§-Past paper

NSWMaths Standard 22023

HSC Maths Standard 2 2023

Worked solutions to every question in the 2023 HSC Mathematics Standard 2 exam. Multiple-choice answers with a one-line reason, and a 'Show worked solution' model answer for each Section II question, aligned to the official NESA marking guidelines.

Marks: 100
Time: 150 min
Authority: NESA
Updated: 2026-06-20

Open the official PDF ->Official marking notes

Every question from the 2023 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 2023 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 38, 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: An amount of $2500 is invested at a simple interest rate of 3% per annum. How much interest is earned in the first two years? A.$ 75 B. $150 C.$ 2575 D. $2652
Answer: B - $I = 2500 \times 0.03 \times 2 = \$150$ .
Q2: In a normal distribution, what is the approximate percentage of scores with a z-score less than 1? A. 50% B. 68% C. 84% D. 97.5%
Answer: C - half the scores (50%) are below the mean and another 34% sit between the mean and $z = 1$ , giving 84%.
Q3: The number of bees leaving a hive was recorded over 14 days at different times of the day; the scatterplot rises strongly. Which Pearson's correlation coefficient best describes the observations? A. $-0.8$ B. $-0.2$ C. $0.2$ D. $0.8$
Answer: D - the points rise together in a fairly tight band, a strong positive association, so $0.8$ .
Q4: A delivery truck valued at $65000 when new depreciates at 22 cents per kilometre. What is its value after 132600 km? A.$ 35828 B. $29172 C.$ 14872 D. $14300
Answer: A - depreciation $= 0.22 \times 132600 = 29172$ , so value $= 65000 - 29172 = \$35828$ .
Q5: Four petrol pumps: 24 L for $46.70, 25 L for$ 48.50, 26 L for $52.30, 27 L for$ 54.80. Which is best value? A. 24 L B. 25 L C. 26 L D. 27 L
Answer: B - 25 L gives $48.50 \div 25 = \$1.94$ per litre, the lowest unit price.
Q6: An item was purchased for $880 including 10% GST. What is the GST included? A.$ 8.00 B. $8.80 C.$ 80.00 D. $88.00
Answer: C - GST is one eleventh of the GST-inclusive price, $880 \div 11 = \$80$ .
Q7: City A is at $34$ degrees S, $151$ degrees E. City B is $72$ degrees north and $25$ degrees west of A. Latitude and longitude of B? A. 16 degrees N, 126 degrees E B. 16 degrees N, 176 degrees E C. 38 degrees N, 126 degrees E D. 38 degrees N, 176 degrees E
Answer: C - $72 - 34 = 38$ degrees N, and $151 - 25 = 126$ degrees E.
Q8: A die and a 4-sector spinner are added to give a score. What is the probability of a score of 7 or more? A. $\tfrac{1}{6}$ B. $\tfrac{1}{4}$ C. $\tfrac{5}{18}$ D. $\tfrac{5}{12}$
Answer: D - of the 24 equally likely outcomes, 10 give a sum of 7 or more, so $\tfrac{10}{24} = \tfrac{5}{12}$ .
Q9: Length and width are measured as 8 cm and 5 cm, to the nearest centimetre. Lower and upper bounds for the area? A. 28 and 54 B. 36 and 42 C. 38.25 and 41.25 D. 33.75 and 46.75
Answer: D - bounds use $7.5 \times 4.5 = 33.75$ and $8.5 \times 5.5 = 46.75$ cm squared.
Q10: $25000 is invested for six years at 8% per annum compounding quarterly. Which expression gives its value? A.$ 25000 \times 1.02^{24} $B.$ 25000 \times 1.02^{6} $C.$ 25000 \times 1.08^{24} $D.$ 25000 \times 1.08^{6}$
Answer: A - quarterly rate $\tfrac{8\%}{4} = 2\%$ over $6 \times 4 = 24$ periods, so $25000 \times 1.02^{24}$ .
Q11: 150 jelly beans in red to blue ratio 2 : 3. Sophie eats 10 of each colour. New ratio? A. 2 : 3 B. 4 : 9 C. 5 : 8 D. 11 : 17
Answer: C - start with 60 red and 90 blue; after eating 10 each, 50 : 80 simplifies to 5 : 8.
Q12: A pipe of radius 12.5 cm holds water whose surface is 20 cm wide. What is the depth? A. 2.5 cm B. 5.0 cm C. 7.5 cm D. 12.5 cm
Answer: B - the half-width 10 and radius 12.5 give a drop of $\sqrt{12.5^2 - 10^2} = 7.5$ below centre, so depth $= 12.5 - 7.5 = 5$ cm.
Q13: An item is discounted 30%, then a further 20% off the reduced price. Total percentage discount? A. 25% B. 44% C. 50% D. 56%
Answer: B - remaining fraction $0.7 \times 0.8 = 0.56$ , so the discount is $44\%$ .
Q14: A network with source A and sink B has a cut of capacity 30. Which statement is correct? A. Max flow is 30 B. Max flow is 35 C. Max flow is 30 or less D. Max flow is 30 or more
Answer: C - the maximum flow can never exceed the capacity of any cut, so it is 30 or less.
Q15: Ashan needs 90 in his sixth test to average 80 over six tests. What was his mean after five tests? A. 78 B. 74 C. 70 D. 65
Answer: A - six tests need a total of $480$ ; without the 90 the first five total $390$ , a mean of $78$ .

