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The 2022 HSC Maths Std 2 Exam was on Monday 15 November. Read on to see how our Maths Team would have answered things.
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The 2022 HSC Maths Std 2 Exam was on Thursday 20 October. Take a look and see how our Maths Team would have answered things.
We wasted no time in compiling our solutions for the this years paper. We hope you find some common ground with your own solutions!
In this post, we will work our way through the 2022 HSC Maths Standard 2 Exam Paper and give you the solutions, written by our Head of Mathematics Mr. Oak Ukrit and his team.
| Question | Answer | Explanation |
| 1 | C | Negatively skewed distributions have values concentrated towards the right. |
| 2 | A | The equation \(y=-2x+2\) is a line. The y-intercept is 2, which eliminates options B and C as they have negative y-intercepts. The gradient is -2 which is negative, leaving only option A as the correct answer. |
| 3 | B | The critical path is the longest path taken to complete the project. Path ACI takes 3 + 13 + 2 = 18. Path ADG takes 3 + 16 + 3 = 22. Path BEH takes 2 + 9 + 6 = 17. Path BEFG takes 2 + 9 + 2 + 3 = 16 Out of these, path ADG takes the longest time, hence is the correct answer. |
| 4 | D | Let the number of fish in the lake be \(x\). We can equate the proportion of tagged fish in the lake and the 40-fish sample: \(\begin{align*} |
| 5 | D | The new dataset in ascending order is: 5, 13, 16, 17, 17, 21, 24. The median remains to be 17. The original mean is \(\frac{13+16+17+17+21+24}{6} = 18\) and the new mean is \(\frac{5+13+16+17+17+21+24}{7} = \frac{113}{7}\) \(\approx 16.14 \quad \text{(2 decimal places)}\). |
| 6 | B | One-third of a day is \(\frac{24}{3} = 8\) hours. 8 hours is equivalent to \(8 \times 60 = 480\) minutes. The 20 minutes to one-third of a day ratio \(20:480 = 1:24\). |
| 7 | C | Each day, Tian is paid \(9 \times \$20.45 + \$16.20\). For 5 days, this amounts to \(5 \times (9 \times \$20.45+\$16.20) = $1001.25\). |
| 8 | C | \(180^\circ – 135^\circ = 45^\circ\) |
| 9 | A | From the plot, the curve lies above \(h=300\) for values between \(t = 6\) & \(t = 10\). Hence the object is above \(300m\) for \(10-6 = 4\) seconds. |
| 10 | B | For each share, Alex makes \($3.40-$2.60 = $0.80\) profit. On 800 shares, this is equal to \(800 \times \$0.80 = \$640\) profit. After deducting the fee, Alex makes a total profit of \(\$640 – \$24.95 = \$615.05\). |
| 11 | D | Since the investment is compounded half-yearly, the rate to discount by is \(4\% \div 2 = 2\%\) or \(0.02\), and the number of periods is \(10 \times 2 = 20\). Using the present value formula, \(PV = \frac{150000}{(1+0.02)^{20}}\). |
| 12 | B | When \(H = 0\), the equation becomes \(M = 20 + 3 \times 0 = 20\). This means the predicted mark of a student will be \(20\) for zero hours of study per week, which corresponds to option B. |
| 13 | B | The probability the random variable is less than \(1.94\) is equal to \(0.9738\) from the provided table. The probability the random variable is less than \(0\) is \(0.5\), as the standard normal distribution is symmetrical. Hence, the probability of between \(0\) and \(1.94\) is equal to \(0.9738-0.5 = 0.4738\). |
| 14 | A | \( \begin{align*} y &= \frac{ax-b}{2} \\ 2y &= ax – b \\ ax &= 2y + b \\ x &= \frac{2y + b}{a} \end{align*} \) |
| 15 | C | The box plot displays the number of downloads at the \(25\)th \((Q_1)\), \(50\)th \((Q_2\) or median), and \(75\)th \((Q_3)\) percentiles. From the cumulative frequency graph, this corresponds to 3, 6 and 7 downloads. The only correct option is C. |
(a)
| \(220-25 = 195\) |
(b)
| \( 50\% \times 195 = 97.5\\ \) \( 85\% \times 195 = 165.75\\\) Tom would benefit most from heart rates between \(97.5\) bpm and \(165.75\)bpm. |
(a)
