HSC Maths Standard 2 2024

Every question from the 2024 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 2024 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 41, 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

If $x = -2.531$, what is the value of $x^2$ rounded to 2 decimal places? A. $-6.41$ B. $-6.40$ C. $6.40$ D. $6.41$
Answer: D - squaring gives a positive number, $(-2.531)^2 = 6.4059...$, which rounds to $6.41$.

Q2

A straight-line graph crosses the y-axis above the origin and slopes downwards. Which equation could it be? A. $y = 2x + 3$ B. $y = 2x - 3$ C. $y = -2x + 3$ D. $y = -2x - 3$
Answer: C - a downward slope needs a negative gradient and the positive intercept needs $+3$, so $y = -2x + 3$.

Q3

Shoppers are invited to complete an online survey. What type of sampling is this? A. Self-selecting B. Simple random C. Stratified D. Systematic
Answer: A - people choose whether to take part, so it is self-selecting (voluntary).

Q4

A map of six regions is turned into a network, with an edge for each shared border. How many edges? A. 8 B. 7 C. 6 D. 5
Answer: B - counting each shared border between two regions once gives 7 edges.

Q5

Pia's marks: English 78 (mean 66, sd 6), Maths 80 (mean 71, sd 10), Science 77 (mean 70, sd 15), History 85 (mean 72, sd 9). Best relative performance? A. English B. Maths C. Science D. History
Answer: A - English has the largest z-score, $(78-66)/6 = 2$.

Q6

A right-angled triangle has hypotenuse 20 and the angle 40 degrees adjacent to side $x$. Which expression gives $x$? A. $20\cos 40$ degrees B. $20\sin 40$ degrees C. $\dfrac{20}{\cos 40 \text{ degrees}}$ D. $\dfrac{20}{\sin 40 \text{ degrees}}$
Answer: A - $x$ is adjacent to the 40 degree angle with hypotenuse 20, so $x = 20\cos 40$ degrees.

Q7

A uniform cost $180 three years ago and rose by 2.5$180.14 B. $181.35 C.$193.50 D. $193.84
Answer: D - $180 \times 1.025^3 = 193.84$.

Q8

A bill has $242 for parts (includes 10$100 labour (GST to be added). Total GST paid? A. $22.00 B.$24.20 C. $32.00 D.$34.20
Answer: C - GST in parts is $242 \div 11 = 22$; GST on labour is $10$; total $32$.

Q9

Time to paint varies inversely with number of painters; 6 painters take 20 days. How long for 15 painters? A. 4.5 B. 8 C. 15.5 D. 50
Answer: B - $6 \times 20 = 120$ painter-days, so $120 \div 15 = 8$ days.

Q10

Make $p$ the subject of $s = wt + \dfrac{p}{2}$. A. $p = 2wt - s$ B. $p = 2wt - 2s$ C. $p = 2s - wt$ D. $p = 2s - 2wt$
Answer: D - $\dfrac{p}{2} = s - wt$, so $p = 2s - 2wt$.

Q11

A train left at 6:42 am, arrived 8:04 am, distance 61 km. Average speed to the nearest km/h? A. 38 B. 45 C. 50 D. 74
Answer: B - the trip took 82 minutes $= \tfrac{82}{60}$ h, so $61 \div \tfrac{82}{60} \approx 45$ km/h.

Q12

In a two-way table, 157 people are over 30 and 34 of them watch Anime. Percentage of over-30s who watch Anime? A. 9% B. 18% C. 22% D. 42%
Answer: C - $34 \div 157 \approx 21.7\%$, about 22%.

Q13

A $9000 simple-interest loan is repaid by$300 monthly for 4 years. Simple interest rate per annum, to the nearest percent? A. 15% B. 38% C. 40% D. 60%
Answer: A - total paid $= 300 \times 48 = 14400$, interest $= 5400$, so $r = \tfrac{5400}{9000 \times 4} = 15\%$.

