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Statistics, STD1 S3 2025 HSC 15

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

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

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

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

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

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

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

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

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

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d.   \(\text{Gradient}=-1\)
 

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

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

Show Worked Solution

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

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

b.    \(\text{Dependent variable:}\)

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


♦♦♦ Mean mark (b) 21%.

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)


♦♦ Mean mark (c) 45%.
d. 

 

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

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


♦♦♦ Mean mark (d) 30%.

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

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


♦♦♦♦ Mean mark (e) 14%.

Measurement, STD1 M4 2025 HSC 12

Vijay’s heart rate before and after his morning run is shown.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \text{Before run} \rule[-1ex]{0pt}{0pt} & \text{72 beats per minute} \\
\hline
\rule{0pt}{2.5ex} \text{After run} \rule[-1ex]{0pt}{0pt} & \text{126 beats per minute} \\
\hline
\end{array}

What is the percentage increase in Vijay’s heart rate?    (2 marks)

Show Answers Only

\(75\%\)

Show Worked Solution
\(\text{% increase}\) \(=\Big[\dfrac{\text{Change in HR}}{\text{Original HR}}\Big]\times 100\%\)
  \(=\Big[\dfrac{126-72}{72}\Big]\times 100\%\)
  \(=75\%\)

♦♦ Mean mark 45%.

Measurement, STD1 M1 2025 HSC 11

Consider the composite shape shown.
 

What is the area of the shape in cm² ?   (3 marks)

Show Answers Only

\(21\ \text{cm}^2\)

Show Worked Solution

\(\text{Height of triangle}=3+2=5\ \text{cm}\)

\(\text{Area}\) \(=\text{Area of triangle}+\text{Area of rectangle}\)
  \(=\dfrac{1}{2}\times 6\times 5+3\times 2\)
  \(=15+6=21\ \text{cm}^2\)
♦♦ Mean mark 30%.

Algebra, STD1 A3 2025 HSC 5 MC

A baker makes and sells cakes.

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

What profit will the baker make by selling 6 cakes?

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

\(A\)

Show Worked Solution

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

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

  
\(\Rightarrow A\)


♦♦ Mean mark 53%.

Financial Maths, STD2 F4 2025 HSC 36

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

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

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

Show Answers Only

\($1476.40\)

Show Worked Solution

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

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

\(S\) \(=V_0(1-r)^n\)
\(44\ 000\) \(=55\ 000(1-r)^1\)
\(\dfrac{44\ 000}{55\ 000}\) \(=1-r\)
\(1-r\) \(=0.8\)
\(r\) \(=1-0.8=0.20\)
♦ Mean mark 47%.

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

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

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

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

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

Algebra, STD2 A4 2025 HSC 30

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

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

 
Write TWO linear equations that represent the relationship below.

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

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

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

Show Worked Solution

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

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

♦♦ Mean mark 35%.

Statistics, STD2 S1 2025 HSC 28

The heights of students in a class were recorded.

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

 

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

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

Show Answers Only

Show Worked Solution

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

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

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

♦ Mean mark 50%.

Measurement, STD2 M1 2025 HSC 26

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

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

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

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

Show Answers Only

a.   \(34.68 \ \text{cm}^2\)

b.   \(582.64 \ \text{cm}^2\)

c.   \(0.490 \%\)

Show Worked Solution

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

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

 

b.    \(\text{Toy has 5 sides.}\)

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

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

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

\(\therefore \ \text{Area of curved surface}\) \(=1300-(408+240+69.36)\)
  \(=582.64 \ \text{cm}^2\)
♦ Mean mark (b) 48%.

c.   \(\text{Convert cm to mm:}\)

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

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

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

♦♦ Mean mark (c) 34%.

Vectors, EXT1 V1 2025 HSC 9 MC

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

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

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

\(B\)

Show Worked Solution

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

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

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

♦ Mean mark 45%.
\(\abs{\underset{\sim}{a}+\underset{\sim}{b}}^2\) \(=(\underset{\sim}{a}+\underset{\sim}{b})(\underset{\sim}{a}+\underset{\sim}{b})=49\)
\(49\) \(=\underset{\sim}{a} \cdot \underset{\sim}{a}+2 a \cdot \underset{\sim}{b}+\underset{\sim}{b} \cdot \underset{\sim}{b}\)
\(49\) \(=\abs{\underset{\sim}{a}}^2+2 a \cdot b+\abs{\underset{\sim}{b}}^2\)
\(49\) \(=9+2 \underset{\sim}{a} \cdot \underset{\sim}{b}+25\)
\(2 \underset{\sim}{a} \cdot \underset{\sim}{b}\) \(=15\)
\(\underset{\sim}{a} \cdot \underset{\sim}{b}\) \(=\dfrac{15}{2}\)

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

 
\(\Rightarrow B\)

Trigonometry, EXT1 T3 2025 HSC 5 MC

How many distinct solutions are there to the equation  \(\cos 5 x+\sin x=0\)  for  \(0 \leq x \leq 2 \pi\) ?

