HSC Maths Standard 2 2025

Every question from the 2025 HSC Mathematics Standard 2 exam, with full worked solutions. Section II solutions are tucked behind a Show worked solution toggle, so you can attempt a question first and reveal the model answer when you are ready.

How to use this page

• Questions are from the 2025 HSC Mathematics Standard 2 exam, copyright NSW Education Standards Authority (NESA). Open the official PDF (button above) for the original stimulus diagrams, networks and graphs.
• Answers are original model responses by ExamExplained (Claude Opus 4.8), written to the official marking guidelines, not copied from NESA's sample answers.
• Each Section II solution shows the mark split and a short Marker's note from the HSC marking feedback.

Structure and timing

100 marks in 150 minutes is about 1.5 minutes per mark.

• Section I (15 marks): 15 multiple-choice. Allow about 25 minutes.
• Section II (85 marks): Questions 16 to 40, short and extended response. Allow about 2 hours and 5 minutes, in proportion to the marks. Show relevant reasoning and calculations, and round only at the final step.

Section I - Multiple choice

Q1

Consider the network diagram (vertices A, B, C, D, E, F). Which vertex has degree 4? A. A B. B C. C D. D
Answer: B - vertex B is the one joined to four edges, so its degree is 4.

Q2

Which graph could represent $y = \dfrac{4}{x}$? A, B, C or D.
Answer: A - $y = \tfrac{4}{x}$ is a hyperbola in the first and third quadrants, so option A's two-branch shape fits.

Q3

A network shows the distances in kilometres along roads connecting towns. What is the value of the largest weighted edge included in the minimum spanning tree? A. 7 B. 8 C. 9 D. 10
Answer: C - building the cheapest tree that joins every town without a cycle, the heaviest edge that must be kept is 9.

Q4

Frankie takes four weeks of annual leave. His weekly pay is $350 and his leave loading is$17.5%$of four weeks pay. What is his total leave pay? A.$245.00 B. $411.25 C.$1461.25 D. $1645.00
Answer: D - four weeks pay is $4 \times 350 = 1400$; loading is $0.175 \times 1400 = 245$; total $1400 + 245 = 1645$.

Q5

Which is arranged from largest to smallest? A, B, C or D (values $3.2 \times 10^{-1}$, $6.2 \times 10^{-3}$, $4.5 \times 10^{-4}$).
Answer: D - $3.2 \times 10^{-1} = 0.32$ is largest, then $6.2 \times 10^{-3} = 0.0062$, then $4.5 \times 10^{-4} = 0.00045$.

Q6

For $n = \dfrac{m - p}{q}$, which shows $p$ as the subject? A. $p = nq - m$ B. $p = m - nq$ C. $p = n + q - m$ D. $p = m - n - q$
Answer: B - multiplying by $q$ gives $nq = m - p$, so $p = m - nq$.

Q7

A school has 960 students; 240 are chosen by stratified sampling. If 200 are in Year 12, how many Year 12 students should be chosen? A. 4 B. 5 C. 50 D. 60
Answer: C - the sample takes $\tfrac{240}{960} = \tfrac14$ of each group, so $\tfrac14 \times 200 = 50$.

Q8

A spinner of 4 colours is spun 100 times; a frequency histogram is shown. Which spinner is most likely to give those results? A, B, C or D.
Answer: A - the spinner whose sector sizes match the relative frequencies (the colour with the tallest bar has the largest sector) is option A.

Q9

The ratio of model car to actual car is $1 : 64$. The actual car is 4.9 m long. The model length in cm, to 1 decimal place? A. 3.1 B. 7.7 C. 13.1 D. 59.1
Answer: B - $4.9 \text{ m} = 490 \text{ cm}$, and $490 \div 64 = 7.65... \approx 7.7$ cm.

Q10

Cost $C$ in cents: charge 37c per kWh used $U$, pay 5c per kWh produced $P$, plus 71c per day. Which formula? A. $C = 71 + 37U - 5P$ B. $C = 71 + 37U + 5P$ C. $C = 71 - 37U - 5P$ D. $C = 71 - 37U + 5P$
Answer: A - the daily fee and usage add to the cost, while electricity produced is paid back (subtracted), giving $C = 71 + 37U - 5P$.