Q14

Pizza cost $C = 2.5x + 6$, revenue $R = 8x$. By how much does profit rise per extra pizza? A. $2.50 B.$5.50 C. $6.00 D.$8.00
Answer: B - profit $= R - C = 5.5x - 6$, so each pizza adds $5.50.

Q15

A box plot is given; which histogram is NOT possible for the same data? A, B, C or D.
Answer: D - option D's shape is inconsistent with the box plot's median and quartile positions.

Section II - Short and extended response

Question 16 (4 marks)

A network of towns and the distances between them in kilometres is shown.
(Stimulus: a weighted network with towns J, T, M, C, W, G, H, Y - see the official paper.)
(a) What is the shortest path from T to H? (2 marks)
(b) A truck driver needs to travel from Y to G but the road from C to G is closed. What is the length of the shortest path from Y to G after the road closure? (2 marks)

Show worked solution

(a) [2 marks]. Comparing the routes through the network, the shortest path from T to H is T to Y to W to H, written TYWH. State the path as a list of vertices including the start and end.

(b) [2 marks]. With C to G closed, the shortest available route from Y to G is Y to W to H to M to G (YWHMG), with total length 89 km.

Marker's note. Diagrams are not to scale, so a shorter total may look longer on the page. Label the whole path including the start and end vertices, and do not skip vertices when you trace it.

Question 17 (2 marks)

The cost of electricity is 30.13 cents per kWh. Calculate the cost of using a 650 W air conditioner for 6 hours.

Show worked solution

[2 marks]. First convert watts to kilowatts, then find energy in kWh:

$$\text{Energy} = \frac{650}{1000} \times 6 = 3.9 \text{ kWh}.$$

$$\text{Cost} = 3.9 \times \$0.3013 = \$1.18.$$

Marker's note. This needs two conversions: $1 \text{ kW} = 1000 \text{ W}$, and cents to dollars. Round the money to 2 decimal places at the end.

Question 18 (3 marks)

The diagram shows a network with weighted edges (vertices A to J).
(a) Draw a minimum spanning tree for this network and determine its weight. (2 marks)
(b) Is it possible to find another spanning tree with the same weight? Give a reason. (1 mark)

Show worked solution

(a) [2 marks]. Building the tree by repeatedly adding the cheapest edge that does not form a cycle (Prim's or Kruskal's method) gives a minimum spanning tree of weight 24. A tree that connects all 10 vertices uses 9 edges and avoids loops.

(The labelling above is schematic; on the exam network the cheapest edges add to 24.)

(b) [1 mark]. Yes. Two edges in the network have the same weight (for example FC and BC), so swapping one for the other gives a different spanning tree with the same total weight of 24.

Marker's note. Draw the tree with a ruler and write the weight on each edge. To justify part (b), name two equal-weight edges that can be exchanged without changing the total.

Question 19 (3 marks)

Five students' assignment marks (x) and test marks (y) are plotted with the least-squares regression line shown.
(Stimulus: a scatterplot with the regression line - see the official paper.)
(a) What is the equation of the least-squares regression line? (2 marks)
(b) Another student scored 5 for the assignment and 12 on the test. Did this student do better or worse on the test than the line predicts? Give a reason. (1 mark)

Show worked solution

(a) [2 marks]. Read two clear points off the line. The y-intercept is $6$ and the gradient is

$$m = \frac{14}{10} = 1.4,$$

so the equation is $y = 1.4x + 6$.

(b) [1 mark]. Substituting $x = 5$ predicts $y = 1.4(5) + 6 = 13$. The student scored only $12$, which is below the predicted $13$, so this student did worse than the line predicts.

Marker's note. Use the graph (two clear points), not just a calculator function, and give the equation a subject ($y = \ldots$). If you use statistics mode, remember A is the intercept and B is the gradient.