  1. 5
  2. 6
  3. 9
  4. 10
Show Answers Only

\(D\)

Show Worked Solution

\(\cos\, 5 x+\sin\, x=0\ \ \Rightarrow \ \ \cos\,5x=- \sin\,x \)

♦♦ Mean mark 34%.

\(\text{A freehand sketch of both graphs:}\)
 

\(\Rightarrow D\)

Calculus, EXT1 C2 2025 HSC 14d

The function  \(f(x)\)  is defined by  \(f(x)=\cos ^{-1}(\sin x)\)  in the domain \((0, \pi)\).

Find  \(f^{\prime}(x)\)  for those values of \(x\) where it is defined.   (3 marks)

Show Answers Only

\(f^{\prime}(x)=-1 \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)\)

\(f^{\prime}(x)=1 \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)\)

Show Worked Solution
\(f(x)\) \(=\cos ^{-1}(\sin x)\)
\(f^{\prime}(x)\) \(=-\dfrac{\cos x}{\sqrt{1-\sin ^2 x}}\)
  \(=-\dfrac{\cos x}{\abs{\cos x}}\)
  \(= \pm 1\)

♦♦ Mean mark 38%.

\(\text{Consider limitations:}\)

\(1-\sin ^2 x \neq 0 \ \Rightarrow \ \sin x \neq \pm 1 \ \Rightarrow \ x \neq \dfrac{\pi}{2}\)

\(\text{In lst quadrant:} \ \ -\dfrac{\cos x}{\abs{\cos x}}=-1\)

\(\text{In 2nd quadrant:}\ \  -\dfrac{\cos x}{\abs{\cos x}}=1\)

\(f^{\prime}(x)=-1 \ \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)\)

\(f^{\prime}(x)=1 \ \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)\)

Vectors, EXT1 V1 2025 HSC 14b

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

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

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

Show Answers Only

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

Show Worked Solution

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

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

♦♦ Mean mark 34%.

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

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

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

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

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

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

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

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

\(\text{Equating ranges:}\)

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

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

Combinatorics, EXT1 A1 2025 HSC 14a

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

Show Answers Only

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

♦♦ Mean mark 33%.

Combinatorics, EXT1 A1 2025 HSC 13e

  1. The Pascal's triangle relation can be expressed as
    1. \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}.\) (Do NOT prove this.)
  2. Show that \(\displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\).   (1 mark)

  3. Hence, or otherwise, prove that
  4. \(\displaystyle\binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\).   (2 marks)
Show Answers Only

i.    \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)\)

\(\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

\(\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}\)

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)
 

ii.    \(\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\)
 

\(\text{Using part (i)}:\)

\(\operatorname{LHS}\) \(=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots\)
  \(\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]\)
  \(=1-1+\displaystyle \binom{2051}{2001}\)
  \(=\displaystyle \binom{2051}{2001}\)
Show Worked Solution

i.    \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)\)

\(\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

\(\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}\)

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

♦ Mean mark (a) 43%.

ii.    \(\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\)
 

\(\text{Using part (i)}:\)

\(\operatorname{LHS}\) \(=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots\)
  \(\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]\)
  \(=1-1+\displaystyle \binom{2051}{2001}\)
  \(=\displaystyle \binom{2051}{2001}\)
♦♦ Mean mark (b) 37%.

Vectors, EXT1 V1 2025 HSC 8 MC

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

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

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

Which of the following must be true?

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

\(A\)

Show Worked Solution

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

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

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

♦♦ Mean mark 39%.

\(\Rightarrow A\)

Calculus, EXT1 C3 2025 HSC 7 MC

A slope field is shown.
 

Which of the following could be the differential equation represented by the slope field?

  1. \(\dfrac{d y}{d x}=x^2\)
  2. \(\dfrac{d y}{d x}=x^2+C, C \neq 0\)
  3. \(\dfrac{d y}{d x}=x^3\)
  4. \(\dfrac{d y}{d x}=x^3+C, C \neq 0\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{For all \(x<0\), gradients are positive (from graph):}\)

\(\text{Eliminate C and D.}\)

\(\text{At \(x=0\), gradient = 0 (from graph):}\)

\(\text{Eliminate B.}\)

\(\Rightarrow A\)

♦ Mean mark 41%.

Calculus, EXT1 C3 2025 HSC 6 MC

Given that \(a\) is a non-zero constant, which of the following integrals is equal to zero?