Q11

Skin thickness varies inversely with surface area of a sphere. If the radius is doubled, the thickness is: A. Divided by 2 B. Multiplied by 2 C. Divided by 4 D. Multiplied by 4
Answer: C - surface area $4\pi r^2$ is multiplied by 4 when $r$ doubles, and inverse variation then divides the thickness by 4.

Q12

A game in Quito (UTC $-5$) starts 3:40 pm Tuesday. What is the time in Sydney (UTC $+10$)? A. 10:40 am Tuesday B. 8:40 pm Tuesday C. 12:40 am Wednesday D. 6:40 am Wednesday
Answer: D - Sydney is 15 hours ahead of Quito, so 3:40 pm Tue $+ 15$ h $=$ 6:40 am Wednesday.

Q13

A ten-sided die has a 1 more likely than the other faces (2 to 10 equally likely). In 153 rolls a 1 came up 72 times. Best estimate for $P(10)$? A. $\tfrac{1}{17}$ B. $\tfrac{1}{11}$ C. $\tfrac{1}{10}$ D. $\tfrac{1}{9}$
Answer: A - relative frequency of a 1 is $\tfrac{72}{153}$, leaving $\tfrac{81}{153}$ shared by nine equal faces, so $P(10) = \tfrac{81}{153 \times 9} = \tfrac{1}{17}$.

Q14

M and P are equidistant from R. The bearing of M from R is $017$ degrees and of P from R is $107$ degrees. The bearing of P from M is: A. Between $000$ and $090$ degrees B. Exactly $090$ degrees C. Between $090$ and $180$ degrees D. Exactly $180$ degrees
Answer: C - triangle RMP is isosceles with a $90$ degree angle at R, so the equal base angles are $45$ degrees each, putting the bearing of P from M between $090$ and $180$ degrees.

Q15

Minimum daily temperatures are normally distributed with the mean equal to the standard deviation. What percentage of recorded minimums was above zero degrees? A. 16% B. 50% C. 68% D. 84%
Answer: D - zero is one standard deviation below the mean, so the area above it is $50\% + 34\% = 84\%$.

Section II - Short and extended response

Question 16 (2 marks)

Paint is sold in two sizes at a local shop.

• Four-litre cans at $90 per can
• Ten-litre cans at $205 per can

Mina needs to purchase 80 litres of paint. Calculate the amount of money Mina will save by purchasing only ten-litre cans of paint rather than only four-litre cans of paint.

Show worked solution

[2 marks]. Ten-litre cans: $80 \div 10 = 8$ cans, so $8 \times \$205 = \$1640$. Four-litre cans: $80 \div 4 = 20$ cans, so $20 \times \$90 = \$1800$.

$$\text{Saving} = 1800 - 1640 = \$160.$$

Marker's note. Cost both options for the whole 80 litres, then show the difference. The saving is for all 80 litres, not for one litre.

Question 17 (3 marks)

The scatter plot shows a bivariate dataset, where $x$ is the independent variable and $y$ is the dependent variable. The points $(0, 14)$ and $(5, 4)$ lie on the line of best fit.
(Stimulus: a scatterplot on axes from 0 to 6 in $x$ and 0 to 18 in $y$ - see the official paper.)
Plot the points $(0, 14)$ and $(5, 4)$ on the graph and hence find the equation of the line of best fit.

Show worked solution

[3 marks]. With the two given points the gradient is

$$m = \frac{4 - 14}{5 - 0} = \frac{-10}{5} = -2.$$

The point $(0, 14)$ is the y-intercept, so $c = 14$ and

$$y = -2x + 14.$$

Marker's note. Plot both points with a ruled line, read the y-intercept from $(0, 14)$, and note the negative gradient (the line slopes down left to right). Write the answer in the form $y = mx + c$.

