Skip to main content
ExamExplained
Maths Standard 2 study scene
Past papers
§-Past paper
NSWMaths Standard 22022

HSC Maths Standard 2 2022

Walkthrough of the 2022 HSC Mathematics Standard 2 exam: what the paper assessed, how to pace 100 marks in 150 minutes, six original exam-style worked questions on annuities and present value, bearings, flow networks, the normal distribution, inverse variation and composite solids, and common errors from NESA's 2022 marking feedback.

Marks
100
Time
150 min
Authority
NESA
Updated
Open the official PDF Official marking notes

What this paper assessed

The 2022 HSC Mathematics Standard 2 exam sampled every strand of the Year 11 and Year 12 Standard 2 course. Questions ranged from quick one-step skills in the multiple-choice section to multi-part problems in Section II that asked students to combine two or more ideas.

  • Financial mathematics: wages with an allowance, profit on a share trade after brokerage, present value under half-yearly compounding, tiered sales commission, saving towards a goal with a future value of an annuity table, comparing a lump-sum investment with regular quarterly deposits, declining-balance versus straight-line depreciation, and a reducing-balance loan set out in a spreadsheet, including the effect of an extra lump-sum repayment.
  • Measurement: time and ratio conversions, capture-recapture estimation, compass and true bearings, a two-triangle right-angled trigonometry problem, the area of a non-right-angled triangle used to find the capacity of a prism in litres, time zones using UTC offsets, the trapezoidal rule combined with a semicircle, the cosine and sine rules in a bearings context, and the surface area of a composite solid.
  • Networks: reading a critical path from an activity network, drawing a weighted network from a table and finding its minimum spanning tree, and a flow network where students found cut capacities, justified a maximum flow using a minimum cut and decided which path to upgrade.
  • Statistics: skewness from a histogram shape, how the mean and median respond to a new score, interpreting the intercept and gradient of a least-squares line, the standard normal table, cumulative frequency graphs and box plots, z-scores with the empirical rule, Pareto charts and cumulative percentages, plotting points and a line of best fit by eye, a four-mark bivariate "describe and interpret" question, and a normal distribution problem that combined a given table fact with symmetry.
  • Algebra and probability: matching a linear equation to its graph, reading a quadratic model from its graph, changing the subject of a formula, substituting into and solving a linear cost formula, inverse variation (finding the constant and graphing the curve), a tree diagram with expected wins, and a ratio problem where part of a mixture is replaced.

Structure and timing

The cover page sets 10 minutes of reading time and 2 hours and 30 minutes of working time for 100 marks. A NESA-approved calculator is allowed and a reference sheet is supplied at the back of the paper.

Section Questions Marks Suggested time
Section I: multiple choice 1 to 15 15 about 25 minutes
Section II: short and extended response 16 to 38 85 about 2 hours and 5 minutes

The overall rate is 150÷100=1.5150 \div 100 = 1.5 minutes per mark. Section I is allowed slightly more than that (25÷15≈1.725 \div 15 \approx 1.7 minutes per mark) and Section II slightly less (125÷85≈1.5125 \div 85 \approx 1.5 minutes per mark).

A practical split for Section II:

  • Questions 16 to 27 (41 marks): about 60 minutes. These are mostly one- and two-step skills; do not let any single part run past 3 minutes.
  • Questions 28 to 38 (44 marks): about 65 minutes. These are the multi-step questions (flow networks, trapezoidal rule, cosine and sine rules, loan spreadsheets, normal distribution). Budget about 6 to 8 minutes for each 4 or 5 mark question.
  • Keep the last 5 minutes to check units, rounding instructions and whether each answer is realistic.

Worked practice questions (exam-style)

Worked example

Question 1 (5 marks): Mei deposits $1500 at the end of every six months for 7 years into an account earning 5% per annum, compounded half-yearly. Her brother Jun wants to have exactly the same amount after 7 years, but he will make a single deposit today into an account earning 4.2% per annum, compounded monthly. Part of a future value of an annuity of $1 table is shown: for a rate per period of 0.025, the factors are 7.54743 (7 periods), 16.51895 (14 periods) and 27.18327 (21 periods); for 0.05 they are 8.14201, 19.59863 and 35.71925. (a) Find the future value of Mei's deposits. (b) Find the single deposit Jun must make today. (c) Who earns more interest, and by how much?

Step 1: Mei's deposits form an annuity, so use the table

Half-yearly rate r=0.052=0.025r = \dfrac{0.05}{2} = 0.025 and number of periods n=7×2=14n = 7 \times 2 = 14. The factor for r=0.025r = 0.025, n=14n = 14 is 16.5189516.51895.