  1. \(\displaystyle \int_{-a}^a x\, \cos ^{-1}(x) d x\)
  2. \(\displaystyle\int_{-a}^a x^2\, \cos ^{-1}(x) d x\)
  3. \(\displaystyle\int_{-a}^a x\, \tan ^{-1}(x) d x\)
  4. \(\displaystyle\int_{-a}^a x^2\, \tan ^{-1}(x) d x\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Since the limits are symmetrical about 0:}\)

\(\text{Integral will equal zero if function is odd.}\)

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

\(f(x)=x^2\, \tan^{-1}(x)\)

\(f(-x)=(-x)^2\, \tan^{-1}(-x)=-x^2\, \tan ^{-1}(x)=-f(x)\ \text{(odd)}\)

\(\Rightarrow D\)

♦ Mean mark 47%.

Measurement, STD1 M5 2025 HSC 26

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

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

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

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

Show Answers Only

\($6105.60\)

Show Worked Solution

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

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

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

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

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

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

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

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

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

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


♦♦ Mean mark 37%.

Networks, STD2 N3 2025 HSC 22

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

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

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

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

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

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

Show Answers Only

a.    \(\text{Capacity} =62\)

b.i.   \(x=30\) 

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

b.iii. \(ACEG\)

Show Worked Solution

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

♦ Mean mark (a) 51%.
b.i.    \(\text{Inflow into} \ C\) \(=\text{Outflow from} \ C\)
  \(x\) \(=5+13+12\)
    \(=30\)

 

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

\(\text{Full capacity occurs when a pipe’s capacity (top diagram) equals}\)

\(\text{its flow in the second diagram.}\)
 

b.iii.  \(ACEG\)

\(\text{These pipes are not at full capacity and can increase their flow.}\)

♦ Mean mark (b.i.) 38%.
♦♦♦ Mean mark (b.ii.) 15%.
♦♦♦ Mean mark (b.iii.) 6%.

Statistics, STD2 S4 2025 HSC 25

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

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

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

\(y=64.3-0.7 x\)

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

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

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

Show Answers Only

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

b.    \(\text{Slope}=-0.7\)

This means that for each added minute of watching television per day, a participant, on average, will exercise for 0.7 minutes less.

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

If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.

c.    Jo is expected to exercise for 34.9 minutes.

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

The model predicts a negative value for time spent exercising, which is not possible.

Show Worked Solution

a.    Form: Linear

Direction: Negative

Mean mark (a) 51%.

b.    \(\text{Slope}=-0.7\)

This means that for each added minute of watching television per day, a participant, on average, will exercise for 0.7 minutes less.

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

If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.

♦♦♦ Mean mark (b) 20%.

c.    \(\text{At} \ \ x=42:\)

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

\(\therefore\) Jo is expected to exercise for 34.9 minutes.
 

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

The model predicts a negative value for time spent exercising, which is not possible.

♦ Mean mark (d) 45%.

Networks, STD2 N3 2025 HSC 19

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

 

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

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

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

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

Show Answers Only

a.           

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

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

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

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

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

Show Worked Solution

a.   

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

  
b.   

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

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

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

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

♦ Mean mark (c) 46%.

Measurement, STD2 M6 2025 HSC 14 MC

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

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

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

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

\(C\)

Show Worked Solution

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

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

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

\(\Rightarrow C\)

♦ Mean mark 41%.

Measurement, STD1 M5 2025 HSC 10 MC

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

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

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

\(B\)

Show Worked Solution

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

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

 \(\Rightarrow B\)


♦♦ Mean mark 38%.

Financial Maths, STD2 F1 2025 HSC 4 MC

Frankie takes four weeks of annual leave. His weekly pay is $350 and his annual leave loading is \( 17 \dfrac{1}{2} \)% of four weeks pay.

What is Frankie's total pay for this period of annual leave?

  1. $245.00
  2. $411.25
  3. $1461.25
  4. $1645.00
Show Answers Only

\(D\)

Show Worked Solution

\(\text{4 weeks pay}\ = 4 \times 350 = $1400\)

\(\text{Annual leave loading}\ =17.5\% \times 1400 = $245\)

\(\text{Total pay}\ = 1400+245=$1645\)

\(\Rightarrow D\)

♦ Mean mark 46%.

Networks, STD1 N1 2025 HSC 6 MC

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

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

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

\(C\)

Show Worked Solution

\(\text{Minimum spanning tree:}\)
 

\(\text{Using Kruskal’s algorithm:}\)

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

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

\(\Rightarrow C\)

♦♦ Mean mark 33%.

Calculus, 2ADV C4 2011 HSC 4d v1

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

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

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

 

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

 

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

Functions, 2ADV F2 2025 HSC 30

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

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

Find the value of \(k\).   (3 marks)

Show Answers Only

\(k=6\)

Show Worked Solution

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

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

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

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

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

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

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

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

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

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

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

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

♦ Mean mark 39%.

Trigonometry, 2ADV T1 2025 HSC 29

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

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

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

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

Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.   \(h=3.03 \ \text{km}\)

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

Show Worked Solution

a.    \(\text{In}\ \triangle TOY:\)

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

 

b.   \(\text{In}\ \triangle TOF:\)

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

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

\(\angle OYF =45^{\circ}\)
 

♦♦♦ Mean mark (b) 25%.

