Mathematics Standard 2 HSC Reference Sheet — Explained
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The NESA Mathematics Standard 2 reference sheet travels with you into every exam — but having a formula in front of you is not the same as knowing when to reach for it. Below, the reference-sheet formulas are unpacked alongside the essential assumed-knowledge formulas you still need to know: symbols decoded, a worked example, and practice to cement your understanding.
In short: the NESA Mathematics Standard 2 reference sheet comes with you into every exam — but knowing a formula is printed isn't the same as knowing when to use it. This guide explains each formula on the sheet, grouped by topic (measurement, financial maths, statistical analysis, algebra and relationships, networks), so you can apply the right one with confidence.
Measurement
14What each symbol means
$A$ — area of sector; $r$ — radius; $\theta$ — angle at centre in degrees.
When to use it
Use when a question asks for the area of a 'pizza-slice' shaped region and gives the radius and the central angle in degrees.
Units:
Area is in square units (e.g. cm², m²).
Worked sample
Find the area of a sector with radius $8$ cm and central angle $45^\circ$.
$A = \dfrac{45}{360} \times \pi \times 8^2 = \dfrac{1}{8} \times 64\pi \approx 25.1$ cm².
Your turn:
A sector has radius $12$ m and central angle $120^\circ$. Find its area to 1 dp.
$A = \dfrac{120}{360} \times \pi \times 12^2 = \dfrac{1}{3} \times 144\pi \approx 150.8$ m².
If the angle is given in radians, use $A = \tfrac{1}{2}r^2\theta$ instead — the Standard 2 sheet uses degrees.
What each symbol means
$l$ — arc length; $r$ — radius; $\theta$ — central angle in degrees.
When to use it
Use when you need the curved edge (arc) of a sector, not the straight sides.
Units:
Length units (e.g. cm, m).
Worked sample
Find the arc length of a sector with $r = 5$ cm and $\theta = 72^\circ$.
$l = \dfrac{72}{360} \times 2\pi \times 5 = 0.2 \times 10\pi \approx 6.28$ cm.
Your turn:
A sector has radius $9$ m and central angle $80^\circ$. Find the arc length to 2 dp.
$l = \dfrac{80}{360} \times 2\pi \times 9 = \dfrac{2}{9} \times 18\pi \approx 12.57$ m.
The perimeter of a sector = arc length + 2 radii. Don't forget to add the two straight edges if asked for the full perimeter.
What each symbol means
$c$ — hypotenuse (longest side, opposite the right angle); $a, b$ — the two shorter sides.
When to use it
Any right-angled triangle where you know two sides and need the third. Identify the hypotenuse first — it is always opposite the right angle.
Units:
Same length units as the sides.
Worked sample
A right triangle has legs $6$ cm and $8$ cm. Find the hypotenuse.
$c^2 = 6^2 + 8^2 = 36 + 64 = 100$, so $c = 10$ cm.
Your turn:
The hypotenuse is $13$ m and one leg is $5$ m. Find the other leg.
$a^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $a = 12$ m.
To find a leg rather than the hypotenuse, rearrange to $a^2 = c^2 - b^2$. Always subtract the other leg² from the hypotenuse².
What each symbol means
$\theta$ — the reference angle; opp — side opposite $\theta$; adj — side adjacent to $\theta$; hyp — hypotenuse.
When to use it
Right-angled triangles only. Use when one angle and one side are known (or two sides and you need an angle).
Worked sample
In a right triangle, $\theta = 35^\circ$ and the hypotenuse $= 10$ cm. Find the opposite side.
$\text{opp} = 10\sin 35^\circ \approx 10 \times 0.574 \approx 5.74$ cm.
Your turn:
In a right triangle, the opposite side is $7$ m and the hypotenuse is $25$ m. Find $\theta$.
$\theta = \sin^{-1}\!\left(\dfrac{7}{25}\right) = \sin^{-1}(0.28) \approx 16.3^\circ$.
Label all three sides relative to the angle you're using before choosing a ratio. Mislabelling opp/adj is the most common error.
What each symbol means
$a, b, c$ — side lengths; $A, B, C$ — angles opposite those sides respectively.
When to use it
Non-right triangles. Use when you know an angle and its opposite side, plus one more angle or side. Cannot be used if all three given values are sides.