Section II - Short and extended response

Question 16 (2 marks)

The graph shows Peta's heart rate, in beats per minute, during the first 60 minutes of a marathon.
(Stimulus: a heart-rate-versus-time line graph - see the official paper.)
(a) What was Peta's heart rate 20 minutes after she started her marathon? (1 mark)
(b) Peta started the marathon at 10 am. At what time would her heart rate first reach 140 beats/minute? (1 mark)

Show worked solution

(a) [1 mark]. Reading up from 20 minutes to the curve, Peta's heart rate was 120 beats/minute.

(b) [1 mark]. The graph first reaches 140 beats/minute at 30 minutes, so the time is 10:30 am.

Marker's note. Read the graph carefully and, in (b), give the actual clock time the rate is reached, not the number of minutes that have passed.

Question 17 (3 marks)

The table shows some flight distances (to the nearest 10 km) between Australian cities (Adelaide, Brisbane, Darwin, Hobart, Perth, Sydney).
(Stimulus: a part-completed distance table and a network diagram - see the official paper.)
(a) Use the information in the table to complete the network diagram, labelling each edge with its distance. (2 marks)
(b) Mahsa wants to travel from Hobart to Darwin, changing planes only once. Using the network diagram, calculate how many kilometres she will travel by plane. (1 mark)

Show worked solution

(a) [2 marks]. Transfer each listed distance onto the matching edge: Adelaide to Hobart 1170, Adelaide to Perth 2120, Brisbane to Darwin 2850, Brisbane to Sydney 750, Darwin to Perth 2650, Darwin to Sydney 3150, Hobart to Sydney 1040, Perth to Sydney 3270. Mark each off the table as you draw it so none is missed.

(b) [1 mark]. With one stop, the route Hobart to Sydney to Darwin gives

1040 + 3150 = 4190 \text{ km}.

Marker's note. Label every edge from the table, and highlight the chosen path before adding the two legs so the working is clear.

Question 18 (2 marks)

The histogram shows a summary of scores on a test.
(Stimulus: a frequency histogram - see the official paper.)
Provide TWO features of the histogram that indicate that the data comes from a normal distribution.

Show worked solution

[2 marks]. Two features:

The histogram is symmetrical about the centre (a bell shape), so the mean, median and mode are about equal.
The histogram is unimodal (a single peak in the middle), as a normal distribution requires.