Question 18 (2 marks)

A table of future value interest factors for an annuity of $1 is shown (rates 1.5%, 3%, 4.5%, 6% against periods 5, 10, 20, 40).
(Stimulus: a future-value annuity table - see the official paper.)
The prize in a lottery is an annuity of $5000 a year for 10 years, invested at 4.5% per annum compounding annually. What will be the value of the prize at the end of 10 years?

Show worked solution

[2 marks]. Read the factor for 4.5% and 10 periods (12.288) and multiply by the yearly payment:

$$\$5000 \times 12.288 = \$61\,440.$$

Marker's note. Use the supplied table, not a formula. Pick the factor at the matching rate and number of periods, then multiply by the $5000 annual payment.

Question 19 (5 marks)

The activities and durations (days) for a project are shown in a network diagram (A 3, B 4, C 3, D 5, E 5, F 7, G 3, H 5).
(Stimulus: an activity network with eight activities A to H - see the official paper.)
(a) Complete the table showing the immediate prerequisites for each activity (rows for B, E, F). (2 marks)
(b) Find the critical path for this project AND state the minimum duration for the project. (2 marks)
(c) 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 worked solution

(a) [2 marks]. Reading the arrows into each activity:

Activity	Immediate prerequisite(s)
B	(none)
E	C, D
F	E

(b) [2 marks]. The longest path through the network is BDEFH, with minimum duration

$$4 + 5 + 5 + 7 + 5 = 26 \text{ days}.$$

(c) [1 mark]. No. Activity A is not on the critical path and has a float of 3 days, so adding 2 days to A is absorbed by its float and the critical path is unchanged.

Marker's note. Transfer the prerequisites from the diagram into the table, identify the longest (critical) path and write down its duration. For (c), an increase only matters if it exceeds the activity's float.

Question 20 (3 marks)

The graph of $h = t^2 - 8t + 12$ is shown.
(Stimulus: an upward parabola crossing the t-axis at 2 and 6 - see the official paper.)
(a) Find the values of $t$ and $h$ at the turning point of the graph. (2 marks)
(b) The graph shows $h = 12$ when $t = 0$. What is the other value of $t$ for which $h = 12$? (1 mark)

Show worked solution

(a) [2 marks]. The turning point sits on the axis of symmetry, halfway between the roots $t = 2$ and $t = 6$:

$$t = \frac{2 + 6}{2} = 4, \qquad h = 4^2 - 8(4) + 12 = -4.$$

So the turning point is $(4, -4)$.

(b) [1 mark]. By symmetry about $t = 4$, the value $t = 0$ is matched by $t = 8$, so $h = 12$ again when $t = 8$.

Marker's note. The turning point is the vertex, not an intercept: use the axis of symmetry to find $t$, then substitute back for $h$. Part (b) follows from the symmetry of the parabola about $t = 4$.

Question 21 (3 marks)

A reverse-cycle air conditioner uses 2.5 kW for cooling and 3.2 kW for heating. Electricity costs 29 cents per kWh.
(a) Find the cost, in dollars and cents, of cooling the house for 6 hours. (1 mark)
(b) The cost of operating the air conditioner to heat the house during winter last year was $640. There are 92 days in winter. Find the number of hours, to 1 decimal place, that the air conditioner was used on average per day. (2 marks)

Show worked solution

(a) [1 mark].

$$\text{Cost} = 2.5 \times 6 \times \$0.29 = \$4.35.$$

(b) [2 marks]. Cost per day $= 640 \div 92 = \$6.956...$; cost to run one hour for heating $= 3.2 \times 0.29 = \$0.928$. So

$$\text{Hours per day} = \frac{6.956...}{0.928} = 7.49... \approx 7.5 \text{ hours}.$$

Marker's note. Convert cents to dollars and keep the cooling and heating powers separate. Do not simply divide the total cost by the number of days; bring in the heating power and the price per kWh.