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

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

 

\(h\)

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

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

\(\dfrac{\sin\angle FOY}{4}\) \(=\dfrac{\sin 45^{\circ}}{3.03}\)  
\(\sin \angle F O Y\) \(=\dfrac{4 \times \sin 45^{\circ}}{3.03}=0.93347\)  
\(\angle FOY\) \(=180-\sin^{-1}(0.93347)=180-68.98 \ldots = 111^{\circ} \ \text{(angle is obtuse)}\)  
♦♦♦ Mean mark (c) 12%.

\(\angle FO\text{S}^{\prime}=111-90=21^{\circ}\)

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

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

Trigonometry, 2ADV T1 2025 HSC 28

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

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

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

Show Answers Only

a.   \(\text{See Worked Solutions}\)

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

Show Worked Solution

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

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

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

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

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

\(25 \pi\) \(=50 \theta-50\, \sin \theta\)
\(50 \theta\) \(=25 \pi+50 \, \sin \theta\)
\(\theta\) \(=\dfrac{\pi}{2}+\sin \theta\)
♦♦ Mean mark (a) 34%.

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

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

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

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

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

♦♦♦ Mean mark (b) 7%.

Calculus, 2ADV C4 2025 HSC 27

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

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

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

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

Show Answers Only

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

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

 

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

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

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

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

a.  

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

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

Mean mark (b) 51%.

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

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

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

♦♦♦ Mean mark (c) 25%.
\(\dfrac{9}{8}\) \(>\dfrac{3}{4 \ln 2}\)  
\(36\, \ln 2\) \(>24\)  
\(\ln 2\) \(>\dfrac{2}{3}\)  
\(e^{\small \dfrac{2}{3}}\) \(>2\)  
\(e\) \(>2^{\small \dfrac{3}{2}}\)  
\(e\) \(>2 \sqrt{2}\)  

Calculus, 2ADV C3 2025 HSC 26

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

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

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

a.   \(\text{See Worked Solutions}\)

b.   \(\text{See Worked Solutions}\)

Show Worked Solution

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

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

♦ Mean mark (a) 41%.

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

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

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

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

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

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

 

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

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

 \(A(x)\)  \(=\dfrac{1}{4}\left(\sqrt{3} x^2+\dfrac{10\,000}{\pi}-\dfrac{600 x}{\pi}+\dfrac{9 x^2}{\pi}\right)\)
   \(=\dfrac{1}{4}\left(\sqrt{3}+\dfrac{9}{\pi}\right) x^2-\dfrac{150}{\pi} x+\dfrac{2500}{\pi}\)
♦♦♦ Mean mark (b) 28%.

\(\text{Consider limits on} \ x:\)

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

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

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

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

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

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

Calculus, 2ADV C4 2025 HSC 25

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

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

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

a.   \(\text{See Worked Solutions}\)

b.   \(2025 \pi\)

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

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

 

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

  

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

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

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

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

♦♦ Mean mark (c) 32%.

Calculus, 2ADV C3 2025 HSC 24

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

♦ Mean mark 41%.

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

   \(y=e\, \ln e=e\)

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

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

   \(e=\dfrac{1}{2e} \times e^2+c\)

   \(c=\dfrac{1}{2} e\)

Statistics, STD2 S5 2025 HSC 40

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

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

Show Answers Only

a.   \(x=89.8 \ \text{kg}\)

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

\(2\) \(=\dfrac{x_m-76.2}{6.8}\)  
\(x_m\) \(=76.2+2 \times 6.8=89.8 \ \text{kg}\)  
♦♦ Mean mark (a) 35%.

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

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

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

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

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

♦♦♦ Mean mark (b) 6%.

Statistics, 2ADV S3 2025 HSC 23

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

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

Show Answers Only

a.   \(x=89.8 \ \text{kg}\)

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

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

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

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

♦♦ Mean mark (b) 35%.

Statistics, 2ADV S3 2025 HSC 21

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

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

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

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

Show Answers Only

a.    
     

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

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

b.   \(Q_1=1.284 \)

Show Worked Solution

a.    
         

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

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

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

♦ Mean mark (a) 48%.

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

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

Financial Maths, STD2 F5 2025 HSC 34

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

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

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

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

Show Answers Only

\($306.78\)

Show Worked Solution

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

\(\text{Lin’s investment:}\)

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

♦ Mean mark 43%.

\(\text{Yemi’s investment:}\)

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

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

Probability, 2ADV S1 2025 HSC 19

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

The probability that the team wins when:

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

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

Show Answers Only

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

Show Worked Solution

♦ Mean mark 49%.