Marker's note. Use the right language - symmetrical, unimodal, bell-shaped - and note that for normal data the mean, median and mode are approximately equal.

Question 19 (4 marks)

A network of running tracks connects points A to H; each edge shows the running time in minutes.
(Stimulus: a weighted network of running tracks - see the official paper.)
(a) Which path could a typical runner take to run from A to D in the shortest time? (2 marks)
(b) A spanning tree of the network is shown. Is it a minimum spanning tree? Give a reason for your answer. (2 marks)

Show worked solution

(a) [2 marks]. Comparing the routes, the shortest path from A to D is A to B to F to G to D (ABFGD). State the path as a list of vertices; it need not pass through every node.

(b) [2 marks]. No, it is not a minimum spanning tree. Vertex C is joined to the tree by edge BC, but a cheaper edge is available to connect C, so swapping it in lowers the total weight.

Marker's note. Include every vertex on your path and give numerical support for the shortest time. To justify (b), identify the edge that should be replaced by a cheaper one; drawing the correct minimum spanning tree also earns the reasoning.

Question 20 (3 marks)

On another planet a ball is launched vertically, with height modelled by $h = -6t^2 + 24t$ , where $t$ is in seconds. The graph is shown.
(Stimulus: the parabola of $h$ against $t$ - see the official paper.)
(a) Based on the graph, what is the maximum height reached by the ball? (1 mark)
(b) Based on the graph, at what TWO times is the ball at $\tfrac{3}{4}$ of its maximum height? (2 marks)

Show worked solution

(a) [1 mark]. The peak of the parabola is at $t = 2$ seconds, giving a maximum height of 24 m.

(b) [2 marks]. Three quarters of the maximum is

\frac{3}{4} \times 24 = 18 \text{ m}.

Reading across at

h = 18

, the curve is at 18 m at $t = 1$ second and $t = 3$ seconds.

Marker's note. Read the maximum from the turning point, then draw a horizontal line at $h = 18$ and drop verticals to the time axis to read both times. Keep your readings inside the axes.

Question 21 (5 marks)

Provider A charges 25 cents per kWh plus a fixed $40 per month. Provider B charges 35 cents per kWh with no fixed charge; its graph is given.
(Stimulus: a part-completed charges table and Provider B's graph - see the official paper.)
(a) Complete the table showing Provider A's monthly charges for 0, 400 and 1000 kWh. (1 mark)
(b) On the grid, graph Provider A's charges from the table in part (a). (1 mark)
(c) Use the two graphs to determine the number of kilowatt hours per month for which Provider A and Provider B charge the same amount. (1 mark)
(d) A customer uses an average of 800 kWh per month. Which provider, A or B, would be the cheaper option and by how much? (2 marks)

Show worked solution

(a) [1 mark]. Provider A charges $40 + 0.25 \times \text{kWh}$ :

Electricity used (kWh)	0	400	1000
Monthly charge ($)	40	140	290

(b) [1 mark]: Plot $(0, 40)$ , $(400, 140)$ and $(1000, 290)$ and join them with a straight line using a ruler.
(c) [1 mark]: The two lines cross at 400 kWh (where both charge $140).
(d) [2 marks]: At 800 kWh: Provider A charges $40 + 0.25 \times 800 = \$240$ ; Provider B charges $0.35 \times 800 = \$280$ . So Provider A is cheaper, by
$280 - 240 = \$40.$

Marker's note. Plot the table points accurately and rule a straight line. In (d), read both parts of the question - which provider and by how much - and answer both.

Question 22 (3 marks)

The braking distance of a car (metres) is directly proportional to the square of its speed (km/h): braking distance $= k \times (\text{speed})^2$ . The braking distance at 50 km/h is 20 m.
(a) Find the value of $k$ . (2 marks)
(b) What is the braking distance when the speed is 90 km/h? (1 mark)

Show worked solution

(a) [2 marks]. Substitute the known values and solve for $k$ :

20 = k \times 50^2, \qquad k = \frac{20}{2500} = 0.008.