Question 22 (5 marks)

A network of pipes with one cut is shown; each edge gives the capacity in L/min.
(Stimulus: a flow network with source A, sink G, and a marked cut - see the official paper.)
(a) What is the capacity of the cut shown? (1 mark)
(b) The diagram shows a possible flow for this network of pipes.
(Stimulus: the same network with a flow assigned to each pipe - see the official paper.)
(i) What is the value of $x$? Give a reason for your answer. (2 marks)
(ii) Which of the pipes in the flow are at full capacity? (1 mark)
(iii) The maximum flow for this network is 50 L/min. Which path of pipes could have an increase in flow of 2 L/min to achieve the maximum flow? (1 mark)

Show worked solution

(a) [1 mark]. The cut crosses three pipes flowing from the source side to the sink side, with capacities 26, 24 and 12:

$$\text{Capacity of cut} = 26 + 24 + 12 = 62 \text{ L/min}.$$

(b)(i) [2 marks]

At vertex C the inflow must equal the outflow. The flow leaves C as $5 + 13 + 12 = 30$, so the inflow $x = 30$ L/min.

(b)(ii) [1 mark]

The pipes carrying flow equal to their capacity are DE, CF, DG and FG.

(b)(iii) [1 mark]

The path ACEG still has spare capacity along every pipe, so it can carry an extra 2 L/min to reach the maximum flow of 50 L/min.

Marker's note. Count only the pipes crossing the cut from source towards sink. Use conservation of flow (inflow $=$ outflow) at a vertex for $x$, and look for a path with spare capacity on every edge for the increase.

Question 23 (2 marks)

Company A and Company B both issue an annual dividend per share:

• Company A: current share price $25.43, annual dividend per share$4.92
• Company B: current share price $2.13, annual dividend per share 45c

Based on the dividend yield, which company would be better to invest in?

Show worked solution

[2 marks]. Dividend yield $=$ dividend $\div$ price:

$$\text{A} = \frac{4.92}{25.43} \times 100 \approx 19\%, \qquad \text{B} = \frac{0.45}{2.13} \times 100 \approx 21\%.$$

Company B has the higher yield, so Company B is the better investment.

Marker's note. Compute both yields and compare; a higher dividend yield is the better return. Finish with a clear concluding statement naming the chosen company.

Question 24 (2 marks)

The population of snails in a garden is approximately 90. One night Bobbie collected 18 snails, tagged each and released them. The next night 20 snails were captured. Approximately how many of the snails in the second sample are expected to have a tag?

Show worked solution

[2 marks]. The fraction of tagged snails in the garden is $\tfrac{18}{90} = \tfrac15$. Applying that fraction to the second sample of 20:

$$\frac{18}{90} \times 20 = 4 \text{ snails}.$$

Marker's note. This is capture-recapture: the proportion tagged in the whole garden equals the expected proportion tagged in the new sample. Multiply $\tfrac{18}{90}$ by 20.

Question 25 (6 marks)

Participants recorded their average daily minutes watching television ($x$) and exercising ($y$); the averages are graphed.
(Stimulus: a scatterplot of exercise minutes against television minutes - see the official paper.)
(a) Describe the bivariate dataset in terms of its form and direction. (2 marks)
(b) The equation of the least-squares regression line is $y = 64.3 - 0.7x$. Interpret the values of the slope and y-intercept in the context of this dataset. (2 marks)
(c) Jo spends an average of 42 minutes per day watching television. Use the equation to determine how many minutes on average Jo is expected to exercise each day. (1 mark)
(d) Explain why it is NOT appropriate to extrapolate the regression line to predict the average exercise minutes for someone who watches an average of 2 hours of television per day. (1 mark)

Show worked solution

(a) [2 marks]

Form: linear. Direction: negative (as television time rises, exercise time falls).

(b) [2 marks]

Slope: for every extra minute per day watching television, the time spent exercising decreases by 0.7 minutes per day. Y-intercept: someone who watches no television is expected to exercise for 64.3 minutes per day.

(c) [1 mark]

$$y = 64.3 - 0.7 \times 42 = 34.9 \text{ minutes}.$$

(d) [1 mark]. Two hours is 120 minutes, which gives $y = 64.3 - 0.7 \times 120 = -19.7$ minutes. A negative exercise time is impossible, and 120 minutes is outside the range of the data, so the prediction is unreliable.