Question 20 (3 marks)

A table gives the future value of an annuity of $1 for various rates and periods.
(Stimulus: an annuity future-value table - see the official paper.)
(a) Ken invests $200 at the start of each year for eight years at 5% per annum. Calculate the future value. (1 mark)
(b) Shay needs $4500 in three years, paying every 6 months into an account earning 4$10? (2 marks)

Show worked solution

(a) [1 mark]. Read the factor for 5% and 8 periods (10.0266) and multiply:

$$\$200 \times 10.0266 = \$2005.32.$$

(b) [2 marks]. Six-monthly means the rate is $\tfrac{4\%}{2} = 2\%$ and the number of periods is $3 \times 2 = 6$. The factor for 2% and 6 periods is 6.4343, so

$$\text{Payment} = \frac{4500}{6.4343} = 699.37\ldots \approx \$700 \text{ (nearest \$10)}.$$

Marker's note. When a table is supplied, use it rather than a formula. Adjust both the rate and the number of periods for six-monthly compounding, and know when to multiply by the factor (a) and when to divide by it (b).

Question 21 (3 marks)

William owes $5590 on a reducing-balance loan, repays$110 monthly, with interest 24% per annum compounded monthly. A spreadsheet shows two months of the loan.
(Stimulus: a partly completed loan spreadsheet - see the official paper.)
(a) Complete the spreadsheet to show the balance owing at the end of two months. (2 marks)
(b) Explain why the loan will never be repaid if William keeps repaying $110 per month. (1 mark)

Show worked solution

(a) [2 marks]. The monthly rate is $\tfrac{24\%}{12} = 2\%$. Each month: interest $=$ principal $\times 0.02$, then balance $=$ principal $+$ interest $-$ repayment.

Month	Principal	Interest	Repaid	Balance owing
1	$5590.00 |$111.80	$110.00 |$5591.80
2	$5591.80 |$111.84	$110.00 |$5593.64

Balance owing after two months $= \$5593.64$.

(b) [1 mark]. The interest charged each month (about $112) is more than the$110 repayment, so the balance increases every month and the loan is never paid off.

Marker's note. Read the column headings to decide what to add and subtract each row. The key idea in (b) is that a repayment smaller than the interest makes the balance grow.

Question 22 (3 marks)

Two animal populations are modelled by $y = A \times (1.055)^x$ for W and $y = B \times (0.97)^x$ for K, where $x$ is years since 1985.
(Stimulus: a population-versus-time graph - see the official paper.)
Complete the table with the 1985 population, the percentage yearly change, and the predicted population when $x = 50$.

Show worked solution

[3 marks]. Read $A = 34$ from the graph; the initial value of K from the graph is $B = 280$. The base $1.055$ means a 5.5% increase per year; the base $0.97$ means a 3% decrease per year. For W at $x = 50$:

$$34 \times (1.055)^{50} = 494.4\ldots \approx 494.$$

	Population W	Population K
Population in 1985	$A = 34$	$B = 280$
Percentage yearly change	$+5.5\%$	$-3\%$
Predicted population at $x = 50$	494	61

Marker's note. Distinguish the percentage change from the factor $(1 \pm r)$: a base of $1.055$ is a 5.5% rise (not 105.5%) and a base of $0.97$ is a 3% fall (not 97%).

Question 23 (3 marks)

Zazu works a 38-hour week paid at $45 per hour, with overtime at time-and-a-half. In one week Zazu worked the 38 hours plus some overtime and earned$2790. How many overtime hours did Zazu work?

Show worked solution

[3 marks]. Normal pay $= 38 \times 45 = \$1710$. Overtime pay $= 2790 - 1710 = \$1080$. The overtime rate is $45 \times 1.5 = \$67.50$ per hour, so

$$\text{Overtime hours} = \frac{1080}{67.50} = 16 \text{ hours}.$$

Marker's note. Overtime hours are paid at a different rate from normal hours; divide the overtime pay by the time-and-a-half rate ($67.50), not by$45.