(b) [1 mark]. Using the same rule at 90 km/h:

\text{braking distance} = 0.008 \times 90^2 = 0.008 \times 8100 = 64.8 \text{ m}.

Marker's note. No unit conversion is needed - the definitions are given in the question. Rearrange to $k = 20 \div 50^2$ carefully and enter it accurately on the calculator.

Question 23 (2 marks)

One hundred tickets are sold in a raffle offering two prizes. Hazel buys five tickets. A ticket is drawn for the first prize, then a second ticket from the rest for the other prize. What is the probability that Hazel wins both prizes?

Show worked solution

[2 marks]. The draws are without replacement, so the second probability uses 99 remaining tickets:

P(\text{both}) = \frac{5}{100} \times \frac{4}{99} = \frac{20}{9900} = \frac{1}{495}.

Marker's note. Recognise this is a without-replacement situation and multiply the two probabilities rather than adding them. A tree diagram can help organise the working.

Question 24 (5 marks)

The diagram shows the cross-section of a wall across a creek, with a base width of 8.0 m and heights 1.9 m, 2.7 m and 1.7 m at equal spacings.
(Stimulus: a cross-section with three vertical measurements - see the official paper.)
(a) Use two applications of the trapezoidal rule to estimate the area of the cross-section of the wall. (2 marks)
(b) The wall has a uniform thickness of 0.80 m. The weight of 1 m cubed of concrete is 3.52 tonnes. How many tonnes of concrete are in the wall? Give the answer to two significant figures. (3 marks)

Show worked solution

(a) [2 marks]. Each strip has width $8.0 \div 2 = 4.0$ m. Applying the trapezoidal rule to each strip and adding:

A = \frac{4.0}{2}(1.9 + 2.7) + \frac{4.0}{2}(2.7 + 1.7) = 9.2 + 8.8 = 18 \text{ m}^2.

(b) [3 marks]. Volume of the wall (a prism):

V = 18 \times 0.8 = 14.4 \text{ m}^3.

Weight of concrete:

14.4 \times 3.52 = 50.688 \approx 51 \text{ tonnes (2 significant figures)}.

Marker's note. Use the stipulated trapezoidal rule with the correct strip height (the full base divided by two). Find volume as area times thickness, then weight, and round to two significant figures (not two decimal places) at the end.

Question 25 (5 marks)

A table of future value interest factors for an annuity of $1 is given (rates 1.5% to 6%; periods 5 to 40).
(Stimulus: an annuity future-value factor table - see the official paper.)
(a) Micky wants to save $450000 over the next 10 years at 6% per annum compounding annually. How much should Micky contribute each year, to the nearest dollar? (2 marks)
(b) Instead, Micky decides to contribute $8535 every three months for 10 years to an annuity paying 6% per annum compounding quarterly. How much will Micky have at the end of 10 years? (3 marks)

Show worked solution

(a) [2 marks]. The factor for 6% and 10 periods is 13.181, so the future value equals the contribution times the factor:

\text{Contribution} = \frac{450\,000}{13.181} = \$34\,140 \text{ (nearest dollar)}.

(b) [3 marks]. Quarterly compounding means the rate is $\tfrac{6\%}{4} = 1.5\%$ over $10 \times 4 = 40$ periods. The factor for 1.5% and 40 periods is 54.268, so

\text{Amount} = 8535 \times 54.268 = \$463\,177.38.

Marker's note. Identifying the factor already uses the rate and periods, so do not apply them again. In (b), convert to a 1.5% rate over 40 periods and multiply $8535 by the matching factor.