Marker's note. Use the correct statistical language for form and direction, and interpret the slope and intercept in context with their values. Substitute carefully in (c), and in (d) note that 120 minutes lies outside the data and gives a negative (impossible) result.

Question 26 (6 marks)

A toy has a curved (shaded) top surface, a uniform cross-section and a rectangular base. Measurements (cm): width 40.0, depth 6.0, and heights 10.2 and 3.8 across the cross-section.
(Stimulus: a solid toy with a curved top and a labelled cross-section - see the official paper.)
(a) Use two applications of the trapezoidal rule to find an approximate area of the cross-section of the toy. (2 marks)
(b) The total surface area of the plastic toy is 1300 cm squared. What is the approximate area of the curved surface? (2 marks)
(c) The measurements are given to the nearest millimetre. What is the percentage error of the measurement of 10.2 cm? Give your answer correct to 3 significant figures. (2 marks)

Show worked solution

(a) [2 marks]. With strip width $h = 5.1$ and the three heights 6, 3.8 and 0:

$$A \approx \frac{5.1}{2}(6 + 3.8) + \frac{5.1}{2}(3.8 + 0) = 34.68 \text{ cm}^2.$$

(b) [2 marks]. The total surface is the two cross-section ends, the two flat rectangular faces and the curved top:

$$1300 = (2 \times 34.68) + (10.2 \times 40) + (6 \times 40) + \text{curved},$$

$$1300 = 717.36 + \text{curved}, \qquad \text{curved} = 582.64 \text{ cm}^2.$$

(c) [2 marks]. To the nearest millimetre the absolute error is half of 0.1 cm $= 0.05$ cm:

$$\text{Percentage error} = \frac{0.05}{10.2} \times 100\% = 0.490\% \text{ (3 significant figures)}.$$

Marker's note. Use $h = 5.1$ in the trapezoidal rule and keep the units consistent. The absolute error is half the smallest unit (0.05 cm), and the answer in (c) must be rounded to 3 significant figures.

Question 27 (3 marks)

A credit card has an interest-free period of 45 days from and including the date of purchase. Interest then compounds daily at 13.74% per annum from the day after the interest-free period. Concert tickets costing $392 were purchased and paid in full on the 68th day. What was the total interest charged?

Show worked solution

[3 marks]. Interest is charged for $68 - 45 = 23$ days at a daily rate of $\tfrac{13.74\%}{365}$:

$$A = 392 \left(1 + \frac{0.1374}{365}\right)^{23} = \$395.41.$$

$$\text{Interest} = 395.41 - 392 = \$3.41.$$

Marker's note. Find the number of days that attract interest (23), convert the annual rate to a daily rate, and use daily compounding. The interest is the amount owing minus the original $392.

Question 28 (3 marks)

The heights of students in a class are displayed in a cumulative frequency graph. The shortest student is 130 cm and the tallest is 180 cm.
(Stimulus: a cumulative frequency graph (ogive) of heights - see the official paper.)
Construct a box-plot for this class.

Show worked solution

[3 marks]. Reading the quartiles from the ogive gives the five-number summary: minimum 130, $Q_1 = 135$, median $= 140$, $Q_3 = 160$, maximum 180.

Marker's note. Read the quartiles from the cumulative frequency graph at the 25%, 50% and 75% points, then draw the box and whiskers to scale on the given axis. State the five-number summary.

Question 29 (2 marks)

A bag contains 7 blue lollies and 9 yellow lollies. One lolly is selected at random and eaten, then a second lolly is selected from those remaining. Find the probability that the two lollies selected are the same colour.

Show worked solution

[2 marks]. Selection is without replacement, so the bag has 15 lollies for the second draw:

$$P(\text{both blue}) = \frac{7}{16} \times \frac{6}{15}, \qquad P(\text{both yellow}) = \frac{9}{16} \times \frac{8}{15}.$$

$$P(\text{same colour}) = \frac{42}{240} + \frac{72}{240} = \frac{114}{240} = \frac{19}{40}.$$

Marker's note. Without replacement, the second denominator drops to 15. Multiply along each branch, add the two same-colour outcomes, and simplify the final fraction.

