2024 HSC Maths Standard 2 Exam Solutions

Here are the 2024 HSC Maths Standard 2 Exam Solutions written by our Maths Team. Read on to see how you went.

Written by:
Oak Ukrit
Blue maths tools with formulas in the background

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Check out our comprehensive 2024 HSC Maths Standard 2 exam solutions, where we break down this year’s paper. Compare your answers and see how they align with ours.

You can read the 2024 Maths Standard 2 paper here.

2024 HSC Maths Standard 2 Exam Solutions

Multiple-Choice Section

QuestionAnswerExplanation
1D\begin{align*}
x^2 &=(-2.531)^2 \\
& \approx 6.40596 \\
&\approx 6.41
\end{align*}
2CWe can see that the line has a positive \(y\)-intercept, so \(c=3\). The line also has a downward slope so \(m<0\), which means \(m=-2\). Therefore:

\begin{align*}
y=-2x+3
\end{align*}

3AThere is no mention of random selection or use of categories.
Since the invited shoppers have a choice about whether they particpate, it would be self-selecting.
4BThere are 7 borders between the regions. Therefore, the network would have seven edges.
5AFirst, we need to find the \(z\)-score for Pia’s mark in each subject:

English:
\begin{align*}
z &= \frac{78-66}{6} \\ & = 2
\end{align*}
Mathematics:
\begin{align*}
z &= \frac{80-71}{10} \\ & = 0.9
\end{align*}
Science:
\begin{align*}
z &= \frac{77-70}{15} \\ & = \frac{7}{15}
\end{align*}
History:
\begin{align*}
z &= \frac{85-72}{9} \\ & = \frac{13}{9}
\end{align*}

Since the English \(z\)-score is the greatest, Pia performed best in English.

6AWe can use the cosine ratio since we have the adjacent and hypotenuse.

\begin{align*}
\cos (40^\circ) &= \frac{x}{20} \\
x &= 20\cos (40^\circ)
\end{align*}

7DThe increase is being compounded annually so we can use the compound interest formula.

\begin{align*}
\text{Price} &= 180(1.025)^3 \\ &=\$193.84 \text{ (nearest cent)}
\end{align*}

8CWe can find the GST for the Labour and Parts separately:

Parts:
\begin{align*}
110\% &\text{ is } \$242 \\
110\% \div 11 &\text{ is } \$242 \div 11 \\
10\% &\text{ is } \$22 \\
\mathrm{GST} &\text{ is } \$22
\end{align*}

Labour:

\begin{align*}
\text{GST} &= 10\% \times \$100 \\
\text{} &= 10
\end{align*}

\begin{align*}
\text{Total GST} &= \$10 + \$22 \\
\text{} &= \$32
\end{align*}

9B\begin{align*}
6 \text{ painters} &: 20 \text{ days} \\
6 \div 2 \text{ painters} &: 20 \times 2 \text{ days} \\
3 \text{ painters} &: 40 \text{ days} \\
3 \times 5 \text{ painters} &: 40 \div 5 \text{ days} \\
15 \text{ painters} &: 8 \text{ days} \\
\end{align*}
10D\begin{align*}
s-wt & =\frac{p}{2} \\
p & =2s-2wt
\end{align*}
11B\begin{align*}
\text{No. minutes} &= 18+60+4\\
\text{} &= 82 \\
\text{No. hours} &= 82 \div 60\\
\text{} &= \frac{41}{30} \\
\text{Speed} &= 61 \div \frac{41}{30}\\
\text{} &\approx 45 \text{ km/h}
\end{align*}
12CAccording to the table there were \(157\) over \(30\) year olds in total, and \(34\) of them watch Anime.

\begin{align*}
\text{Percentage} &= \frac{34}{157}\times 100\% \\
\text{} &\approx 21.656\% \\
\text{} &\approx 22\%
\end{align*}