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

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

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

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

Statistics, 2ADV S2 2025 HSC 14

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

 

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

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

\(y=64.3-0.7 x\)

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

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

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

Show Answers Only

a.    Form: Linear, Direction: Negative
 

b.    \(\text{Slope}=-0.7\)

This means that for each added minute of watching television per day, a participant, on average, will exercise for 0.7 minutes less.

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

If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.
 

c.    \(\text{34.9 minutes}\)
 

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

The model predicts a negative value for time spent exercising, which is not possible.

Show Worked Solution

a.    Form: Linear

Direction: Negative
 

b.    \(\text{Slope}=-0.7\)

This means that for each added minute of watching television per day, a participant, on average, will exercise for 0.7 minutes less.

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

If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.

♦ Mean mark (b) 41%.
 
c.    \(\text{At} \ \ x=42:\)

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

\(\therefore\) Jo is expected to exercise for 34.9 minutes.
 

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

The model predicts a negative value for time spent exercising, which is not possible.

Probability, STD2 S2 2025 HSC 13 MC

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

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

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

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

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

\(A\)

Show Worked Solution

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

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

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

 
\(\Rightarrow A\)

♦ Mean mark 53%.

Calculus, 2ADV C1 2019 HSC 14d v1

The equation of the tangent to the curve  `y = ae^(2x)+bx`  at the point where  `x = 0`  is  `y = 3x +2`.

Find the values of  `a`  and  `b`.  (3 marks)

Show Answers Only

`b = -1,\ \ a = 2`

Show Worked Solution
`y ` `= ae^(2x)+bx`
`(dy)/(dx)` `= 2ae^(2x)+b`

 
`text(When)\ \ x = 0,\ \ (dy)/(dx) = 3`

`2a + b` `= 3\ …\ (1)`

 
`text(The point)\ (0, 2)\ text(lies on)\ y:`

`a(1) +b(0)` `=2`
`a` `= 2\ …\ (2)`

  

`text(Substitute into)\ (1)`

`2(2)+b` `= 3`
`b` `= -1`

Calculus, 2ADV C1 EO-Bank 3 MC

At which point on the curve  \(y = x^{2}-6x + 8\)  can a normal be drawn such that it is inclined at 45\(^{\circ}\) to the positive \(x\)-axis?

  1. \((1,3)\)
  2. \((2,0)\)
  3. \(\left(\dfrac{5}{2}, -\dfrac{3}{4}\right)\)
  4. \((5,-7)\)
Show Answers Only

\(C\)

Show Worked Solution

\(y = x^{2}-6x + 8\)

\(y^{′} = 2x-6\)

If the normal is inclined at \(45^{\circ}\) to the positive x-axis then:

\(m_{\text{normal}} = \tan 45^{\circ} = 1\)

\(\text{Since } m_{\text{tangent}} \times m_{\text{normal}} = -1,\)

\(\therefore m_{\text{tangent}} = -1.\)
 

Find \(x\) when \(y^{′} = -1:\)

\(2x-6\) \(=-1\)  
\(2x\) \(=5\)  
\(x\) \(=\dfrac{5}{2}\)  

 
Find \(y:\)

\(y\) \(= \left(\dfrac{5}{2}\right)^{2}-6\left(\dfrac{5}{2}\right) + 8\)  
  \(=\dfrac{25}{4}-15 + 8\)  
  \(= -\dfrac{3}{4}\)  

 
\(\Rightarrow C\)

Calculus, 2ADV C1 EO-Bank 11 MC v1

Two functions, \(f\) and \(g\), are continuous and differentiable for all  \(x\in R\). It is given that  \(f(-1)=7,\ g(-1)=5\)  and  \(f^{′}(-1)=-4,\ g^{′}(-1)=-2\).

The gradient of the graph  \(y=\dfrac{f(x)}{g(x)}\)  at the point where  \(x=-1\)  is

  1. \(-\dfrac{6}{49}\)
  2. \(\dfrac{6}{49}\)
  3. \(\dfrac{6}{25}\)
  4. \(-\dfrac{6}{25}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Using the Quotient Rule when}\ \ x=-1:\)

\(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)\) \(=\dfrac{g(x)f^{′}(x)-f(x)g^{′}(x)}{g(x)^2}\)
  \(=\dfrac{g(-1)f^{′}(-1)-f(-1)g^{′}(-1)}{g(-1)^2}\)
  \(=\dfrac{5 \times -4-7 \times -2}{5^2}\)
  \(=-\dfrac{6}{25}\)

 
\(\Rightarrow D\)

Calculus, 2ADV C1 2023 HSC 7 MC v1

It is given that  \(y=f(g(x))\), where  \(f(2)=5\), \(f^{′}(2)=3\), \(g(4)=2\)  and  \(g^{′}(4)=-2\).

What is the value of \(y^{′}\) at  \(x=4\)?