Question 26 (5 marks)

Kim builds a 0.5 m wide path around a garden at the back of a house, as shown.
(Stimulus: a plan of the garden, house and path - see the official paper.)
(a) Find the area of the path. (2 marks)
(b) The concrete mix is crushed rock, sand and cement in the ratio 4 : 2 : 1 by weight. Kim needs 2.1 tonnes of mix. Calculate how many 15 kg bags of cement Kim needs to buy. (3 marks)

Show worked solution

(a) [2 marks]. The path is the outer rectangle minus the inner garden. The outer region is $3 \text{ m} \times 8 \text{ m}$ and the garden inside the path is $(3 - 0.5) \times (8 - 0.5 - 0.5) = 2.5 \times 7$ :

\text{Area of path} = (3 \times 8) - (7 \times 2.5) = 24 - 17.5 = 6.5 \text{ m}^2.

(b) [3 marks]. Cement is 1 part out of $4 + 2 + 1 = 7$ :

\text{Cement} = \frac{1}{7} \times 2.1 = 0.3 \text{ tonnes} = 300 \text{ kg}.

Number of 15 kg bags:

\frac{300}{15} = 20 \text{ bags}.

Marker's note. Break the composite area into simple rectangles and subtract. In (b), use the unitary method to find the cement's share of the ratio, then convert tonnes to kilograms before dividing by the bag size.

Question 27 (4 marks)

Y is on a bearing of 120 degrees and 15 km from X. C is 40 km from X and due west of Y. P lies on the line CY and is due south of X.
(Stimulus: a bearings diagram with X, Y, C and P - see the official paper.)
(a) Find the distance from X to P. (2 marks)
(b) What is the bearing of C from X, to the nearest degree? (2 marks)

Show worked solution

(a) [2 marks]. Since P is due south of X and Y is on a bearing of 120 degrees, the angle PXY is $180 - 120 = 60$ degrees. In right triangle PXY with hypotenuse XY = 15,

XP = 15 \times \cos 60 \text{ degrees} = 7.5 \text{ km}.

(b) [2 marks]. In right triangle CXP, with XP = 7.5 adjacent and XC = 40 as the hypotenuse,

\cos(\angle CXP) = \frac{7.5}{40}, \qquad \angle CXP = 79 \text{ degrees (nearest degree)}.

C is west of south of X, so the bearing of C is

180 + 79 = 259 \text{ degrees}.

Marker's note. Transfer the given information to the diagram and use right-angled trigonometry, identifying the adjacent side and hypotenuse correctly. The bearing of C should fall between 180 and 270 degrees.

Question 28 (3 marks)

A plumber leases equipment valued at $60000. Salvage value can be found by straight-line depreciation ($ 3500 per annum) or declining-balance depreciation (12% per annum). Under which method would the salvage value be lower at the end of 3 years? Justify your answer with appropriate mathematical calculations.

Show worked solution

[3 marks]. Straight-line method after 3 years:

S = 60\,000 - 3500 \times 3 = \$49\,500.

Declining-balance method after 3 years:

S = 60\,000 \times (1 - 0.12)^3 = 60\,000 \times 0.88^3 = \$40\,888.32.

Since

40\,888.32 < 49\,500

, the declining-balance method gives the lower salvage value.

Marker's note. Apply both depreciation formulas from the Reference Sheet correctly and compare the two salvage values to justify the conclusion. Remember salvage value is the value remaining, not the amount lost.

Question 29 (4 marks)

A table gives monthly repayments per $1000 borrowed for various rates and terms.
(Stimulus: a monthly repayment table - see the official paper.)
(a) A couple borrows $520000 to buy a house at 8% per annum over 25 years. How much does the couple repay in total for this loan? (3 marks)
(b) Chris borrows some money at 7% per annum, repaying over 15 years at $3596 per month. How much money does Chris borrow? (1 mark)

Show worked solution

(a) [3 marks]. The table factor for 8% over 25 years is 7.72 per $1000. The monthly repayment is

\frac{520\,000}{1000} \times 7.72 = \$4014.40.