Question 30 (2 marks)

It costs $465 to register a passenger car and$350 to register a motorcycle. Let $P$ be the number of passenger cars and $B$ the number of motorcycles. Write TWO linear equations representing:

• There are 11 times as many passenger cars as motorcycles.
• The total registration fees for passenger cars and motorcycles is $494 million.

Show worked solution

[2 marks]. From the two statements:

$$P = 11B,$$

$$465P + 350B = 494\,000\,000.$$

Marker's note. Use the supplied pronumerals $P$ and $B$. The first relationship is a simple multiple; the second multiplies each fee by the count and sums to the total in dollars.

Question 31 (3 marks)

The 2024-2025 resident income tax table is given. Alex's tax payable was $47 420 (excluding the Medicare levy). What was Alex's taxable income?
(Stimulus: the 2024-2025 income tax rate table - see the official paper.)

Show worked solution

[3 marks]. Since $47\,420$ lies between $31\,288$ and $51\,638$, Alex is in the $135\,001$ to $190\,000$ bracket: tax $= 31\,288 + 0.37(\text{income} - 135\,000)$.

$$47\,420 = 31\,288 + 0.37(\text{income} - 135\,000),$$

$$16\,132 = 0.37(\text{income} - 135\,000), \qquad \text{income} - 135\,000 = 43\,600.$$

$$\text{Taxable income} = 43\,600 + 135\,000 = \$178\,600.$$

Marker's note. Choose the correct bracket from the size of the tax, convert 37 cents to 0.37, and work backwards. Keep the fixed amount ($31 288) separate from the marginal part.

Question 32 (3 marks)

Solid spheres are placed inside a square-based pyramid. The base has side length 14 cm and the height is $h$ cm. Each sphere has radius 1.5 cm. After 30 spheres are placed, the empty space remaining is 634 cm cubed. What is the height $h$, to the nearest centimetre?
(Stimulus: a square-based pyramid containing solid spheres - see the official paper.)

Show worked solution

[3 marks]. Volume of 30 spheres:

$$30 \times \frac{4}{3}\pi(1.5)^3 = 424.115... \text{ cm}^3.$$

The pyramid holds the spheres plus the empty space:

$$V_{\text{pyramid}} = 424.115... + 634 = 1058.115... \text{ cm}^3.$$

Using $V = \tfrac13 \times \text{base area} \times h$ with base area $14 \times 14 = 196$:

$$1058.115 = \frac{1}{3} \times 196 \times h, \qquad h = \frac{1058.115 \times 3}{196} = 16.2... \approx 16 \text{ cm}.$$

Marker's note. The pyramid volume is the spheres' volume plus the empty space. Multiply one sphere's volume by 30, add 634, then solve the pyramid formula for $h$.

Question 33 (5 marks)

A used car has a sale price of $24 200. In addition: transfer of registration$50, and stamp duty at $3 for every$100 (or part thereof) of the sale price. Kat borrows the total amount payable. Simple interest at 6.8% per annum is charged, and the loan is repaid in equal monthly repayments over 3 years. Calculate Kat's monthly repayment.

Show worked solution

[5 marks]. Stamp duty: $24\,200 \div 100 = 242$ lots of $100, so$242 \times $3 = $726$.

Amount borrowed:

$$726 + 24\,200 + 50 = \$24\,976.$$

Simple interest over 3 years:

$$I = 24\,976 \times 0.068 \times 3 = \$5095.10.$$

Total to repay:

$$24\,976 + 5095.10 = \$30\,071.10.$$

Monthly repayment over $3 \times 12 = 36$ months:

$$30\,071.10 \div 36 = \$835.31.$$

Marker's note. Calculate stamp duty on the sale price, add the registration transfer, then charge interest on the whole amount borrowed (not just the car price). Divide the total owing by 36 months.

Question 34 (3 marks)

A table of future value interest factors for an annuity of $1 is given.
(Stimulus: a future-value annuity table for rates 0.005 to 0.06 and periods 7, 28, 56, 84 - see the official paper.)
Lin invests a lump sum of $21 000 for 7 years at 6% per annum compounding monthly. Yemi wants the same future value using an annuity, depositing a fixed amount at the end of each month for 7 years at 6% per annum compounding monthly. Using the table, determine how much Yemi needs to deposit each month.