13A\begin{align*}
\text{Amount paid} &= 300 \times 4 \times 12 \\ &= \$14 400
\end{align*}
\begin{align*}
I &= 14400-9000 \\ &= \$5400 \\
\end{align*}
\begin{align*}
I &= Prn \\
5400 &= 9000 \times r \times 4 \\
5400 &= 36000 r \\
r&=\frac{5400}{36000}\times 100\% \\
r &= 15\% \text{p.a.}
\end{align*}
14B\begin{align*}
\text{Profit} &=R – C \\
\text{} &= 8x-(2.5x+6) \\
\text{} &= 5.5x-6
\end{align*}
Therefore, the profit increases by \($5.50\) for each additional pizza sold.
15DFrom the box plot: \(Q_1 – Min = Max – Q_4\). Similarly, \(Q_2 – Q_1 = Q_4 – Q_3\). Additionally, \(Q_2 – Q_1\) is approximately double \(Q_1 – Min\). In each histogram, there are 16 scores. From the box plot, we can find the position of each statistic in the five-number summary:

\begin{align*}
\text{Min:} & \text{ 1st score}, \\
Q_1: & \text{ 4th & 5th score}, \\
Q_2: & \text{ 8th & 9th score}, \\
Q_3: & \text{ 12th & 13th score}, \\
\text{Max:} & \text{ 16th score}.
\end{align*}

We label the bottom axis as follows to calculate the quartiles of each graph.

The values for the five-number summaries are as follows:

\begin{align*}
\text{A:} & \, \text{Min } = 1, \, Q_1 = 2, \, Q_2 = 4, \, Q_3 = 6, \, \text{Max } = 7, \\
\text{B:} & \, \text{Min } = 1, \, Q_1 = 2, \, Q_2 = 4, \, Q_3 = 6, \, \text{Max } = 7, \\
\text{C:} & \, \text{Min } = 1, \, Q_1 = 2, \, Q_2 = 4, \, Q_3 = 6, \, \text{Max } = 7, \\
\text{D:} & \, \text{Min } = 1, \, Q_1 = 3, \, Q_2 = 4, \, Q_3 = 5, \, \text{Max } = 7.
\end{align*}

For D: \(Q_3 – Q_2 = 1\) and \(Max – Q_3 = 2\). This does not reflect the properties of the box plot. Thus, we arrive at our solution \(D\).

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2024 HSC Maths Standard 2 Exam Solutions – Long Response Section

Question 16

(a).

\begin{align*}
TYWH&=65
\end{align*}

(b).

Try different paths to find the shortest.

\begin{align*}
YWCMG&=15+25+25+25 \\&=90 \\
YWHMG&=15+20+29+25 \\&=89 \\
\end{align*}

The shortest path is \(89km\).

Question 17

\begin{align*}
\text{Total kWh} &= 0.65\text{ kW} \times 6 \text{ h} \\
&= 3.9 \text{ kWh}
\end{align*}

\begin{align*}
\text{Total Cost} &=3.9 \times30.13 \\
&= 117.507\text{ cents} \\
&\approx \$1.18
\end{align*}

Question 18

(a).

Minimum Spanning Tree:

Graph diagram with weighted edges for 2024 HSC Maths Standard 2 exam solutions

Weight\(\, = 24\)

(b).

Yes, since \(BC=2\), we can create another spanning tree where we include \(BC\) instead of \(FC\).

Question 19

(a).

We can see from the graph that the \(y\)-intercept is \(6\), and the line passes through \((10,20)\).
\begin{align*} m &= \frac{20-6}{10-0} \\ m&=1.4 \end{align*}
\begin{align*} y &= mx+b \\ y&=1.4x+6 \end{align*}

(b).

Worse, because the point \((5, 12)\) lies below the line.

Question 20

(a).

\(\text{No. Periods}= 8, \, r=5\%\), so according to the table a \($1\) annuity will give \(FV=\$10.0266\).
For a \(\$200\) annuity:

\begin{align*}
FV &= 200 \times 10.0266\\
&= \$2005.32
\end{align*}

(b).

\begin{align*}
r &= 4\% \div 2 \\
&= 2\% \\
n &= 3\times 2 \\
&=6
\end{align*}

According to the table a \($1\) annuity will give \(FV=\$6.4343\).
To get \(FV=\$4500\):

\begin{align*}
\text{Payment} &= \frac{4500}{6.4343}\\
&= 699.38 \\
&\approx \$700
\end{align*}