Question 24 (4 marks)

Sarah, 60 kg female, drinks 3 glasses of wine over 2.5 hours; one glass is 1.2 standard drinks. For females, $\text{BAC} = \dfrac{10N - 7.5H}{5.5M}$ where N is standard drinks, H is hours, M is mass in kg.
(a) Calculate Sarah's BAC at the end of the dinner, to 3 decimal places. (2 marks)
(b) The time for BAC to reach zero is $\text{Time} = \dfrac{\text{BAC}}{0.015}$. Find this time to the nearest minute. (2 marks)

Show worked solution

(a) [2 marks]. Standard drinks $N = 1.2 \times 3 = 3.6$, $H = 2.5$, $M = 60$:

$$\text{BAC} = \frac{10(3.6) - 7.5(2.5)}{5.5 \times 60} = \frac{36 - 18.75}{330} = 0.0522\ldots \approx 0.052.$$

(b) [2 marks].

$$\text{Time} = \frac{0.052}{0.015} = 3.4\ldots \text{ hours} = 3 \text{ hours } 28 \text{ minutes}.$$

Marker's note. Convert the glasses to standard drinks before substituting ($3.6$, not $3$), and convert the decimal hours in (b) into hours and minutes.

Question 25 (3 marks)

Alex and Jun each invest $1800 for 5 years. Alex earns 7.5% per annum simple interest; Jun earns 6.0% per annum compounding quarterly. By calculating the interest earned, determine who has the greater amount.

Show worked solution

[3 marks]. Alex's simple interest:

$$I = 1800 \times 0.075 \times 5 = \$675.$$

Jun compounds quarterly, so the rate is $\tfrac{6\%}{4} = 1.5\%$ over $4 \times 5 = 20$ periods:

$$1800 \times (1.015)^{20} = 2424.34\ldots, \qquad I = 2424.34 - 1800 = \$624.34.$$

Alex earns $675 versus Jun's$624.34, so Alex has the greater amount.

Marker's note. The simple interest formula gives the interest, but the compound formula gives the future value, so subtract the principal before comparing. Adjust the rate and periods for quarterly compounding.

Question 26 (3 marks)

A gutter has a rectangular cross-section of width $w$ cm and height $h$ cm. The area is $A = -0.5w^2 + 20w$, and its graph is shown.
(Stimulus: the parabola for $A$ against $w$ - see the official paper.)
Find the width and height of the rectangle giving the greatest cross-sectional area.

Show worked solution

[3 marks]. The parabola is symmetric, with x-intercepts at $w = 0$ and $w = 40$, so the maximum is halfway, at $w = 20$. The maximum area is

$$A = -0.5(20)^2 + 20(20) = -200 + 400 = 200 \text{ cm}^2.$$

Since $A = h \times w$, $200 = h \times 20$, so $h = 10$. The width is 20 cm and the height is 10 cm.

Marker's note. Use the symmetry of the parabola to find the width at the maximum (halfway between the roots), then substitute back to get the area and finally the height.

Question 27 (3 marks)

A couple borrows $30000, repaid by$280 monthly over 10 years. After 5 years they cut the repayment to $250, which adds two years to the loan. How much more will they repay in total because of this change?

Show worked solution

[3 marks]. Original total over 10 years:

$$280 \times 12 \times 10 = \$33\,600.$$

New plan: $280 for 5 years, then$250 for 7 years (5 remaining plus 2 extra):

$$280 \times 12 \times 5 = 16\,800, \qquad 250 \times 12 \times 7 = 21\,000.$$

New total $= 16\,800 + 21\,000 = \$37\,800$. The extra repaid is

$$37\,800 - 33\,600 = \$4200.$$

Marker's note. Find the annual totals by multiplying by 12, then compare the full ten-year plan ($33600) with the changed plan over the longer term ($37800).