Total repaid over 25 years:

4014.40 \times 12 \times 25 = \$1\,204\,320.

(b) [1 mark]. The factor for 7% over 15 years is 8.99 per $1000, so

\text{Amount borrowed} = \frac{3596}{8.99} \times 1000 = \$400\,000.

Marker's note. The factors are per $1000 borrowed, so divide the loan by 1000 first. Multiply the monthly repayment by 12 then by the number of years for the total. In (b), divide by the factor (do not multiply) and scale by 1000.

Question 30 (3 marks)

A supermarket receipt shows a total of $124.87. The GST shown is$ 3.86. GST, at 10%, is only charged on some items. What was the value of the items which did NOT have GST charged?

Show worked solution

[3 marks]. The GST of $3.86 is 10% of the value of the taxable goods, so the value of those goods (before GST) is

3.86 \times 10 = \$38.60.

The total is made up of the taxable goods, their GST, and the GST-free goods. So the GST-free goods are

124.87 - 38.60 - 3.86 = \$82.41.

Marker's note. The receipt total includes the taxable goods, their GST and the GST-free goods. Subtract both the $38.60 and the$ 3.86 from the total; GST is not charged on every item.

Question 31 (4 marks)

A network diagram shows the time in hours for tasks A to K that must be completed between functions.
(Stimulus: an activity network with task times - see the official paper.)
(a) Find the TWO critical paths. (2 marks)
(b) The centre wants to decrease the length of each critical path by 3 hours by hiring more staff for ONE task. For which task should they hire more staff, and how long should that task take to ensure all tasks can be completed in 14 hours? (2 marks)

Show worked solution

(a) [2 marks]. Using forward and backward scanning, the two paths of greatest and equal length are the critical paths: H, I, G, C and H, I, K.

(b) [2 marks]. Both critical paths share tasks H and I. Task H is only 2 hours, so it cannot be cut by 3. The shared task to shorten is task I: reducing it by 3 hours makes it

7 - 3 = 4 \text{ hours},

which brings both critical paths down to 14 hours.

Marker's note. Use EST and LST from forward and backward scanning; two equal critical paths have the same length. Shorten a task common to both, and note the task chosen had to be longer than 3 hours.

Question 32 (4 marks)

Ali's credit card has no interest-free period; interest is 13.5% per annum compounding daily on the amount owing. Ali made one purchase of $450 and repaid the full amount 21 days later.
(a) Calculate the amount of interest charged on the purchase, assuming interest is charged for the 21 days. (2 marks)
(b) What percentage of the full amount repaid is the interest? Give the answer to two decimal places. (2 marks)

Show worked solution

(a) [2 marks]. The daily rate is $\tfrac{13.5\%}{365}$ over 21 days:

\text{Full amount} = 450 \left(1 + \frac{0.135}{365}\right)^{21} = \$453.51.

The interest is the future value minus the principal:

453.51 - 450 = \$3.51.

(b) [2 marks]. As a percentage of the full amount repaid:

\frac{3.51}{453.51} \times 100 = 0.77\%.

Marker's note. Use compound (not simple) interest, dividing the annual rate by 365. The formula gives the future value, so subtract the principal to find the interest only.

Question 33 (4 marks)

A shape APQBCD is a rectangle ABCD with an arc PQ on side AB. The rectangle has BC = 3.6 m and CD = 8.0 m. The arc PQ has centre O, radius 2.1 m and angle POQ = 110 degrees.
(Stimulus: a rectangle with a circular arc cut into one side - see the official paper.)
What is the perimeter of the shape APQBCD? Give your answer correct to one decimal place.

Show worked solution

[4 marks]. The arc PQ replaces part of side AB. Arc length:

\text{arc } PQ = \frac{110}{360} \times 2\pi \times 2.1 = 4.0317 \text{ m}.

