution, there are two occasions when Anna's and Billie's carriages are at the same heights. At what two heights does this occur? Give your answer correct to 2 decimal places.    (4 marks)
Show Answers Only

a.   \(c=\ \text{centre of motion}\ = \dfrac{39+3}{2} = 21\)

\(k=\ \text{amplitude}\ = \dfrac{39-3}{2}=18\)
 

b.  \(T= 48\ \text{seconds}\)

c.  \(h_1=4.37\ \text{m,}\ h_2=37.63\ \text{m}\)

Show Worked Solution

a.   \(c=\ \text{centre of motion}\ = \dfrac{39+3}{2} = 21\)

\(k=\ \text{amplitude}\ = \dfrac{39-3}{2}=18\)
 

b.  \(A(t)=21-18\,\cos\Big( \dfrac{\pi t}{24}\Big)\ \Rightarrow\ \ n=\dfrac{\pi}{24} \)

\(T=\dfrac{2\pi}{n} = 2\pi \times \dfrac{24}{\pi} = 48\ \text{seconds}\)
 

c.   \(\text{Billie’s carriage is 6 seconds behind Anna’s.}\)

\(\text{Angle between the 2 carriages}\ = \dfrac{\pi \times 6}{24} = \dfrac{\pi}{4} \)

\(\text{By inspection:}\)
 

\(\text{Heights are the same:}\)

\(h_1=21-18\cos(\dfrac{\pi}{8}) = 4.370… = 4.37\ \text{m (2 d.p.)}\)

\(h_2=21-18\cos(\dfrac{7\pi}{8}) = 37.629… = 37.63\ \text{m (2 d.p.)}\)

♦♦ Mean mark (c) 38%.

Calculus, 2ADV C3 2024 HSC 17

In a particular electrical circuit, the voltage \(V\) (volts) across a capacitor is given by

\(V(t)=6.5\left(1-e^{-k t}\right)\),

where \(k\) is a positive constant and \(t\) is the number of seconds after the circuit is switched on.

  1. Draw a sketch of the graph of \(V(t)\), showing its behaviour as \(t\) increases.   (2 marks)
     

  1. When \(t=1\), the voltage across the capacitor is 2.6 volts.
  2. Find the value of \(k\), correct to 3 decimal places.   (2 marks)
  1. Find the rate at which the voltage is increasing when \(t=2\), correct to 3 decimal places.   (2 marks)

Show Answers Only

a.   
       

b.   \(k=0.511\)

c.  \(1.195\ \text{V/s}\)

Show Worked Solution

a.   
       
♦ Mean mark (a) 49%.

b.   \(V=6.5(1-e^{-kt})\)

\(\text{When}\ \ t=1, V=2.6:\)

\(2.6\) \(=6.5(1-e^{-k})\)  
\(1-e^{-k}\) \(=0.4\)  
\(e^{-k}\) \(=0.6\)  
\(-k\) \(=\ln(0.6)\)  
\(k\) \(=0.5108…\)  
  \(=0.511\ \text{(3 d.p.)}\)  

 
c.   \(V=6.5-6.5e^{-kt}\)

\(\dfrac{dV}{dt}=6.5ke^{-kt}\)

\(\text{Find}\ \dfrac{dV}{dt}\ \text{when}\ \ t=2:\)

\(\dfrac{dV}{dt}\) \(=6.5 \times 0.511 \times e^{-2 \times 0.511}\)  
  \(=1.1953…\)  
  \(=1.195\ \text{V/s (3 d.p.)}\)  

Probability, 2ADV S1 2024 HSC 18

In a game, the probability that a particular player scores a goal at each attempt is 0.15.

  1. What is the probability that this player does NOT score a goal in the first two attempts?   (1 mark)

  2. Determine the least number of attempts that this player must make so that the probability of scoring at least one goal is greater than 0.8.   (2 marks)

Show Answers Only

a.   \(0.7225\)

b.   \(n=10\)

Show Worked Solution

a.   \(P(G)=0.15, \ \ P(\bar{G})=0.85\)

\(P(\bar{G}\bar{G}) = 0.85^2=0.7225\)
 

b.   \(\text{2 attempts:}\ P(\text{at least 1 goal})=1-P(\bar{G}\bar{G})=1-0.85^{2}\)

\(\text{3 attempts:}\ P(\text{at least 1 goal})=1-0.85^{3}\)

\(\text{n attempts:}\ P(\text{at least 1 goal})=1-0.85^{n}\)

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

\(1-0.85^{n}\) \(\gt 0.8\)  
\(0.85^{n}\) \(\lt 0.2\)  
\(n \times\ln(0.85)\) \(\lt \ln(0.2)\)  
\(n\) \(\gt \dfrac{\ln(0.2)}{\ln(0.85)}\)  
  \(\gt 9.9…\)  

 
\(\therefore\ \text{Least}\ n=10\)

♦ Mean mark (b) 39%.

CHEMISTRY, M2 EQ-Bank 10

A sample of nitrogen gas at a pressure of 400 kPa and a temperature of 60.0°C occupies a volume of 35.0 L.

  1. Calculate the volume of this nitrogen gas at 298 K and 120 kPa.   (3 marks)
  1. Calculate the number of moles of nitrogen in this sample.   (2 marks)
  1. Calculate the mass of this nitrogen sample.   (2 marks)
Show Answers Only

a.    \(104.4\ \text{L}\)

b.    \(5.06\ \text{mol}\)

c.    \(141.8\ \text{g}\)

Show Worked Solution

a.    Using the Combined Gas Law: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)

\(T_1 = 60^{\circ}\text{C}=333\ \text{K}\)

\(V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{400 \times 35 \times 298}{333 \times 120}=104.4\ \text{L}\)

 

b.    Using the Ideal Gas Law: \(PV=nRT\) 

\(n=\dfrac{PV}{RT}=\dfrac{400 \times 35}{8.314 \times 333}=5.06\ \text{mol}\)

 

c.    Nitrogen gas \(\ce{(N2)}\) has a molar mass of \(28.02\ \text{g mol}^{-1}\)

\(\ce{m(N2)}= n \times MM = 5.06 \times 28.02 = 141.8\ \text{g}\)