2021 HSC Maths Standard 2 Exam Paper Solutions

The Matrix 2021 HSC Maths Standard 2 Exam Paper Solutions are here!

2021 HSC Maths Standard 2 Exam Paper Solutions

Have you seen the 2021 HSC Mathematics Standard 2 exam Paper yet?

In this post, we will work our way through the 2021 HSC Maths Standard 2 exam paper and give you the solutions, written by our Head of Mathematics Oak Ukrit and his team.

Multiple-choice section

Question	Answer	Explanation
1	A	For option A, the perimeter is \(4 \times 8 = 32\) cm. For option B, the perimeter is \(2 \left(3 + 11\right) = 28\) cm. For option C, the perimeter is \(3 \times 10 = 30\) cm. For option D, the perimeter is \(3 + 4 + 4 + 9 = 20\) cm.
2	D	Vertex A has degree 4, vertex B is degree 3, vertex C has degree 4, vertex D has degree 2, and vertex E has degree 3. So the sum of the degrees of all the vertices is \(4 + 3 + 4 + 2 + 3 = 16\).
3	C	\(25\) appears twice, all the others appear once only, so the mode is \(25\).
4	D	Its salvage value is \begin{align*} S &= V_0 \left(1 – r\right)^n \\ &= \$2467 \left(1 – 0.15\right)^3 \\ &= \$1515 \end{align*}
5	B	His hourly wage after four years will be \begin{equation*} \$21.50 \left(1 + \frac{0.021}{1}\right)^{1 \times 4} = \$21.50 \times \left(1.021\right)^4 \end{equation*}
6	A	\begin{align*} a &= \frac{22}{7} \\ &= 3.14285 \dots \\ &= 3.14 \end{align*}
7	C	Options A and D show that there are close to 100 downloads on day 1, which is not consistent with the graph given, so they are wrong. The graph given shows that the number of downloads increases from day 1 to day 10, and it decreases after that. Option B shows that the number of downloads every day is roughly the same, and that is inconsistent with the graph given. The graph in option C is consistent with all the features shown in the graph given.
8	D	\begin{align*} z &= \frac{x – \mu}{\sigma} \\ 2 &= \frac{x – 63}{8} \\ 16 &= x – 63 \\ x &= 79 \end{align*}
9	C	After subtracting \(8\) from the number, we get \(x – 8\). After multiply the result by \(3\), we get \(3 \left(x – 8\right)\).\(2\) more than \(x\) is \(x + 2\). Since these two results are the same, we obtain \begin{equation*} 3 \left(x – 8\right) = x + 2 \end{equation*}
10	A	When \(x = 0\),\begin{align*} y &= 10 \left(0.8\right)^0 \\ &= 10 \left(1\right) \\ &= 10 \end{align*} so options B and D are wrong.Since \(0.8 < 1\), \(y = 10 \left(0.8\right)^x\) decreases as \(x\) increases, which means that option A is correct.
11	D	The probability that Kim and Sam choose chocolates with different centres is \begin{align*} P(\text{one P and one C}) &= \frac{3}{8} \times \frac{5}{7} + \frac{5}{8} \times \frac{3}{7} \\ &= \frac{15}{28} \end{align*}
12	B	\begin{align*} \text{area} &\approx \frac{6}{2} \left(1.2 + 2\right) + \frac{6}{2} \left(2 + 1.4\right) \\ &= 19.8 \; \text{km}^2 \end{align*}
13	B	Let \(C\) be the number of cleaners, $t$ be the amount of time (in hours) it takes to clean the warehouse, and [/altex]k[/latex] be a constant. Then \begin{align*} C &= \frac{k}{t} \\ 8 &= \frac{k}{60} \\ k &= 480 \end{align*} \begin{equation*} ∴ C = \frac{480}{t} \end{equation*} When \(C = 10\), \begin{align*} 10 &= \frac{480}{t} \\ t &= \frac{480}{10} \\ t &= 48 \end{align*}
14	D	Since \(270^\circ + 25^\circ = 295^\circ\), the true bearing of \(A\) from \(B\) is \(295^\circ\).
15	B	\begin{align*}\text{number of professional runners} &= 11400 \times \frac{3}{3 + 16} \\ &= 1800 \end{align*} and \begin{align*} \text{total number of amateurs} &= 11400 \times \frac{16}{3 + 16} \\ &= 9600 \end{align*}Also, \begin{align*} \text{number of amateurs who completed the race} &= 9600 – 600 \\ &= 9000 \end{align*}So, \begin{align*} \text{no.\ of professional runners who completed the race} : \text{no.\ of amateurs who completed the race} &= 1800 : 9000 \\ &= 1 : 5 \end{align*}