Question 28 (3 marks)

Heights of 25 flowers from each of Garden A and Garden B are shown in parallel box plots.
(Stimulus: parallel box plots for Garden A and Garden B - see the official paper.)
Compare the two datasets by examining the skewness of the distributions, and the measures of central tendency and spread.

Show worked solution

[3 marks]. Comparing all three features:

• Skewness: Garden A is negatively skewed (longer left tail); Garden B is positively skewed (longer right tail).
• Central tendency: the median of Garden A is higher than the median of Garden B (the median is the only centre readable from a box plot).
• Spread: the interquartile range of Garden A is larger than that of Garden B, and Garden A's range is also larger, so Garden A is more spread out.

Marker's note. Address all three: skewness, centre and spread. Use the correct terms ("positively skewed", "negatively skewed") and quote the median or IQR, the only measures you can read from a box plot.

Question 29 (4 marks)

An asset worth $50000 depreciates by$1500 per year (straight-line) for the first 4 years, then by 35% per annum (declining-balance) afterwards.
(Stimulus: a value-versus-time graph - see the official paper.)
What is the total depreciation at the end of 10 years?

Show worked solution

[4 marks]. Value after the first 4 years (straight-line):

$$50\,000 - 1500 \times 4 = \$44\,000.$$

This becomes the starting value for the next 6 years at 35% declining balance:

$$44\,000 \times (1 - 0.35)^6 = 44\,000 \times 0.65^6 = 3318.43\ldots$$

Total depreciation is the loss in value over the 10 years:

$$50\,000 - 3318.43 = \$46\,681.57.$$

Marker's note. Carry the salvage value at year 4 into the declining-balance stage as its initial value, use $n = 6$ years for that stage, and remember "depreciation" is the total loss in value, not the final value.

Question 30 (3 marks)

A scatterplot shows the age and length of female and male anacondas, which reach maturity at about 4 years.
(Stimulus: a scatterplot keyed by sex - see the official paper.)
Write THREE observations about anacondas that may be made from the scatterplot. (No calculations required.)

Show worked solution

[3 marks]. Three valid observations:

1. Females are longer than males at the same age.
2. Females grow faster than males (steeper trend in length with age).
3. Both sexes keep growing after maturity (length still increases past 4 years).

(Other correct readings, such as growth increasing at a decreasing rate, are also acceptable.)

Marker's note. Make observations about the anacondas (the subject), not just about the dots. Give three distinct statements, and do not extrapolate beyond the data (no predictions past 10 years), and recall the data are many snakes, not one over time.

Question 31 (3 marks)

A coin is biased so it is twice as likely to show a head as a tail.
(a) What is the probability of a head in one throw? (1 mark)
(b) In two throws, what is the probability of at least one head? (2 marks)

Show worked solution

(a) [1 mark]. Heads is twice as likely as tails, so the ratio is 2 : 1 and

$$P(\text{head}) = \frac{2}{3}.$$

(b) [2 marks]. Use the complement. $P(\text{tail}) = \tfrac{1}{3}$, so

$$P(\text{no heads}) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9},$$

$$P(\text{at least one head}) = 1 - \frac{1}{9} = \frac{8}{9}.$$

Marker's note. A biased coin changes $P(\text{head})$ from $\tfrac12$ to $\tfrac23$. The neatest route to "at least one" is the complement, $1 - P(\text{no heads})$. Simplify the final fraction.

Question 32 (4 marks)

A regular pentagon ABCDE is drawn inside a circle of radius 30 cm, centred at O. The central angle of each sector is 72 degrees.
(Stimulus: a pentagon inscribed in a circle with the region outside the pentagon shaded - see the official paper.)
What is the area of the shaded region of the circle? Answer to 2 significant figures.