Question 21

(a).

\begin{align*}
r &= 0.24\div 12\\
r &=0.02
\end{align*}
\begin{align*}
\text{E2} &= 5590 +111.80-110\\
\text{} &=\$5591.80
\end{align*}
\begin{align*}
\text{B3} &=\$5591.80
\end{align*}
\begin{align*}
\text{C3} &= 5591.80 \times 0.02\\
\text{} &=\$111.84
\end{align*}
\begin{align*}
\text{E3} &= 5591.80 +111.84-110\\
\text{} &=\$5593.64
\end{align*}

Table showing principal, interest, payments, and balance owing for 2024 HSC Maths Standard 2 exam solutions

(b).

The interest is more than the repayments, so each month the principal will increase.

Question 22

The vertical intercept of the function for Population \(K\) is \(280\).

\begin{align*}
B &= 280 \end{align*}

Percentage change in the yearly population:
Population \(W\):

\begin{align*}
1.055-1 &= 0.055\\
&= 5.5\%\\
\end{align*}
Population \(K\):
\begin{align*}
1-0.97 &= 0.03\\
&= 3\%\\
\end{align*}
When \(x=50\):
\begin{align*}
y &= 35 \times 1.055^{50}\\
&\approx 494\\
\end{align*}

Table showing population data and predictions for 2024 HSC Maths Standard 2 exam solutions

Question 23

\begin{align*}
2790&=38\times 45 + x \times 45 \times 1.5\\
2790&= 1710 + 67.5x\\
67.5x &= 1080 \\
x &\approx 16
\end{align*}
Zazu worked 16 hours of overtime.

Question 24

(a).

\begin{align*}
N &= 3 \times 1.2 \\
&= 3.6 \\
H &= 2.5 \\
M &= 60
\end{align*}
\begin{align*}
BAC &= \frac{10\times 3.6 – 7.5 \times 2.5}{5.5 \times 60} \\
&= 0.052 \\
\end{align*}

(b).

\begin{align*}
\text{Time} &= \frac{0.052}{0.015} \\
&\approx 3.4666… \\
&\approx 3 \text{ hours } 28 \text{ minutes}
\end{align*}

Question 25 

Alex’s Interest:

\begin{align*}
I &= 1800 \times 0.075 \times 5\\
&= \$675
\end{align*}
Jun’s Interest:
\begin{align*}
r &= 0.06 \div 4 \\
&= 0.015 \\
n &= 5 \times 4 \\
&= 20
\end{align*}
\begin{align*}
A &= 1800(1.015)^{20}\\
&= \$2424.34 \\
I &= 2424.34 -1800 \\
&= \$624.34
\end{align*}
Therefore, Alex will have the greater amount because he earned the most interest.

Question 26

The greatest area is at the maximum turning point of the parabola, which occurs at the midpoint of \(w=0\) and \(w=40\). So the greatest area occurs when \(w=20\text{ cm}\).

\begin{align*}
A &= -0.5(20)^2+20\times 20 \\
&=200\text{ cm}^2 \\
\end{align*}
For the cross-section:
\begin{align*}
A&= wh \\
200&=20\times h \\
h &= 10\text{ cm}
\end{align*}
A width of \(20\text{ cm}\) and a height of \(10\text{ cm}\) give the greatest area.

Question 27

Original plan:
\begin{align*}
\text{Total Repayments } &= 280 \times 12 \times 10 \\
&=\$33600
\end{align*}
New plan:
\begin{align*}
\text{Total Repayments } &= 280 \times 12 \times 5 +250\times 12 \times (5+2) \\
&=\$37800
\end{align*}
\begin{align*}
\text{Difference } &= 37800 – 33600 \\
&=\$4200
\end{align*}
They will pay \($4200\) more using the new plan.

Question 28

Garden A is negatively skewed, has a median of approximately \(171\), and an interquartile range of \(176-154=22\).
Garden B is positively skewed, has a median of approximately \(152\), and an interquartile range of \(158-149=9\).

Therefore, Garden B has consistently lower heights since both the IQR and median are lower than Garden A. Also, for Garden B most of the scores are in the lower range since it is positively skewed.