| Looking at the diagram below, there are 6 ways out of 9 in which Bob ends up with a higher card than Amy. Hence, the probability that Bob wins \(\frac{6}{9} = \frac{2}{3}\). |
(b)
| \(\frac{2}{3} \times 30 = 20\) times. |
| The \(z\)-score of 90 is: $$z = \frac{90-60}{15} = 2$$ By the empirical (68-95-99.7) rule, there are \(\frac{100\%-95\%}{2} = 2.5\%\) marks higher than \(90\). |
(a)
| The value of A can be found by adding the frequency of delivery fee complaints to the cumulative frequency of the previous category (stock shortage). $$A = 98 + 62 = 160$$ The value of B can be found by dividing the cumulative frequency of the ‘Damaged item’ category by the total number of complaints. $$B =\frac{192}{200}\times 100=96$$ |
(b)
| The line graph on the Pareto chart corresponds to the cumulative percentage, with the values represented by the axis on the right of the chart. The cumulative percentage reaches a value of \(80%\) within the ‘Delivery Fee’ so \(80%\) of the complaints are stock shortages and delivery fees. |
(a)
| The weighted network diagram is: |
(b)
| The minimum spanning tree is a spanning tree connecting all the vertices together, with the smallest possible total edge weight. By connecting vertices S, B, C, Y, M, the length of this minimum spanning tree is \(275+150+60+530=1015\). |
| \begin{align*} \text{Commission} &= 800000 \times 2\% + (1500000 – 800000) \times 1.5\% \\ &= 16000 + 10500 \\ &= 26500 \end{align*} |
(a)
| \begin{align*} \text{Cost} &= 100 \times 1943 + 20180 \\ &= 194300 + 20180 \\ &= 214480 \end{align*} |
(b)
| \begin{align*} C &= 100n + an + 20180 \\ 97040 &= 100n + 26 n + 20180 \\ 97040 – 20180 &= 100n + 26n \\ 76860 &= 126n \\ n &= \frac{76860}{126} \\ n &= 610 \quad \text{laptops} \end{align*} |
(a)
| The scatterplot is shown below with the new points and line of best fit in red: |
(b)
| \(9\) hours |
(a)
| Since \(M\) varies inversely with \(T\), we can formulate this as \(M = \frac{k}{T}\). Substitute \(M = 12 and T = 15\) into the equation to find \(k\): \begin{align*} |
(b)
To complete the table, the values of \(M\) can be found substituting each \(T\) value into \(M=\frac{180}{T}\) from part (a).
|
| The relationship can then be graphed by plotting each of the points from the table. The line connecting the points should be curved, as the relationship \(M=\frac{180}{T}\) is fo the form \(y=\frac{k}{x}\), which is a hyperbola. Note: The question specifies “from T = 5 degrees Celsius to T = 30 degrees Celsius”, so the graph should not extend past these values. |
(a)
| \begin{align*} \text{Yearly deposit} \; &= \frac{15000}{4.184} \\ &= $3585.09 \end{align*} |
(b)
| \begin{align*} \text{Interest} \; &= 15000-4 \times 3585.09 \\ &= \$660 \quad \text{(nearest dollar)} \end{align*} |
(a)
| Firstly, consider \(\triangle ABC\): $$\begin{align*} |
(b)
| \begin{align*} \sin \theta &= \frac{AB}{93} \\ &= \frac{35 \tan 41^\circ}{93} \\ \theta &= \sin^{-1} \left(\frac{35 \tan 41^\circ}{93}\right) \\ &= 19^\circ 6′ \quad \text{(nearest minute)} \end{align*} |
(a) i
| \begin{align*} \text{Depreciation rate} \; &= 1 – 0.80 \\ &= 0.2 \; \text{or} \; 20\%\\ \end{align*} |
(a) ii
| \begin{align*} S &= 50000 \times (0.80)^3 \\ &= $25600 \end{align*} |
(b)
| From the reference sheet, the straight-line depreciation formula is \[S = V_0 – Dn\] We have \(S = 25600\), \(V_0 = 50000\) and \(D = 12.2\% \times 50000 = 6100\). So our formula becomes \begin{align*} 25600 &= 50000-6100n \\ \\ -24400 &=-6100n \\ \\ n &= 4 \\ \\ ∴ \text{In 4 years} \end{align*}
|
| \begin{align*} \text{Area of triangular base} &= \frac{1}{2} \times 25 \times 25 \times \sin 150^\circ \\ &= 156.25 \; \mathrm{m}^2 \end{align*}\begin{align*} \mathrm{Volume} &= \text{Area of triangular base} \times 50 \\ &= 156.25 \times 50 \\ &= 7812.5 \mathrm{m}^3 \\ &= 7 812 500 \mathrm{L} \end{align*} |