Show worked solution

[3 marks]. Monthly rate $r = \tfrac{6\%}{12} = 0.005$ over $n = 7 \times 12 = 84$ months. Lin's future value:

$$21\,000 \times (1.005)^{84} = \$31\,927.76.$$

The table factor at $r = 0.005$, $n = 84$ is 104.07393, so Yemi's monthly deposit is

$$\frac{31\,927.76}{104.07393} = \$306.78.$$

Marker's note. Convert the rate and periods to the monthly compounding basis ($r = 0.005$, $n = 84$). Find Lin's future value first, then divide by the matching table factor to get the annuity deposit.

Question 35 (3 marks)

The triangle PTA is shown. PA is 75 m and PT is 51 m. The angle of depression from T to A is 36 degrees, and the angle PTA is obtuse.
(Stimulus: a triangle PTA with the angle of depression marked at T - see the official paper.)
Find the length of TA, correct to 2 decimal places.

Show worked solution

[3 marks]. The angle of depression gives angle TAP $= 36$ degrees. By the sine rule, with the angle opposite PT,

$$\frac{\sin(\angle PTA)}{75} = \frac{\sin 36 \text{ degrees}}{51}, \qquad \angle PTA = 59.8... \text{ degrees}.$$

Since angle PTA is obtuse, take $\angle PTA = 180 - 60 = 120$ degrees, so

$$\angle TPA = 180 - 120 - 36 = 24 \text{ degrees}.$$

Then for TA, opposite the 24 degree angle,

$$\frac{TA}{\sin 24 \text{ degrees}} = \frac{51}{\sin 36 \text{ degrees}}, \qquad TA = \frac{51 \times \sin 24 \text{ degrees}}{\sin 36 \text{ degrees}} = 35.29 \text{ m}.$$

Marker's note. This is the ambiguous case of the sine rule: the acute solution gives about 60 degrees, but the obtuse condition means $\angle PTA = 120$ degrees. Then use the sine rule again to find TA.

Question 36 (4 marks)

The graph shows the salvage value of a car over 5 years, based on the declining-balance method. By what amount will the car's value depreciate during the 10th year?
(Stimulus: a declining-balance salvage-value graph starting at $55 000 - see the official paper.)

Show worked solution

[4 marks]. From the graph the value falls from $55\,000 to$44,000 in the first year, so the depreciation rate is

$$r = \frac{55\,000 - 44\,000}{55\,000} = 0.2 = 20\%.$$

Value after 9 years and after 10 years:

$$55\,000(1 - 0.2)^9 = \$7381.98, \qquad 55\,000(1 - 0.2)^{10} = \$5905.58.$$

Depreciation during the 10th year:

$$7381.98 - 5905.58 = \$1476.40.$$

Marker's note. Find the declining-balance rate from the first year of the graph, then value the car at the end of years 9 and 10. The depreciation during the 10th year is the drop between those two values, not the value at year 10.

Question 37 (4 marks)

A park consists of two equilateral triangles; the shaded triangle is a grassed section. Measurements (metres): outer triangle side 9, with segments of 3 marked along each side.
(Stimulus: a large equilateral triangle with a smaller shaded triangle inside - see the official paper.)
How long will it take to mow the grassed section if it takes 5 minutes to mow 20 m squared? Give your answer to the nearest minute.

Show worked solution

[4 marks]. Each side of the grassed triangle joins a point 3 m along one side to a point on the next, with the included angle 60 degrees. By the cosine rule its side length is

$$c^2 = 9^2 + 3^2 - 2 \times 9 \times 3 \cos 60 \text{ degrees}, \qquad c = 7.937 \text{ m}.$$

Area of the grassed (equilateral) triangle:

$$A = \frac{1}{2} \times 7.937 \times 7.937 \times \sin 60 \text{ degrees} = 27.28 \text{ m}^2.$$

Mowing time at 5 minutes per 20 m squared:

$$\frac{5}{20} \times 27.28 = 6.81 \text{ minutes} \approx 7 \text{ minutes}.$$

Marker's note. Use the cosine rule to find a side of the inner triangle (included angle 60 degrees), then the $\tfrac12 ab\sin C$ area formula. Finish with the mowing ratio and round to the nearest minute.