FV=1500×16.51895≈$24 778.43FV = 1500 \times 16.51895 \approx \$24\,778.43

Step 2: Jun makes one deposit, so use the present value form of compound interest

Monthly rate r=0.04212=0.0035r = \dfrac{0.042}{12} = 0.0035 and number of periods n=7×12=84n = 7 \times 12 = 84.

PV=FV(1+r)n=24 778.43(1.0035)84≈$18 476.26PV = \frac{FV}{(1 + r)^n} = \frac{24\,778.43}{(1.0035)^{84}} \approx \$18\,476.26

Step 3: Compare the interest, not the deposits

Each person ends with the same balance, so the interest is the balance minus the total amount deposited.

Mei deposits 1500×14=$21 000,Interest=24 778.43−21 000=$3778.43\text{Mei deposits } 1500 \times 14 = \$21\,000, \qquad \text{Interest} = 24\,778.43 - 21\,000 = \$3778.43

Jun’s interest=24 778.43−18 476.26=$6302.17\text{Jun's interest} = 24\,778.43 - 18\,476.26 = \$6302.17

6302.17−3778.43=$2523.746302.17 - 3778.43 = \$2523.74

Final answer: (a) $24 778.43\text{\textdollar}24\,778.43; (b) $18 476.26\text{\textdollar}18\,476.26; (c) Jun earns more interest, by $2523.74\text{\textdollar}2523.74, because his whole deposit earns interest for the full 7 years.

Worked example

Question 2 (5 marks): A kayaker paddles 12 km from a jetty P on a true bearing of 060° to a buoy Q. She then paddles 15 km on a true bearing of 170° to an island R. (a) Show that ∠PQR=70∘\angle PQR = 70^\circ. (b) Find the distance PR, correct to 1 decimal place. (c) Find the true bearing of R from P, to the nearest degree.

Step 1: Find the angle at Q

The bearing of P from Q is the back bearing of 060∘060^\circ, which is 060∘+180∘=240∘060^\circ + 180^\circ = 240^\circ. The leg QR leaves Q on 170∘170^\circ. The angle between the two directions at Q is

∠PQR=240∘−170∘=70∘\angle PQR = 240^\circ - 170^\circ = 70^\circ

Step 2: Cosine rule for PR (two sides and the included angle)

PR2=122+152−2×12×15×cos⁡70∘≈245.87PR^2 = 12^2 + 15^2 - 2 \times 12 \times 15 \times \cos 70^\circ \approx 245.87

PR≈15.68≈15.7 kmPR \approx 15.68 \approx 15.7 \text{ km}

Step 3: Sine rule for the angle at P

sin⁡∠QPR15=sin⁡70∘15.68⇒sin⁡∠QPR≈0.899⇒∠QPR≈64∘\frac{\sin \angle QPR}{15} = \frac{\sin 70^\circ}{15.68} \quad\Rightarrow\quad \sin \angle QPR \approx 0.899 \quad\Rightarrow\quad \angle QPR \approx 64^\circ

QR (15 km, opposite P) is shorter than PR (opposite the 70∘70^\circ angle at Q), so ∠QPR\angle QPR is less than 70∘70^\circ and must be acute; the calculator value is the one we want.

Step 4: Convert to a bearing

R is clockwise from the direction PQ, so add the angle to the bearing of Q from P:

060∘+64∘=124∘060^\circ + 64^\circ = 124^\circ

Final answer: (a) ∠PQR=70∘\angle PQR = 70^\circ; (b) PR≈15.7PR \approx 15.7 km; (c) the bearing of R from P is about 124∘T124^\circ\text{T}.

Worked example

Question 3 (5 marks): Water flows through a pipe network from a reservoir S to a town T through pumping stations P, Q and R. The directed pipes and their capacities (in megalitres per day) are: S to P 14, S to Q 10, P to Q 4, P to R 8, P to T 5, Q to R 6, Q to T 7, R to T 15. (a) Find the maximum flow from S to T, justifying your answer with a minimum cut. (b) The pipe from P to T is closed for repairs. Find the new maximum flow and identify a minimum cut that confirms it.

Step 1: List the cuts that separate S from T

Keep S on the source side and T on the sink side, and count only pipes that go from the source side to the sink side. For example:

{S}: 14+10=24,{S,P}: 10+4+8+5=27,{S,P,Q}: 8+5+6+7=26\{S\}: \ 14 + 10 = 24, \qquad \{S, P\}: \ 10 + 4 + 8 + 5 = 27, \qquad \{S, P, Q\}: \ 8 + 5 + 6 + 7 = 26

Checking the remaining cuts shows none is smaller than 24.

Step 2: Confirm a flow of 24 is achievable

Send 14 into P: 5 to T, 8 to R and 1 to Q. Send 10 into Q, so Q receives 11: 7 to T and 4 to R. Then R receives 12 and sends it to T (capacity 15). Every pipe is within capacity and 5+7+12=245 + 7 + 12 = 24 arrives at T. By the maximum-flow minimum-cut theorem, the maximum flow is 24.