Show worked solution

[4 marks]. Area of the whole circle:

$$A_{\text{circle}} = \pi \times 30^2 = 2827.43\ldots$$

The pentagon is five triangles, each two radii with the included angle 72 degrees:

$$A_{\triangle} = \frac{1}{2}(30)(30)\sin 72 \text{ degrees} = 427.98\ldots,$$

$$A_{\text{pentagon}} = 5 \times 427.98 = 2139.88\ldots$$

Shaded region (circle minus pentagon):

$$2827.43 - 2139.88 = 687.5\ldots \approx 690 \text{ cm}^2 \text{ (2 significant figures)}.$$

Marker's note. Use $\tfrac12 ab\sin C$ for each triangle, set out a clear sequence of calculations, and round to 2 significant figures (not 2 decimal places) at the end.

Question 33 (3 marks)

A wombat can run at 40 km/h over short distances. At this speed, how many seconds would it take to run 150 metres?

Show worked solution

[3 marks]. Keep the units consistent. Convert 150 m to km: $0.15$ km. Then

$$\text{Time} = \frac{\text{distance}}{\text{speed}} = \frac{0.15}{40} = 0.00375 \text{ hours}.$$

Convert hours to seconds:

$$0.00375 \times 3600 = 13.5 \text{ seconds}.$$

Marker's note. Match the units before dividing (km with km/h giving hours, or m with m/s giving seconds), and remember the speed-distance-time relationship $\text{time} = \text{distance} \div \text{speed}$.

Question 34 (4 marks)

A container is a cylinder with a half sphere on each end. Three soccer balls of diameter 23 cm fit exactly inside, and the hemispherical ends just touch the end balls.
(Stimulus: a capsule-shaped container with three balls inside - see the official paper.)
What is the total surface area of the container, in square metres to 1 decimal place?

Show worked solution

[4 marks]. The two half spheres together form one sphere of diameter 23 cm (radius 11.5 cm). The cylindrical body holds three balls, so its length is $23 \times 2 = 46$ cm between the hemispheres (the end balls sit in the caps), and the cylinder is open at both ends.

Sphere surface area:

$$4\pi (11.5)^2 = 1661.9\ldots \text{ cm}^2.$$

Curved surface of the cylinder (open ends):

$$2\pi r h = 2\pi (11.5)(46) = 3323.8\ldots \text{ cm}^2.$$

Total:

$$1661.9 + 3323.8 = 4985.7 \text{ cm}^2.$$

Convert to square metres ($1 \text{ m}^2 = 10\,000 \text{ cm}^2$):

$$4985.7 \div 10\,000 = 0.5 \text{ m}^2 \text{ (1 decimal place)}.$$

Marker's note. Recognise the two hemispheres make one sphere and the cylinder is open, not capped. Convert $\text{cm}^2$ to $\text{m}^2$ by dividing by 10000, and give the answer in square metres.

Question 35 (5 marks)

A standard normal table gives the probability that $z$ is less than the listed value (z from 0.6 to 1.4).
(Stimulus: a standard normal probability table and a shaded normal curve - see the official paper.)
Exam scores are normally distributed with mean 58 and standard deviation 15.
(a) By calculating a z-score, find the percentage of scores between 58 and 70. (2 marks)
(b) Explain why the percentage of scores between 46 and 70 is twice your answer to (a). (1 mark)
(c) Using the table, find an approximate minimum score to be in the top 10% of candidates. (2 marks)

Show worked solution

(a) [2 marks].

$$z = \frac{70 - 58}{15} = 0.8.$$

From the table $P(z < 0.8) = 0.7881$. Since 58 is the mean, subtract the lower half:

$$0.7881 - 0.5 = 0.2881 = 28.81\%.$$

(b) [1 mark]. The score 46 has $z = \tfrac{46 - 58}{15} = -0.8$. By the symmetry of the normal curve, the area between 46 and 58 equals the area between 58 and 70. So the band 46 to 70 is twice the band 58 to 70, i.e. twice 28.81%.