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

\(A\)

Show Worked Solution
\(y\) \(=f(g(x))\)  
\(y^{′}\) \(=f^{′}(g(4)) \times g^{′}(4)\)  
  \(=f^{′}(2) \times -2\)  
  \(=3 \times -2\)  
  \(=-6\)  

 
\(\Rightarrow A\)

BIOLOGY, M7 2021 VCE 11

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

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

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

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

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

 

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

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

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

b.    Hypothesis:

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

c.i.  Experiment analysis:

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

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

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

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

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

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

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

b.    Hypothesis:

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

♦♦♦ Mean mark (b) 40%.

c.i.  Experiment analysis:

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

♦♦♦ Mean mark (c)(i) 6%.

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

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

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

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

♦♦ Mean mark (c)(ii) 50%.

CHEMISTRY, M1 EQ-Bank 19

Silicon \(\ce{(Si)}\) is in Group 14 and Period 3 of the periodic table. Sodium \(\ce{(Na)}\) is in Group 1 and Period 3.   (6 marks)

Compare the properties of silicon and sodium with reference to their:

    • Metallic character
    • Electrical conductivity
    • Ion formation
Show Answers Only

Metallic character:

  • Sodium has much greater metallic character than silicon. Sodium is an alkali metal (Group 1) with typical metallic properties such as being shiny, malleable, and ductile.
  • Silicon is a metalloid with properties intermediate between metals and non-metals.

Electrical conductivity:

  • Sodium is an excellent electrical conductor, while silicon is a semiconductor. As a Group 1 metal, sodium has one valence electron that is delocalised in a metallic lattice, allowing it to move freely and conduct electricity very effectively.
  • Silicon, as a metalloid, has moderate electrical conductivity that increases with temperature – its conductivity is much lower than sodium’s but higher than non-metals.

Ion formation:

  • Sodium readily forms positive ions (cations) with a 1+ charge (\(\ce{Na^+}\)) by losing its single valence electron to achieve a stable electron configuration similar to neon. Silicon typically does not form simple ions due to its position as a metalloid.
  • Instead, silicon forms covalent bonds by sharing its four valence electrons with other atoms. While silicon can theoretically form \(\ce{Si^4+}\) or \(\ce{Si^4-}\) ions, the energy required to remove or add four electrons is too high.
Show Worked Solution

Metallic character:

  • Sodium has much greater metallic character than silicon. Sodium is an alkali metal (Group 1) with typical metallic properties such as being shiny, malleable, and ductile.
  • Silicon is a metalloid with properties intermediate between metals and non-metals.

Electrical conductivity:

  • Sodium is an excellent electrical conductor, while silicon is a semiconductor. As a Group 1 metal, sodium has one valence electron that is delocalised in a metallic lattice, allowing it to move freely and conduct electricity very effectively.
  • Silicon, as a metalloid, has moderate electrical conductivity that increases with temperature – its conductivity is much lower than sodium’s but higher than non-metals.

Ion formation:

  • Sodium readily forms positive ions (cations) with a 1+ charge (\(\ce{Na^+}\)) by losing its single valence electron to achieve a stable electron configuration similar to neon. Silicon typically does not form simple ions due to its position as a metalloid.
  • Instead, silicon forms covalent bonds by sharing its four valence electrons with other atoms. While silicon can theoretically form \(\ce{Si^4+}\) or \(\ce{Si^4-}\) ions, the energy required to remove or add four electrons is too high.

CHEMISTRY, M1 EQ-Bank 17

Element \(\ce{X}\) is a shiny solid at room temperature that conducts electricity well. It forms a compound with oxygen with the formula \(\ce{X2O3}\).

Identify the group in the periodic table to which element \(\ce{X}\) most likely belongs. Justify your answer.   (2 marks)

Show Answers Only
  • Element \(\ce{X}\) most likely belongs to Group 13 of the periodic table.
  • The formula \(\ce{X2O3}\) indicates that element \(\ce{X}\) forms ions with a \(+3\) charge (\(\ce{X^3+}\)), since oxygen forms \(\ce{O^2-}\) ions.
  • To balance the charges in a neutral compound, two \(\ce{X^3+}\) ions (total charge: 6+) combine with three \(\ce{O^2-}\) ions (total charge: 6–).
  • Elements in Group 13, such as aluminium, form \(+3\) ions and are metals with the described properties of being shiny solids that conduct electricity well.
Show Worked Solution
  • Element \(\ce{X}\) most likely belongs to Group 13 of the periodic table.
  • The formula \(\ce{X2O3}\) indicates that element \(\ce{X}\) forms ions with a \(+3\) charge (\(\ce{X^3+}\)), since oxygen forms \(\ce{O^2-}\) ions.
  • To balance the charges in a neutral compound, two \(\ce{X^3+}\) ions (total charge: 6+) combine with three \(\ce{O^2-}\) ions (total charge: 6–).
  • Elements in Group 13, such as aluminium, form \(+3\) ions and are metals with the described properties of being shiny solids that conduct electricity well.