Step 3: Remove P to T and recheck the cuts

With P to T gone, the cut {S,P,Q}\{S, P, Q\} now crosses only P to R, Q to R and Q to T:

8+6+7=218 + 6 + 7 = 21

The other cuts become {S}:24\{S\}: 24, {S,P}:10+4+8=22\{S, P\}: 10 + 4 + 8 = 22 and {S,P,Q,R}:7+15=22\{S, P, Q, R\}: 7 + 15 = 22, so 21 is the smallest.

Step 4: Confirm a flow of 21

Send 11 into P: 8 to R and 3 to Q. Send 10 into Q, so Q receives 13: 7 to T and 6 to R. R receives 14 and sends it to T. The total reaching T is 7+14=217 + 14 = 21, with every pipe within capacity.

Final answer: (a) maximum flow = 24 ML per day, because the cut through S to P and S to Q has capacity 24 and a flow of 24 can be achieved; (b) the maximum flow falls to 21 ML per day, confirmed by the cut through P to R, Q to R and Q to T (capacity 8+6+7=218 + 6 + 7 = 21).

Worked example

Question 4 (4 marks): The heights of seedlings in a nursery are normally distributed with a mean of 30 cm and a standard deviation of 4 cm. (a) Use the empirical rule to find the percentage of seedlings with heights between 22 cm and 34 cm. (b) In a tray of 400 of these seedlings, about how many would be shorter than 26 cm? (c) A seedling from a second nursery, where heights have a mean of 26 cm and a standard deviation of 3 cm, is 32 cm tall. Is it relatively taller than a 36 cm seedling from the first nursery? Justify using z-scores.

Step 1: Convert the heights to z-scores

z22=22−304=−2,z34=34−304=1z_{22} = \frac{22 - 30}{4} = -2, \qquad z_{34} = \frac{34 - 30}{4} = 1

Step 2: Build the percentage from the mean outwards

Half of the 95% within z=±2z = \pm 2 lies between z=−2z = -2 and the mean, and half of the 68% within z=±1z = \pm 1 lies between the mean and z=1z = 1:

47.5%+34%=81.5%47.5\% + 34\% = 81.5\%

Step 3: Shorter than 26 cm

z26=26−304=−1z_{26} = \dfrac{26 - 30}{4} = -1. Outside z=±1z = \pm 1 is 100%−68%=32%100\% - 68\% = 32\%, and only half of that is in the lower tail, so 16% are shorter than 26 cm.

0.16×400=64 seedlings0.16 \times 400 = 64 \text{ seedlings}

Step 4: Compare z-scores

zsecond=32−263=2,zfirst=36−304=1.5z_{\text{second}} = \frac{32 - 26}{3} = 2, \qquad z_{\text{first}} = \frac{36 - 30}{4} = 1.5

The second-nursery seedling is 2 standard deviations above its mean, compared with 1.5 for the first-nursery seedling.

Final answer: (a) 81.5%; (b) about 64 seedlings; (c) yes, the 32 cm seedling is relatively taller (z=2z = 2 compared with z=1.5z = 1.5).

Worked example

Question 5 (4 marks): The number of days, DD, needed to paint a mural varies inversely with the number of painters, PP. A team of 6 painters takes 10 days. (a) Find an equation relating DD and PP. (b) How long would 8 painters take? (c) What is the smallest number of painters needed to finish in 4 days or fewer?

Step 1: Write the inverse variation and find kk

D=kP,10=k6⇒k=60D = \frac{k}{P}, \qquad 10 = \frac{k}{6} \quad\Rightarrow\quad k = 60

D=60PD = \frac{60}{P}

Step 2: Substitute P=8P = 8

D=608=7.5 daysD = \frac{60}{8} = 7.5 \text{ days}

Step 3: Rearrange for PP with D=4D = 4

4=60P⇒P=604=154 = \frac{60}{P} \quad\Rightarrow\quad P = \frac{60}{4} = 15

A check on reasonableness: more painters should mean fewer days, and 15 painters taking 4 days fits that pattern (the product is still 60 painter-days).

Final answer: (a) D=60PD = \dfrac{60}{P}; (b) 7.5 days; (c) 15 painters.

Worked example

Question 6 (4 marks): A solid toy is made from a cylinder with radius 5 cm and height 12 cm, topped by a cone with the same radius and a perpendicular height of 12 cm. (a) Find the total surface area of the toy, correct to 1 decimal place. (b) The toy maker also produces a hollow version of the same shape to use as a container. Find its capacity in litres, correct to 2 decimal places.