(c) [2 marks]. The top 10% lies above the 90th percentile. The table value closest to 0.90 is $0.9032$ at $z = 1.3$. Then

$$x = \mu + z\sigma = 58 + 1.3 \times 15 = 77.5,$$

so the approximate minimum score is about 77 (78 is also acceptable).

Marker's note. A z-score is not a probability. In (b), use the symmetry of the curve, noting $z = -0.8$ for 46. In (c), match a cumulative probability near 0.90 to its z-score, then convert back to a raw score.

Question 36 (4 marks)

Two vertical flagpoles BE and CD stand on sloping ground, with angles 106 degrees and 27 degrees marked and AB = 20 m, with a 53.8 m measurement shown.
(Stimulus: a side view of two flagpoles on a slope - see the official paper.)
(a) What is the height of flagpole BE, to 1 decimal place? (2 marks)
(b) What is the height of flagpole CD, to 1 decimal place? (2 marks)

Show worked solution

(a) [2 marks]. In the triangle containing BE, apply the sine rule:

$$\frac{BE}{\sin 27 \text{ degrees}} = \frac{53.8}{\sin 106 \text{ degrees}},$$

$$BE = \frac{53.8 \times \sin 27 \text{ degrees}}{\sin 106 \text{ degrees}} = 25.4 \text{ m}.$$

(b) [2 marks]. Angle EBC $= 180 \text{ degrees} - 106 \text{ degrees} = 74$ degrees. In the right triangle with the 20 m base,

$$\tan 74 \text{ degrees} = \frac{20}{BX}, \qquad BX = \frac{20}{\tan 74 \text{ degrees}} = 5.73\ldots$$

So the height of CD is

$$CD = 25.4 - 5.73 = 19.7 \text{ m (1 decimal place)}.$$

Marker's note. Use the sine rule for BE (one side and its opposite angle), then trigonometric ratios in the right-angled triangle for CD. Track which lengths you have actually found before subtracting.

Question 37 (2 marks)

Sakura flies from Sydney (UTC +10) to Rio de Janeiro (UTC -3); the flight takes 20 hours and arrives at 3 pm on Wednesday 20 July. On what day and at what time does Sakura leave Sydney?

Show worked solution

[2 marks]. Work in Rio time first, then convert to Sydney. Subtract the flight time from the arrival:

$$3 \text{ pm Wed} - 20 \text{ hours} = 7 \text{ pm Tuesday 19 July (Rio time)}.$$

Sydney is 13 hours ahead of Rio ($+10$ versus $-3$), so add 13 hours:

$$7 \text{ pm Tue} + 13 \text{ hours} = 8 \text{ am Wednesday 20 July (Sydney time)}.$$

Sakura leaves Sydney at 8 am on Wednesday 20 July.

Marker's note. The time difference between UTC +10 and UTC -3 is 13 hours. Break the problem into two steps (subtract flight time, then convert zones) and keep careful track of the day as you cross midnight.

Question 38 (3 marks)

A cake is a cylinder (diameter 30 cm, height 6 cm) topped by a cone. The cylinder-to-cone volume ratio is 5 : 1. The cake is cut into equal slices of volume 212 cm cubed.
(Stimulus: a cylinder with a cone on top - see the official paper.)
How many equal slices can be cut from the cake?

Show worked solution

[3 marks]. Volume of the cylinder (radius 15 cm):

$$\pi \times 15^2 \times 6 = 4241.15\ldots \text{ cm}^3.$$

This is 5 parts; the whole cake (cylinder + cone) is 6 parts:

$$V_{\text{cake}} = \frac{6}{5} \times 4241.15 = 5089.38\ldots \text{ cm}^3.$$

Number of 212 cm cubed slices:

$$\frac{5089.38}{212} = 24.0\ldots \approx 24 \text{ slices}.$$

Marker's note. Use the correct cylinder volume formula and apply the ratio with the unitary method to get the total volume before dividing by the slice volume.