Worked sample
In $\triangle ABC$, $A = 40^\circ$, $B = 75^\circ$, $a = 8$ m. Find $b$.
$b = \dfrac{8 \sin 75^\circ}{\sin 40^\circ} = \dfrac{8 \times 0.966}{0.643} \approx 12.0$ m.
Your turn:
In $\triangle ABC$, $a = 5$, $A = 30^\circ$, $B = 80^\circ$. Find $b$ (1 dp).
$b = \dfrac{5\sin 80^\circ}{\sin 30^\circ} = \dfrac{5 \times 0.985}{0.5} \approx 9.8$.
Always check whether the ambiguous case applies (angle-side-side given and angle is acute). In Standard exams a diagram usually makes it clear which triangle is intended.
What each symbol means
$a, b, c$ — side lengths; $C$ — the angle between sides $a$ and $b$ (opposite side $c$).
When to use it
Non-right triangles. Use when you know two sides and the included angle (SAS), or all three sides (SSS) and need an angle.
Worked sample
In $\triangle ABC$, $a = 6$, $b = 10$, $C = 50^\circ$. Find $c$ (1 dp).
$c^2 = 36 + 100 - 2(6)(10)\cos 50^\circ = 136 - 77.1 = 58.9$, so $c \approx 7.7$.
Your turn:
Three sides: $a = 5$, $b = 7$, $c = 9$. Find angle $C$ (nearest degree).
$\cos C = \dfrac{25 + 49 - 81}{70} = \dfrac{-7}{70} \approx -0.1$, so $C \approx 96^\circ$.
To find an angle, rearrange to $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$. A negative cosine means $C$ is obtuse — that is fine, do not panic.
What each symbol means
$a, b$ — two side lengths; $C$ — the angle between them.
When to use it
Non-right triangles where you know two sides and the included angle but not the height. If you know the base and perpendicular height, $A = \tfrac{1}{2}bh$ is simpler.
Units:
Square units.
Worked sample
Find the area of $\triangle ABC$ with $a = 7$, $b = 5$, $C = 60^\circ$.
$A = \tfrac{1}{2}(7)(5)\sin 60^\circ = 17.5 \times 0.866 \approx 15.2$ square units.
Your turn:
Find the area with $a = 8$ cm, $b = 11$ cm, $C = 110^\circ$ (1 dp).
$A = \tfrac{1}{2}(8)(11)\sin 110^\circ = 44 \times 0.940 \approx 41.4$ cm².
$\sin C$ is positive for both acute and obtuse angles, so the formula works whether the included angle is less than or greater than $90^\circ$.
What each symbol means
$r$ — radius of the sphere.
When to use it
Use to find the total surface area of a complete sphere. For a hemisphere, halve it and add the circular base: $SA = 2\pi r^2 + \pi r^2 = 3\pi r^2$.
Units:
Square units.
Worked sample
Find the surface area of a sphere with radius $6$ cm (1 dp).
$SA = 4\pi(6)^2 = 144\pi \approx 452.4$ cm².
Your turn:
A sphere has surface area $200\pi$ cm². Find its radius.
$r^2 = \dfrac{200\pi}{4\pi} = 50$, so $r = \sqrt{50} \approx 7.07$ cm.
Watch out for hemispheres — the question often asks for the total surface area including the flat circular face.
What each symbol means
$r$ — radius.
When to use it
Volume of any sphere or ball. For a hemisphere divide the result by 2.
Units:
Cubic units.
Worked sample
Find the volume of a sphere with radius $3$ cm (2 dp).
$V = \dfrac{4}{3}\pi(3)^3 = \dfrac{4}{3}\pi \times 27 = 36\pi \approx 113.10$ cm³.
Your turn:
A sphere has volume $288\pi$ cm³. Find its radius.
$r^3 = \dfrac{3 \times 288\pi}{4\pi} = 216$, so $r = 6$ cm.
The $r^3$ makes a big difference — if the radius doubles, the volume increases 8-fold. Re-read whether the question gives radius or diameter.
What each symbol means
$r$ — base radius; $h$ — perpendicular height.
When to use it
Volume of any right cone. A pyramid with a circular base. Compare: a cylinder has volume $\pi r^2 h$, so the cone holds exactly one-third of the enclosing cylinder.
