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Here are the 2024 HSC Maths Standard 2 Exam Solutions written by our Maths Team. Read on to see how you went.
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The 2024 HSC Maths Standard 2 Exam is done! Curious to see how our expert Maths Team tackled the questions?
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.
Question | Answer | Explanation |
1 | D | \begin{align*} x^2 &=(-2.531)^2 \\ & \approx 6.40596 \\ &\approx 6.41 \end{align*} |
2 | C | We 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*} |
3 | A | There 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. |
4 | B | There are 7 borders between the regions. Therefore, the network would have seven edges. |
5 | A | First, we need to find the \(z\)-score for Pia’s mark in each subject: English: Since the English \(z\)-score is the greatest, Pia performed best in English. |
6 | A | We can use the cosine ratio since we have the adjacent and hypotenuse. \begin{align*} |
7 | D | The increase is being compounded annually so we can use the compound interest formula. \begin{align*} |
8 | C | We can find the GST for the Labour and Parts separately: Parts: Labour: \begin{align*} \begin{align*} |
9 | B | \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*} |
10 | D | \begin{align*} s-wt & =\frac{p}{2} \\ p & =2s-2wt \end{align*} |
11 | B | \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*} |
12 | C | According to the table there were \(157\) over \(30\) year olds in total, and \(34\) of them watch Anime. \begin{align*} |
13 | A | \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*} |
14 | B | \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. |
15 | D | From 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*} 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*} 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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(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 \(89km\).
\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*}
(a).
Minimum Spanning Tree:
Weight\(\, = 24\)
(b).
Yes, since \(BC=2\), we can create another spanning tree where we include \(BC\) instead of \(FC\).
(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.
(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*}
(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*}
(b).
The interest is more than the repayments, so each month the principal will increase.
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*}
\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.
(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*}
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.
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.
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.
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.
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*}
(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*}
\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*}
\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.
\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*}
(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 \(77.5\).
(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*}
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*}
\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.
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.
\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*}
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.