Question 29

For the first four years, using the straight-line method we have:

\begin{align*}
S&=50000-1500\times 4\\
&=\$44\,000
\end{align*}

For the next six years, using declining-balance method we have:

\begin{align*}
S&=44000(1-0.35)^6\\
&=\$3318.43
\end{align*}
Total Depreciation:
\begin{align*}
D&=50000-3318.43\\
&=\$46681.57
\end{align*}

Question 30

  1. The length of male anacondas increases slower after 4 years of age.
  2. Female anacondas are longer than males of a similar age.
  3. Anacondas grow in length for at least 10 years.

Question 31

(a).

\begin{align*}
H:T &= 2:1 \\
P(H) &= \frac{2}{3}
\end{align*}

(b).

\begin{align*}
P(\text{at least }1 \, H) &= 1-P(TT) \\ &=1 – \frac{1}{3}\times \frac{1}{3} \\ &=1 – \frac{1}{9} \\ &= \frac{8}{9}
\end{align*}

Question 32

\begin{align*}
\text{Pentagonal Area} &= \frac{1}{2}\times 30 \times 30 \times \sin(72) \times 5 \\
&=2139.75
\end{align*}

\begin{align*}
\text{Shaded Area} &= \pi \times 30^2 – 2139.75 \\
&\approx 687.56 \\
&\approx 690 \text{ cm}^2
\end{align*}

Question 33

\begin{align*}
40 \text{ km} &: 1\text{ hour} \\
40 000 \text{ m} &: 3600\text{ sec} \\
1 \text{ m} &: 3600 \div 40 000\text{ sec} \\
150 \text{ m} &: 3600 \div 40 000 \times 150\text{ sec} \\
150 \text{ m} &: 13.5\text{ sec}
\end{align*}

Therefore it takes \(13.5\) seconds to run \(150\) metres.

Question 34

\begin{align*}
r &= 23\div 2 \\ &= 11.5
\end{align*}

The two ends make 1 sphere:
\begin{align*}
\text{Sphere }&= 4\times \pi \times 11.5^2 \\
&= 1661.9 \text{ cm}^2
\end{align*}

\begin{align*}
\text{Rectangular side height }&= 2d \\ &= 2 \times 23 \\
&= 46 \text{ cm}
\end{align*}

\begin{align*}
\text{Rectangular side }&= 2\pi \times 11.5 \times 46 \\
&= 3323.8 \text{ cm}^2
\end{align*}

\begin{align*}
\text{Surface Area }&= 1661.9 + 3323.8 \\
&= 4985.7 \text{ cm}^2 \\
&= 4985.7 \div 100^2 \text{ m}^2 \\
&= 0.49857 \text{ m}^2 \\
&\approx 0.5\text{ m}^2
\end{align*}

Question 35

(a).

For \(x=58: z=0\).
For \(x=70:\)

\begin{align*}
z &= \frac{70-58}{15} \\
&= 0.8
\end{align*}

\begin{align*}
P(58<x<70) &= P(0 < z<0.8) \\
&= P(z=0.8)-P(z=0) \\
&= 0.7881 – 0.5 \\
&= 0.2881
\end{align*}

(b).

For \(x=46:\)

\begin{align*}
z &= \frac{46-58}{15} \\
&= -0.8
\end{align*}

\begin{align*}
P(46<x<70) &= P(-0.8 < z<0.8) \\
&= P(-0.8 < z<0) + P(0 < z<0.8) \\
&= 2\times P(0 < z<0.8)
\end{align*}

The area under the normal curve between \(z =-0.8\) and \(z=0\) is equal to the area under the normal curve between \(z =0\) and \(z=0.8\).
Therefore, the probability is equal to double the area under the normal curve between \(z =0\) and \(z=0.8\).

(c).

From the table, when \(P=0.9032: z=1.3\).

\begin{align*}
1.3 &\approx \frac{x-58}{15} \\
19.5 &\approx x-58 \\
x &\approx 77.5
\end{align*}

Therefore, the approximate minimum score is \(77.5\).