Units:
Cubic units.
Worked sample
A cone has base radius $4$ cm and height $9$ cm. Find its volume (2 dp).
$V = \dfrac{1}{3}\pi(4)^2(9) = \dfrac{1}{3} \times 144\pi \approx 150.80$ cm³.
Your turn:
A conical pile of sand has $r = 5$ m and $h = 6$ m. Volume to 1 dp?
$V = \dfrac{1}{3}\pi(25)(6) = 50\pi \approx 157.1$ m³.
Make sure you use the perpendicular height, not the slant height. If given the slant height, use Pythagoras first.
What each symbol means
$A$ — area of the cross-sectional base; $h$ — perpendicular height (or length).
When to use it
Any prism (triangular, rectangular, trapezoidal, etc.) or cylinder. First find the area of the uniform cross-section, then multiply by the length.
Units:
Cubic units.
Worked sample
A trapezoidal prism has parallel sides $4$ m and $6$ m, height of trapezium $3$ m, and length $10$ m. Find the volume.
$A = \tfrac{1}{2}(4+6)(3) = 15$ m². $V = 15 \times 10 = 150$ m³.
Your turn:
A cylinder has radius $5$ cm and length $12$ cm. Volume to 1 dp?
$A = \pi(5)^2 = 25\pi$ cm². $V = 25\pi \times 12 = 300\pi \approx 942.5$ cm³.
Identify the base shape first and write out its area formula before multiplying by $h$. Mixing up area and volume formulas is the most common mistake here.
What each symbol means
$h$ — the distance between the two offsets; $d_f$ — first offset; $d_l$ — last offset.
When to use it
Estimating the area of an irregular shape when you have two parallel offset measurements and the perpendicular distance between them. Common in land-survey and irregular-paddock questions.
Units:
Square units (if measurements are in metres, area is in m²).
Worked sample
A field is measured with two perpendicular offsets $d_f=10$ m and $d_l=8$ m, a distance $h=6$ m apart. Estimate the area.
$A = \dfrac{6}{2}(10+8) = 3\times18 = 54$ m².
Your turn:
Estimate the area with offsets $5$ m and $7$ m, $h=8$ m apart.
$A = \dfrac{8}{2}(5+7) = 4\times12 = 48$ m².
On the Standard 2 sheet, $h$ is the full distance between the two offsets and the divisor is 2 (not 3) — only two offsets are used.
What each symbol means
$\text{precision}$ — the smallest unit of measurement (e.g. 1 mm, 0.1 kg); absolute error = half of that precision.
When to use it
Any question asking for upper or lower bounds of a measurement, or asking how accurate a given measurement is. Upper bound = measurement + absolute error; lower bound = measurement − absolute error.
Units:
Same units as the measurement.
Worked sample
A length is recorded as $8.4$ cm (measured to the nearest 0.1 cm). Find the absolute error and state the upper and lower bounds.
Absolute error $= \tfrac{1}{2} \times 0.1 = 0.05$ cm. Lower bound $= 8.35$ cm; upper bound $= 8.45$ cm.
Your turn:
A mass is measured as $3$ kg to the nearest kilogram. Find the absolute error and the bounds.
Absolute error $= 0.5$ kg. Bounds: $2.5$ kg $\le$ mass $< 3.5$ kg.
The precision is the smallest graduation on the measuring instrument. A reading to the nearest millimetre has precision 1 mm and absolute error 0.5 mm.
What each symbol means
$a, b$ — the two parallel sides; $h$ — the perpendicular distance between them.
When to use it
Any quadrilateral with exactly one pair of parallel sides (a trapezium). Also used in the trapezoidal rule for estimating irregular areas.
Units:
Square units.
Worked sample
A trapezium has parallel sides $5$ m and $9$ m, with a perpendicular height of $4$ m. Find the area.
$A = \dfrac{4}{2}(5+9) = 2 \times 14 = 28$ m².
Your turn:
Parallel sides are $7$ cm and $13$ cm; perpendicular height is $6$ cm. Find the area.
$A = \dfrac{6}{2}(7+13) = 3 \times 20 = 60$ cm².
$h$ must be the perpendicular distance between the parallel sides, not the length of a slanted leg. Check the diagram carefully before substituting.