CHEMISTRY, M1 EQ-Bank 8 MC

Which mixture is best separated into its component parts by fractional distillation?

  1. Iron filings and sulfur powder
  2. Ethanol and water
  3. Oil and vinegar
  4. Ethane \(\ce{C2H6}\) and octane \(\ce{C8H18}\)
Show Answers Only

\(D\)

Show Worked Solution
  • Fractional distillation is a separation technique used to separate miscible liquids that have different boiling points. It is most effective when separating components of a homogeneous liquid mixture.
  • Ethane and octane are both hydrocarbons that mix completely and have sufficiently different boiling points to be separated by this technique.

\(\Rightarrow D\)

CHEMISTRY, M1 EQ-Bank 16

Hexane and water are liquids that are immiscible with each other. Some of their properties are shown in the table.

\begin{array} {|c|c|c|}
\hline  & \text{Boiling point } (^{\circ}\text{C}) & \text{Density } (\text{g mL}^{-1})\\
\hline \text{Hexane} & 68.7 & 0.66 \\
\hline \text{Water} & 100 & 1.00 \\
\hline \end{array}

A chemist finds a bottle containing hexane and water and needs to determine whether she should use a separating funnel or distillation to separate the liquids.

Assess the effectiveness of each technique when separating hexane and water.   (4 marks)

Show Answers Only
  • Hexane and water are immiscible, so they do not mix and, instead, form separate layers.
  • As a result, they can be separated effectively with a separating funnel. The density of hexane is lower than the density of water, so the hexane will be the upper layer and the water will be the lower layer.
  • The liquids could also be separated effectively through distillation.
  • They have a large difference in boiling points. Hexane has a lower boiling point than water (68.7°C). As such, it will be the first fraction to be collected during distillation.
Show Worked Solution
  • Hexane and water are immiscible, so they do not mix and, instead, form separate layers.
  • As a result, they can be separated effectively with a separating funnel. The density of hexane is lower than the density of water, so the hexane will be the upper layer and the water will be the lower layer.
  • The liquids could also be separated effectively through distillation.
  • They have a large difference in boiling points. Hexane has a lower boiling point than water (68.7°C). As such, it will be the first fraction to be collected during distillation.

CHEMISTRY, M1 EQ-Bank 9 MC

A radioisotope \(X\) undergoes alpha decay to form isotope \(Y\).

Isotope \(Y\) then undergoes beta decay to form lead-210.

Which of the following identifies isotope \(X\)?

  1. bismuth-214
  2. polonium-214
  3. radon-222
  4. uranium-238
Show Answers Only

\(A\)

Show Worked Solution
  • The final product is \(\ce{^210_82Pb}\)
  • Before the beta decay, isotope \(Y\) must have the same mass number and an atomic number of 81 as during beta decay a neutron is converted into a proton.
  •     \(\ce{^210_81Tl -> ^210_82Pb + ^0_-1e^-}\)
  • Before the alpha decay, isotope \(X\) must have a mass number of \(214\) and an atomic number of \(83\) as during alpha decay the decaying particle shoots out a \(\ce{^4_2\alpha}\) particle.
  •     \(\ce{^214_83Bi -> ^210_81Tl + ^4_2\alpha}\)

\(\Rightarrow A\)

CHEMISTRY, M8 2021 VCE 30 MC

The \(^1\)H NMR spectrum of an organic compound has three unique sets of peaks: a single peak, seven peaks (septet) and two peaks (doublet).

The compound is

  1. 3-methyl butanoic acid.
  2. 2-methyl propanoic acid.
  3. 2-chloro-2-methylpropane.
  4. 1,2-dichloro-2-methylpropane.
Show Answers Only

\(B\)

Show Worked Solution
  • On a \(^1\)H NMR spectrum:
  •    A single peak → no hydrogens on the adjacent carbon.
  •    A double peak → 1 hydrogen on the adjacent carbon.
  •    A septet peak → 6 hydrogens with the same environment on adjacent carbons.
  • Therefore, there are 3 unique hydrogen environments in this compound.
  • 2-methyl propanoic acid is the only compound with 3 different hydrogen environments.

\(\Rightarrow B\)

♦ Mean mark 47%.

CHEMISTRY, M1 EQ-Bank 6

Describe the process by which emission line spectra are formed.   (4 marks)

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  • Electrons in atoms exist in discrete energy levels.
  • When an atom absorbs energy (e.g. from heat or electricity), an electron is excited to a higher energy level.
  • The excited electron is unstable and will eventually fall back to a lower energy level.
  • As it does so, it releases energy in the form of a photon of light.
  • The energy of the photon corresponds to the difference between the two energy levels, so only specific wavelengths of light are emitted.
  • Passing this light through a spectroscope produces a series of discrete coloured lines known as the emission spectrum.
Show Worked Solution
  • Electrons in atoms exist in discrete energy levels.
  • When an atom absorbs energy (e.g. from heat or electricity), an electron is excited to a higher energy level.
  • The excited electron is unstable and will eventually fall back to a lower energy level.
  • As it does so, it releases energy in the form of a photon of light.
  • The energy of the photon corresponds to the difference between the two energy levels, so only specific wavelengths of light are emitted.
  • Passing this light through a spectroscope produces a series of discrete coloured lines known as the emission spectrum.