Step 1: Identify the exposed surfaces

The toy has a circular base, the curved side of the cylinder and the curved surface of the cone. The top of the cylinder is covered by the cone, so it is not counted.

Step 2: Slant height of the cone

l=52+122=169=13 cml = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 \text{ cm}

Step 3: Add the surface areas

Base=πr2=25π,Cylinder side=2πrh=2π×5×12=120π\text{Base} = \pi r^2 = 25\pi, \qquad \text{Cylinder side} = 2\pi r h = 2\pi \times 5 \times 12 = 120\pi

Cone surface=πrl=π×5×13=65π\text{Cone surface} = \pi r l = \pi \times 5 \times 13 = 65\pi

SA=25π+120π+65π=210π≈659.7 cm2SA = 25\pi + 120\pi + 65\pi = 210\pi \approx 659.7 \text{ cm}^2

Step 4: Volume, then convert to litres

V=πr2h+13πr2h=300π+100π=400π≈1256.6 cm3V = \pi r^2 h + \tfrac{1}{3}\pi r^2 h = 300\pi + 100\pi = 400\pi \approx 1256.6 \text{ cm}^3

Since 1000 cm3=1 L1000 \text{ cm}^3 = 1 \text{ L}, the capacity is 1256.6÷1000≈1.261256.6 \div 1000 \approx 1.26 L.

Final answer: (a) about 659.7 cm2659.7 \text{ cm}^2; (b) about 1.26 L.

Common errors students made

NESA's 2022 marking feedback for Mathematics Standard 2 pointed to these recurring problems (paraphrased):

  • Percentages and commission. Finding a percentage of an amount was a weakness in the heart-rate question, and in the commission question many students did not apply different rates to the different portions of the sale price.
  • Outcomes versus probabilities. In the tree diagram question, students confused an outcome with a probability, forgot that a probability must lie between 0 and 1, and did not check that an expected number of wins could not exceed the number of games played.
  • Normal distribution. Finding a z-score was listed as an area to improve, and the feedback stressed that only half of the 5% outside two standard deviations lies above the mean. Feedback recommended sketching a labelled bell curve and using the reference sheet, especially in the question that combined a table fact with symmetry.
  • Network terminology. Some students drew a spanning tree rather than a minimum spanning tree, or left out the weights or the total length. In the flow question, students mixed up a path and a vertex and could not calculate the capacity of a cut.
  • Rearranging and reasonableness. Rearranging formulas (including trigonometric ratios with the unknown in the denominator) caused errors, and answers that made no sense in context were not questioned. The depreciation question drew answers of thousands of years for a machine to lose value, and some students confused a rate of depreciation with an amount.
  • Lines of best fit. Points were misread from the table, lines were not drawn with a ruler, and some students gave a range rather than a single estimate from their line.
  • Annuities versus lump sums. Students did not distinguish an annuity from compound interest on a single deposit, had trouble reading the correct factor from a table, and did not convert the annual rate and time to match the compounding period.
  • Measurement formulas. Some students found surface area when the question needed volume for a capacity, misused the trapezoidal rule or the semicircle area, mixed up area and volume formulas, and forgot to halve a diameter to get the radius.
  • Bearings and directed rules. Students were unsure how true and compass bearings are written, did not use the inverse sine function correctly, and sometimes used a different rule from the one the question told them to use.
  • Time zones. Students added the flight time incorrectly or did not state both the time and the day.
  • Bivariate interpretation. Responses confused strong, moderate and weak correlation, wrongly linked skewness to the correlation coefficient, mixed up interpolation and extrapolation, and did not describe the data in context. The feedback noted that a four-mark question usually needs four distinct points.
  • Ratios. Simplified ratios were left with decimals instead of being written as whole numbers.

How to use this paper

  1. Sit it under exam conditions. Download the official 2022 paper and give yourself 10 minutes of reading time and 2 hours and 30 minutes of working time with only the reference sheet and an approved calculator.
  2. Mark it against the official criteria. Use the 2022 marking guidelines to see how part marks are awarded. Many Section II questions give a mark for the first correct step, so note where you would have picked up partial credit.
  3. Read the feedback against your mistakes. Open the 2022 marking feedback and match each error you made to one of the problems listed above.
  4. Re-drill by strand. Redo the six worked questions on this page without looking, then find two more questions from the same strand for any topic where you dropped marks, such as annuity tables, flow networks or normal distribution symmetry.
  5. Build a reasonableness habit. After every answer, ask whether it makes sense in context: a probability above 1, a machine depreciating for thousands of years or a bearing over 360° is a signal to go back and check.

Use this paper well

  1. Sit the paper under exam conditions (150 minutes, 100 marks).
  2. Mark yourself against the official NESA marking notes.
  3. Compare against the Maths Standard 2 hub to find the syllabus dot points this paper tested.

Keep going

ExamExplained