The straight chord PQ (removed from AB) by the cosine rule:

PQ = \sqrt{2.1^2 + 2.1^2 - 2(2.1)(2.1)\cos 110 \text{ degrees}} = 3.4404 \text{ m}.

The perimeter is the two sides BC, the side CD, the remaining part of AB (

8.0 - PQ

), and the arc:

P = (3.6 \times 2) + 8.0 + (8.0 - 3.4404) + 4.0317 = 23.8 \text{ m (1 decimal place)}.

Marker's note. The arc is $\tfrac{110}{360}$ of a circle, not a semicircle. Use the arc-length formula for PQ and the cosine rule for the chord, write each formula before substituting, and add every section of the boundary.

Question 34 (6 marks)

Over 10 winter weekdays a university recorded average outside temperature ( $x$ , in degrees C) and total daily gas usage ( $y$ , in MW). The regression line predicts 236 MW at 0 degrees C. The ten temperatures were 0, 0, 0, 2, 5, 7, 8, 9, 9, 10. The total gas usage was 1840 MW. The least-squares line passes through $(\bar{x}, \bar{y})$ .
(Stimulus: a grid for plotting - see the official paper.)
(a) Plot the point $(\bar{x}, \bar{y})$ and the y-intercept of the least-squares regression line on the grid. (3 marks)
(b) What is the equation of the regression line? (2 marks)
(c) In the context of the dataset, identify ONE problem with using the regression line to predict gas usage when the average outside temperature is 23 degrees C. (1 mark)

Show worked solution

(a) [3 marks]: Calculate the two means:
$\bar{x} = \frac{0+0+0+2+5+7+8+9+9+10}{10} = 5, \qquad \bar{y} = \frac{1840}{10} = 184.$

Plot $(\bar{x}, \bar{y}) = (5, 184)$ and the y-intercept $(0, 236)$ .
(b) [2 marks]: The line passes through $(0, 236)$ and $(5, 184)$ , so the gradient is
$m = \frac{184 - 236}{5 - 0} = -10.4,$

and with y-intercept 236 the equation is $y = 236 - 10.4x$ .
(c) [1 mark]: A temperature of 23 degrees C is well outside the data range (0 to 10 degrees), so it is extrapolation. The equation would give $y = 236 - 10.4 \times 23 < 0$ , a negative gas usage, which is not physically possible.

Marker's note. Read the grid scale carefully, recognise the initial value (236) is the y-intercept, and find the gradient from two points. Write an equation, not an expression, and relate the part (c) answer to the context (negative gas usage from extrapolation).

Question 35 (3 marks)

The diagram shows triangle ABC with a 12 m side, a 60 degree angle and a 25 degree angle.
(Stimulus: triangle ABC with the marked sides and angles - see the official paper.)
Calculate the area of the triangle, to the nearest square metre.

Show worked solution

[3 marks]. The third angle is $180 - 60 - 25 = 95$ degrees. Use the sine rule to find the side opposite the 60 degree angle:

\frac{x}{\sin 60 \text{ degrees}} = \frac{12}{\sin 25 \text{ degrees}}, \qquad x = \frac{12 \sin 60 \text{ degrees}}{\sin 25 \text{ degrees}} = 24.59 \text{ m}.

Now apply the area rule with the two sides and their included angle of 95 degrees:

A = \frac{1}{2} \times 12 \times 24.59 \times \sin 95 \text{ degrees} = 146.98 \approx 147 \text{ m}^2.

Marker's note. Find the missing 95 degree angle first, use the sine rule to get a side, then $A = \tfrac12 ab\sin C$ with the correct two sides and included angle.

Question 36 (4 marks)

For males, $\text{BAC} = \dfrac{10N - 7.5H}{6.8M}$ , where N is standard drinks, M is weight in kg and H is hours of drinking. Cameron weighs 75 kg and started with BAC zero. At 9:00 pm, after 3 standard drinks, his BAC was 0.02. Using the formula, estimate the time Cameron began drinking, to the nearest minute.