| 8:20 pm Wednesday +20 hours and 24 minutes = 4:44 pm Thursday So, the plane arrives in New York at 4:44 pm Thursday (Sydney time).New York is 15 hours behind Sydney. 4:44 pm Thursday – 15 hours = 1:44 am Thursday. Therefore, plane arrives in New York at 1:44 am Thursday (New York time). |
(a)
| $$40000\left(1 + \frac{0.012}{12}\right)^{10\times 12} = $45097.17$$ |
(b)
| Quarterly interest rate \(= \frac{0.024}{4} = 0.006 \\\) Number of periods \(= 10 \times 4 = 40 \\\) Future value interest rate (from table) \(= 45.05630 \\\) \begin{align*} \text{Option 2} \; &= $1000 \times 45.05630 \\ &= $45056.30 \end{align*} \begin{align*} \text{Difference} \; &= $45097.17 – $45056.30 \\ &= $40.87 \end{align*} |
(a)
| Note that the flow capacity of a cut only includes flows from the source to the sink. Flow capacity \(= 10 + 20 + 10 = 40\). |
(b)
(b)
| The network diagram with minimum cut in red is shown above. The flow capacity of the minimum cut \( = 15 + 10 + 10 = 35\). By the Maximum-Flow Minimum-Cut theorem, the maximum flow capacity must be \(35\) as well, which is less than 40. |
(c)
| We should select a path that is cut by the minimum cut. To allow for an overall maximum flow of \(40\), the path \(AE\) can be increased by \(5\) to a capacity of \(20\). |
(a)
| \begin{align*} \text{Area of grassed section} \; &= \frac{100}{2}(160+150) + \frac{100}{2}(150+250) \\ &= 35500 \; \mathrm{m}^2 \end{align*} |
(b)
| \begin{align*} \text{Area of rectangle} \; &= 200 \times 250 \\ &= 50000\; \mathrm{m}^2 \end{align*} \begin{align*} \text{Area of semi-circle} \; &= \frac{\pi \times 100^2}{2} \\ &= 5000\pi \; \mathrm{m}^2 \end{align*} \begin{align*} \text{Therefore, Area of lake} \; &= 50000 + 5000\pi – 35500 \\ &= 30208 \; \mathrm{m}^2 \quad \text{(nearest square metre)} \end{align*} |
(a)
| \(\angle AOB = 180^\circ – 135^\circ = 45^\circ \\\) Using the cosine rule, \begin{align*} AB^2 &= 50^2 + 20^2 – 2(50)(20)(\cos 45^\circ) \\ \\ AB &= \sqrt{50^2 + 20^2 – 2(50)(20)(\cos 45^\circ)} \\ \\ &= 38.5 \; \mathrm{km} \quad \text{(1 decimal place)} \end{align*} |
(b)
| Using the sine rule: \begin{align*} \frac{\sin \angle OBA}{20} &= \frac{\sin 45^\circ}{AB} \\ \\ \sin \angle OBA &= \frac{20 \sin 45^\circ}{38.5} \\ \\ \angle OBA &= \sin^{-1}\left(\frac{20 \sin 45^\circ}{38.5}\right) \\ \\ &= 22^\circ \quad \text{(nearest degree)} \end{align*} |
(c)
| Bearing of \(B\) from \(A\) is \(180^\circ + 22^\circ = 202^\circ \mathrm{T}\) |
| Surface area of cylinder’s base \(= \pi \times 2^2 = 4\pi \\\) Surface area of cylinder’s curved surface \(= 2 \pi \times 2 \times 3 = 12\pi \\\) Surface area of hemisphere’s annulus base \(= \pi \times 3^2 – \pi \times 2^2 = 5\pi \\\) Surface area of hemisphere’s curved surface \(= \frac{4\pi \times 3^2}{2} = 18\pi \\\)Therefore, Total surface area = \(4\pi + 12\pi + 5\pi + 18\pi = 122.5 \; \mathrm{cm}^2 \quad \text{(1 decimal place)}\) |
|
(a)
| \(A = $199715 \times \frac{0.072}{12} = $1198.29 \\\) \begin{align*} B &= $199428.29 + $1196.57 – $1485 \\ &= $199139.86 \end{align*} |
(b)
| Without lump sum payment, the total amount repaid is: \(1485 \times 23 \times 12 = $409860 \\\) \($1485 \times 50 + $40000 + \$1485 \times 146 = $331060 \\\) |
| Split the range 820-920 at the mean to produce two sub-ranges: 820-840 and 840-920. For \(840 – 920 \\\): So, \(\frac{68}{2} = 34\%\) lie between \(840-920\). For \(820-840 \\\): Again due to symmetry about the mean, this means \(10%\) lie between \(820-840\). Overall: |
| After removing \(1.2L\) of the mixture, there is \(4.8-1.2 = 3.6L\) remaining. There is \(3.6 \times \frac{1}{4} = 0.9L\) of cordial. Refilling the container by adding another \(1.2L\) of water means there is now \(2.7 + 1.2 = 3.9L\) of water. The final mixture has \(0.9L\) of cordial and \(3.9L\) of water. |
Written by Matrix Maths Team
The Matrix Maths Team are tutors and teachers with a passion for Mathematics and a dedication to seeing Matrix Students achieving their academic goals.
© 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.