Question 38 (3 marks)

A car's fuel efficiency is 30 miles per US gallon. 1 US gallon = 3.8 litres and 1 mile = 1.6 km (both to 2 significant figures). Calculate the car's fuel efficiency in litres per 100 km, correct to 1 decimal place.

Show worked solution

[3 marks]. Convert the distance: $30 \text{ miles} = 30 \times 1.6 = 48$ km, using 3.8 L of fuel. So the car uses 3.8 L per 48 km. Scaling to 100 km:

$$\frac{3.8}{48} \times 100 = 7.916... \approx 7.9 \text{ L}/100 \text{ km}.$$

Marker's note. Do one conversion at a time: change miles to km first, so 30 miles per gallon becomes 3.8 L per 48 km, then scale up to 100 km. Keep the volume and distance separate.

Question 39 (3 marks)

After a dose of medication, the amount remaining is modelled by $y = k a^x$, where $x$ is hours after the dose and $y$ is the amount in mg. The graph shows two points: $(0, 15)$ and $(2, 9)$.
(Stimulus: a decreasing exponential graph passing through $(0, 15)$ and $(2, 9)$ - see the official paper.)
Using the information provided, find the amount of medication that remains when $x = 4$.

Show worked solution

[3 marks]. At $x = 0$, $y = 15$, so $k = 15$. At $x = 2$, $y = 9$:

$$9 = 15a^2, \qquad a^2 = \frac{9}{15} = 0.6, \qquad a = 0.7746...$$

When $x = 4$:

$$y = 15 \times (0.7746...)^4 = 5.4 \text{ mg}.$$

Marker's note. The y-intercept gives $k = 15$. Use the second point to solve for $a$, then substitute $x = 4$. Recognising the exponential's key features on the graph is the main skill here.

Question 40 (5 marks)

(a) In a flock of 12 600 sheep, the ratio of males to females is 1 : 20. The weights of the male sheep are normally distributed with mean 76.2 kg and standard deviation 6.8 kg. In the flock, 15 of the male sheep each weigh more than $x$ kg. Find the value of $x$. (4 marks)
(b) The weights of the female sheep are also normally distributed but have a smaller mean and smaller standard deviation than the male sheep. 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 worked solution

(a) [4 marks]. Number of male sheep: the ratio 1 : 20 means males are $\tfrac{1}{21}$ of the flock,

$$\frac{12\,600}{21} = 600 \text{ males}.$$

The 15 heaviest males are

$$\frac{15}{600} \times 100\% = 2.5\%$$

of the males. In a normal distribution, 2.5% lie above $z = 2$ (two standard deviations above the mean), so

$$x = 76.2 + 2 \times 6.8 = 89.8 \text{ kg}.$$

(b) [1 mark]. There are $12\,600 - 600 = 12\,000$ females. Because the females have a smaller mean and smaller standard deviation, the value $x = 89.8$ kg is more than 2 standard deviations above their mean, so less than 2.5% of females exceed it. That is fewer than $0.025 \times 12\,000 = 300$, so it could not be expected that 300 females weigh more than $x$ kg.

Marker's note. Find the number of males from the ratio, express 15 as a percentage of 600, and match 2.5% to $z = 2$ using the Reference Sheet. In (b), the smaller female mean and spread push $x$ further into the tail, so fewer than 300 females exceed it.

General marker feedback

Stronger responses across the paper: showed clear mathematical reasoning and calculations; read each question carefully so no component was missed; recognised key command words (show, calculate, hence); used the Reference Sheet and the supplied tables rather than reaching for a formula; engaged with the stimulus and referred to it; kept units consistent and noted any required units; rounded only at the final step; and constructed graphs and box-plots neatly with all required information shown.