Show worked solution

[4 marks]. Substitute $N = 3$ , $M = 75$ , BAC $= 0.02$ and solve for $H$ :

0.02 = \frac{10(3) - 7.5H}{6.8 \times 75} = \frac{30 - 7.5H}{510}.

0.02 \times 510 = 30 - 7.5H, \qquad 10.2 = 30 - 7.5H,

7.5H = 30 - 10.2 = 19.8, \qquad H = 2.64 \text{ hours}.

Convert:

0.64 \times 60 = 38

minutes, so

H = 2

hours 38 minutes. Subtracting from 9:00 pm:

9{:}00 \text{ pm} - 2 \text{ h } 38 \text{ min} = 6{:}22 \text{ pm}.

Marker's note. Rearrange the fraction carefully to solve for H, evaluating at each step, then convert the decimal hours to minutes and subtract the time correctly to reach 6:22 pm.

Question 37 (3 marks)

A table of personal income tax rates is given. For $78801 to$ 108400 the tax is $18292 plus X cents for each$ 1 over $78800. A person with taxable income$ 90000 pays 25.8% of that income in tax (excluding levies). What is the value of X in the table?

Show worked solution

[3 marks]. Total tax paid:

90\,000 \times 0.258 = \$23\,220.

The income above $78800 is

90\,000 - 78\,800 = \$11\,200.

The tax on this portion, above the fixed $18292, is

23\,220 - 18\,292 = \$4928.

So X cents per dollar over the $11200 gives

11\,200 \times X = 4928, \qquad X = \frac{4928}{11\,200} = 0.44.

Therefore

X = 44

cents.

Marker's note. Select the correct tax bracket for $90000, find the total tax from 25.8%, then work back to the rate on the amount over$ 78800. Do not add a Medicare levy when the question says to exclude levies.

Question 38 (4 marks)

A standard normal table gives the probability that a $z$ value lies below $z$ , for $z$ from 1.30 to 1.39.
(Stimulus: a standard normal probability table and a shaded normal curve - see the official paper.)
Adult male koala weights are normally distributed with mean 10.40 kg and standard deviation 1.15 kg. In a group of 400 adult male koalas, how many would be expected to weigh more than 11.93 kg?

Show worked solution

[4 marks]. Find the z-score for 11.93 kg:

z = \frac{11.93 - 10.40}{1.15} = 1.33.

From the table,

P(z < 1.33) = 0.9082

, so

P(\text{more than } 11.93) = 1 - 0.9082 = 0.0918.

Expected number out of 400:

0.0918 \times 400 = 36.72 \approx 36 \text{ koalas (37 also accepted)}.

Marker's note. Calculate the z-score with the Reference Sheet formula, read the matching probability from the table, then subtract from 1 because the question asks for "more than". Multiply by 400 for the expected count.

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 words such as show, calculate and justify; used the Reference Sheet and the supplied tables where appropriate; set out legible, well-sequenced working; engaged with the stimulus material and referred to it; checked the answer actually answered the question; rounded only at the final step; constructed graphs neatly with a ruler showing all required information; and noted the units of measurement given in each question.

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Structure and timing

Section I - Multiple choice

Section II - Short and extended response

Question 16 (2 marks)

Question 17 (3 marks)

Question 18 (2 marks)

Question 19 (4 marks)

Question 20 (3 marks)

Question 21 (5 marks)

Question 22 (3 marks)

Question 23 (2 marks)

Question 24 (5 marks)

Question 25 (5 marks)

Question 26 (5 marks)

Question 27 (4 marks)

Question 28 (3 marks)

Question 29 (4 marks)

Question 30 (3 marks)

Question 31 (4 marks)

Question 32 (4 marks)

Question 33 (4 marks)

Question 34 (6 marks)

Question 35 (3 marks)

Question 36 (4 marks)

Question 37 (3 marks)

Question 38 (4 marks)

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