Question 39 (3 marks)

A nine-activity project is shown as a network. A table gives EST and LST for activities A (0, 2), C (0, 1) and I (12, 12). All times are in hours.
(Stimulus: an activity network with nine activities A to I - see the official paper.)
(a) What is the critical path? (1 mark)
(b) The minimum completion time is 19 hours. What is the duration of activity I? (1 mark)
(c) The duration of activity C is 3 hours. What is the maximum time that could occur between the start of activity F and the end of activity H? (1 mark)

Show worked solution

(a) [1 mark]. Activities A and C have float (their EST and LST differ), so they are not critical. The critical path is BEGI.

(b) [1 mark]. Activity I has EST 12 and the project finishes at 19 hours, so

$$\text{duration of I} = 19 - 12 = 7 \text{ hours}.$$

(c) [1 mark]. Activity C has EST 0 and duration 3, so its float lets the gap reach $12 - 3 = 9$ hours; allowing for the structure, the maximum time between the start of F and the end of H is 8 hours.

Marker's note. Allow for float times when choosing the critical path. Use the EST and LST values with forward and backward scanning, and write the final answers in the spaces, not only on the diagram.

Question 40 (3 marks)

A compass radial survey from O shows A, B at bearing 110 degrees, C, and D at bearing 285 degrees, with radial lengths 38 m, 42 m, 40 m and 67.5 m. AC is a straight line.
(Stimulus: a radial survey diagram - see the official paper.)
Find the bearing of C from O, to the nearest degree.

Show worked solution

[3 marks]. In triangle ODA, use the cosine rule with OA = 38, OD = 42 and AD = 67.5:

$$\cos(\angle AOD) = \frac{38^2 + 42^2 - 67.5^2}{2 \times 38 \times 42} = -0.4224\ldots,$$

$$\angle AOD = 115 \text{ degrees (nearest degree)}.$$

The bearing of D is 285 degrees, so $\angle DON = 360 - 285 = 75$ degrees. Since $\angle AOD = \angle DON + \angle AON$,

$$115 = 75 + \angle AON \implies \angle AON = 40 \text{ degrees},$$

so the bearing of A is 040 degrees. Because AC is a straight line through O, C is opposite A:

$$\text{bearing of C} = 040 + 180 = 220 \text{ degrees}.$$

Marker's note. Use the cosine rule first to find an unknown angle, then combine the radial-survey bearings. Remember angles on a straight line through O differ by 180 degrees.

Question 41 (4 marks)

Phoenix deposited a single sum 25 years ago into an account earning 2.4% per annum compounding monthly. A present-value annuity factor table is provided.
(Stimulus: a present-value interest-factor table - see the official paper.)
Withdrawals were $2000 at the end of each month for the first 15 years, then$1200 at the end of each month for the next 10 years. Calculate the minimum sum Phoenix could have deposited to make these withdrawals.

Show worked solution

[4 marks]. The monthly rate is $r = \tfrac{2.4\%}{12} = 0.002$. Value the withdrawals as two present values at deposit.

The $1200 monthly withdrawals run for all 25 years ($n = 300$months); the extra$800 per month ($2000 - 1200$) runs only for the first 15 years ($n = 180$ months). Using the table factors at $r = 0.002$:

$$1200 \times 225.430 = 270\,516,$$

$$800 \times 151.036 = 120\,828.80.$$

Minimum deposit:

$$270\,516 + 120\,828.80 = \$391\,344.80.$$

Marker's note. Split the withdrawals into a stream that runs for the full 25 years and the extra amount that runs only for the first 15 years, and select the matching present-value factors ($n = 300$ and $n = 180$) at $r = 0.002$.

General marker feedback

Stronger responses across the paper: showed clear mathematical reasoning and calculations; read each question carefully so no part was missed; 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 required units; rounded only at the final step; and constructed graphs and networks neatly with a ruler, showing all required information.