Question 36

(a).

\begin{align*}
\frac{BE}{\sin 27^\circ}&= \frac{53.8}{\sin 106^\circ}\\
BE &= \frac{53.8 \sin 27^\circ}{\sin 106^\circ} \\
&\approx 25.4 \text{ m}\\
\end{align*}

(b).

\begin{align*}
\angle XBC &= 180^\circ – 106^\circ \\
&= 74^\circ
\end{align*}

\begin{align*}
\tan 74^\circ &= \frac{20}{BX} \\
BX &= \frac{20}{\tan 74^\circ} \\
&\approx 5.735
\end{align*}

\begin{align*}
CD &= 25.4 – 5.735\\
&\approx 19.7 \text{ m}
\end{align*}

Question 37

To find the Sydney departure time, we need to go back \(20\) hours and forward \(13\) hours, since Sydney is \(13\) hours ahead of Rio de Janeiro.

\begin{align*}
\text{Sydney Time} &= 3\text{ p.m. Wed} – 20 \text{ hours} + 13\text{ hours}\\
&= 3\text{ p.m. Wed} – 7 \text{ hours} \\
&= 8\text{ a.m. Wed, 20 July}
\end{align*}

Question 38

\(r=15, h=6\)

\begin{align*}
V_{cylinder}&=\pi \times 15^2 \times 6\\
&\approx 4241.15\text{ cm}^3
\end{align*}
\begin{align*}
V_{cone}&=4241.15 \div 5\\
&\approx 848.23\text{ cm}^3
\end{align*}
\begin{align*}
V_{total}&=4241.15 + 848.23\\
&=5089.38\text{ cm}^3
\end{align*}
\begin{align*}
\text{No. Slices}&=5089.38 \div 212\\
&\approx 24.007
\end{align*}

Therefore, 24 slices can be cut from the cake.

Question 39

a. For \(B\): \(EST=0, LST=0\). Therefore, the critical path is \(BEGI\).
b. \(19-12 = 7 \text{ hours}\)
c. \(EST\) for \(F\) is \(3\). If \(I\) starts at \(12\) hours then the maximum time will be \(12-3=9\) hours.

Question 40

\begin{align*}
\angle DON &= 360 – 285 \\
&= 75^\circ
\end{align*}

let \(\theta =\angle DOA\), using the cosine rule:

\begin{align*}
\cos \theta &= \frac{38^2+42^2-67.5^2}{2\times 38\times 42} \\
\theta &= \cos ^{-1} \left( \frac{-5393}{12768} \right) \\
&\approx 115^\circ
\end{align*}
\begin{align*}
\angle NOA &= 115^\circ – 75^\circ\\
&= 40^\circ
\end{align*}
\begin{align*}
\text{Bearing} &= 40^\circ+180^\circ\\
&= 220^\circ \text{T}
\end{align*}

Question 41

For the first \(15\) years:

\begin{align*}
n &= 15\times 12 \\
&= 180 \\
r &= 0.024\div 12 \\
&= 0.002
\end{align*}
\begin{align*}
PV &= 151.036 \times 2000 \\
&= \$302072
\end{align*}

To find the present value for the next 10 years, we can subtract the present value of 15 years from the present value of 25 years using \(r=0.002\) and withdrawals of \(\$1200\).

For 25 years:
\begin{align*}
n &= 25\times 12 \\
&= 300
\end{align*}
\begin{align*}
PV &= 225.43 \times 1200 \\
&= \$270516
\end{align*}
For 15 years:
\begin{align*}
n &= 15\times 12 \\
&= 180
\end{align*}
\begin{align*}
PV &= 151.036 \times 1200 \\
&= \$181243.20
\end{align*}
\begin{align*}
\text{Total }PV &= 302072 + 270516 – 181243.20\\
&= \$391344.80
\end{align*}

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Written by Oak Ukrit

Oak is the Head of Mathematics at Matrix Education and has been teaching for over 12 years and has been helping students at Matrix since 2016. He has 1st class honours in Aeronautical Engineering from UNSW where he taught for over 4 years while he was undertaking a PhD. When not plane spotting he enjoys landscape photography.

© Matrix Education and www.matrix.edu.au, 2023. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

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