CHEMISTRY, M1 EQ-Bank 5

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

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

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

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

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

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

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

Show Worked Solution

a.    \(\alpha < \beta < \gamma\)

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

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

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

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

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

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

CHEMISTRY, M1 EQ-Bank 5 MC

A calcium ion, \(\ce{Ca^{2+}}\), has an atomic number of 20 and a mass number of 40. Which of the following statements is correct?

  1. A calcium ion has 20 protons, 20 electrons and 20 neutrons
  2. A calcium ion has 20 protons, 18 electrons and 20 neutrons
  3. A calcium ion has 18 protons, 20 electrons and 20 neutrons
  4. A calcium ion has 20 protons, 18 electrons and 18 neutrons
Show Answers Only

\(B\)

Show Worked Solution
  • Atomic number \(=\) number of protons \(= 20\).
  • Mass number \(=\) protons \(+\) neutrons \(= 40\) → neutrons \(40-20 = 20\).
  • \(\ce{Ca^{2+}}\) means it has lost 2 electrons, so electrons \(= 20-2 = 18\).

\(\Rightarrow B\)

CHEMISTRY, M1 EQ-Bank 3 MC

Thorium-232 undergoes a sequence of radioactive decays. In the first part of this sequence, what is the isotope produced when thorium-232 undergoes one alpha decay followed by one beta decay?

  1. \(\ce{^228_88Ra}\)
  2. \(\ce{^228_89Ac}\)
  3. \(\ce{^232_90Th}\)
  4. \(\ce{^228_90Th}\)
Show Answers Only

\(B\)

Show Worked Solution
  • During Alpha decay, the radio active particle emits a alpha particle \(\ce{^4_2He}\).
  •    \(\ce{^232_90Th -> ^228_88Ra + ^4_2He}\)
  • Beta decay process involves a neutron being converted to a proton with the emission of an electron.
  •    \(\ce{^228_88Ra -> ^228_89Ac + ^0_-1e^-}\)

\(\Rightarrow B\)

CHEMISTRY, M1 EQ-Bank 1

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

  1. Tellurium has two naturally occurring stable isotopes that contribute to its average atomic mass. One of these isotopes, \(\ce{^126Te}\), has a percentage abundance of 45.2%. Calculate the mass number of the other isotope.   (2 marks)
  1. One unstable isotope of tellurium is \(\ce{^130Te}\). Write a nuclear equation for the decay of this isotope when it undergoes beta decay.   (1 marks)
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a.    \(\ce{^129Te}\)

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

Show Worked Solution

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

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

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

CHEMISTRY, M1 EQ-Bank 14

Compare and explain the reactivity of Group 1 (alkali metals) and Group 2 (alkaline earth metals) with water. In your answer, link your explanation to electron configuration, atomic radius, and ionisation energy.   (6 marks)

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  • Group 1 metals (e.g. \(\ce{Li, Na, K}\)) react vigorously with water to form a hydroxide and hydrogen gas.
  • Group 2 metals (e.g. \(\ce{Mg, Ca}\)) also react with water but much less vigorously, especially at the top of the group. For example, magnesium reacts only slowly with cold water.
  • Electron configuration: Group 1 metals have one valence electron, while Group 2 metals have two valence electrons. Losing one electron requires less energy than losing two, making Group 1 metals more reactive.
  • Atomic radius and ionisation energy: Down both groups, the atomic radius increases, shielding increases, and ionisation energy decreases. This means reactivity with water increases down the group.
  • Therefore: Reactivity increases down both groups, but Group 1 metals show higher reactivity with water compared with Group 2 metals in the same period.
Show Worked Solution
  • Group 1 metals (e.g. \(\ce{Li, Na, K}\)) react vigorously with water to form a hydroxide and hydrogen gas.
  • Group 2 metals (e.g. \(\ce{Mg, Ca}\)) also react with water but much less vigorously, especially at the top of the group. For example, magnesium reacts only slowly with cold water.
  • Electron configuration: Group 1 metals have one valence electron, while Group 2 metals have two valence electrons. Losing one electron requires less energy than losing two, making Group 1 metals more reactive.
  • Atomic radius and ionisation energy: Down both groups, the atomic radius increases, shielding increases, and ionisation energy decreases. This means reactivity with water increases down the group.
  • Therefore: Reactivity increases down both groups, but Group 1 metals show higher reactivity with water compared with Group 2 metals in the same period.