Long Response Section

Question 16

Question 17

Question 18

Question 19

Question 20

We are only concerned about the longitude, i.e. how far East or West. Since \(151^\circ – 16^\circ = 135^\circ\),

Question 21

Question 22

Question 23a

The minimum spanning tree is

Question 23b

Question 24a

Question 24b

Question 24c

Question 25

Question 26a

Question 26b

Question 27a

Question 27b

Question 28a

Set calculator to STAT mode, select “A + Bx” and enter data points into table.

The \(y\)-intercept is \(2\) and the gradient is \(3.2\).

Hence, the equation of the least-squares regression line is \(y = 3.2 x + 2\);

Question 28b

The data point (10,45) would be above the regression line in the diagram, so the gradient would increase.

Question 29

Question 30

Question 31

Question 32a

Question 32b

Question 33a

(i)

(ii)

As the height above sea level increases, the average maximum daily temperature decreases since the gradient of the regression line is negative.

Question 33b

Latitude would be a better predictor as the magnitude of the correlation coefficient is greater  (\(-0.897\) compared to \(-0.494\)). This indicates a stronger linear relationship between the two variables.

Question 34

We can formulate the statement “In total the emus and goannas have 76 legs”, as

\begin{equation*}
4x + 2y = 76 \,,
\end{equation*}

which dividing both sides by \(2\) gives

\begin{equation}
2x + y = 38 \tag{1}
\end{equation}

We are given that

\begin{equation}
x + y = 31 \tag{2}
\end{equation}

So \(\text{Eq.}(1) – \text{Eq.}(2)\) gives
\begin{equation*}
x = 7
\end{equation*}

Substituting \(x = 7\) into Eq.(2) gives

\begin{align*}
7 + y &= 31 \\
y &= 24
\end{align*}

Therefore, there are 7 goannas and 24 emus.

Question 35a

Question 35b

Question 36

Question 36a

40 minutes.

Question 36b

ACDEGJK

Question 36c

Question 37

Question 38a

Question 38b

Question 39a

Question 39b

Question 40

Question 41

Let \(X\) be a female’s height in the population and \(Y\) be a male’s height.

\begin{align*}
P(X< 175) &= 97.5\% \\
P(X < 160.6) &= 16\% \\
P(Y < h) &= 84\%
\end{align*}

Our aim is the find \(h\). Use the empirical rule with the information given above.

\begin{align*}
\mu + 2\sigma &= 175 & \text{(1)} \\
\mu – \sigma &= 160.6 & \text{(2)}
\end{align*}

Subtracting equation (1) by (2) gives

\begin{align*}
3 \sigma &= 14.4 \\
∴ \sigma &= 4.8
\end{align*}

Substitute \(\sigma = 4.8\) into (1),

\begin{align*}
\mu + 2(4.8) &= 175 \\
\mu = 165.4
\end{align*}

Hence, for the adult male population, the mean is \(1.05 \mu = 173.67\) and the standard deviation is \(1.1 \sigma = 5.28\).

Since being 84 % taller than the male population is 1 standard deviation above the mean, then we can calculate \(h\) as

\begin{align*}
h &= 1.05 \mu + 1.1 \sigma \\
&= 173.67 + 5.28 \\
&= 178.95 \; \text{cm}
\end{align*}