Financial Mathematics
7What each symbol means
$I$ — interest earned; $P$ — principal (initial amount); $r$ — interest rate per period (as a decimal); $n$ — number of periods.
When to use it
Interest is calculated only on the original principal each period — common in short-term loans, some bonds, and Basic/Standard questions labelled 'simple interest'.
Units:
$ (dollars). Ensure $r$ and $n$ use the same time unit (e.g. both annual, or both monthly).
Worked sample
Find the simple interest on $\$5000$ at $6\%$ p.a. for $3$ years.
$I = 5000 \times 0.06 \times 3 = \$900$.
Your turn:
How much simple interest accrues on $\$12\,000$ at $4.5\%$ p.a. over $2.5$ years?
$I = 12000 \times 0.045 \times 2.5 = \$1350$.
Convert the percentage rate to a decimal before substituting ($6\% = 0.06$). Check whether the rate is given per year while the time is in months — adjust one to match the other.
What each symbol means
$FV$ — future value; $PV$ — present value (principal); $r$ — interest rate per compounding period (as a decimal); $n$ — number of compounding periods.
When to use it
Interest is added to the balance each period so that subsequent interest is earned on a larger amount. Use whenever a question says 'compound interest', 'compounding annually/quarterly/monthly', or 'growing investment'.
Units:
Dollars. Make sure $r$ and $n$ match the same compounding period.
Worked sample
$\$8000$ is invested at $5\%$ p.a. compounding annually for $4$ years. Find the future value.
$FV = 8000(1 + 0.05)^4 = 8000 \times 1.2155 \approx \$9724$.
Your turn:
$\$15\,000$ at $3.6\%$ p.a. compounding monthly for $2$ years. Find $FV$ (nearest cent).
$r = 0.036/12 = 0.003$, $n = 24$. $FV = 15000(1.003)^{24} \approx \$16\,118.09$.
Always divide the annual rate by the number of compounding periods per year, and multiply $n$ by that same number. Forgetting to adjust for monthly compounding is the most common error.
What each symbol means
$PV$ — present value (amount needed now); $FV$ — target future value; $r$ — rate per period; $n$ — number of periods.
When to use it
'How much do I need to invest today to reach $\$X$ in $n$ years?' Rearranges the $FV$ formula by dividing both sides by $(1+r)^n$.
Units:
Dollars.
Worked sample
How much must be invested now at $4\%$ p.a. compounding annually to have $\$20\,000$ in $10$ years?
$PV = 20000(1.04)^{-10} = 20000 / 1.4802 \approx \$13\,514$.
Your turn:
Target: $\$50\,000$ in $5$ years at $6\%$ p.a. compounding annually. Find $PV$.
$PV = 50000(1.06)^{-5} = 50000 / 1.3382 \approx \$37\,363$.
On a calculator use the negative exponent: $1.04^{-10}$. Some students divide by $(1+r)^n$ which is identical — either approach works.
What each symbol means
$S$ — salvage value after $n$ periods; $V_0$ — initial value; $D$ — amount of depreciation per period; $n$ — number of periods.
When to use it
An asset loses the same dollar amount each period. Common for equipment, tools, and vehicles in HSC questions that specify 'straight-line' or 'flat-rate' depreciation.
Units:
Dollars.
Worked sample
A machine costs $\$24\,000$ and depreciates by $\$3000$ per year. Find its value after $5$ years.
$S = 24000 - 3000 \times 5 = 24000 - 15000 = \$9000$.
Your turn:
A vehicle is purchased for $\$32\,000$ and depreciates $\$4500$/year. After how many years is it worth $\$14\,000$?
$14000 = 32000 - 4500n \Rightarrow 4500n = 18000 \Rightarrow n = 4$ years.
The value decreases by a fixed amount each period (linear graph). If the rate is given as a percentage of the original, multiply to get $D$ first: $D = V_0 \times r$.
What each symbol means
$S$ — value after $n$ periods; $V_0$ — initial value; $r$ — depreciation rate per period (as a decimal); $n$ — number of periods.
When to use it
An asset loses a fixed percentage of its current value each period (so the dollar loss shrinks over time). Common for electronic equipment, cars, machinery — any HSC question saying 'declining balance' or 'reducing value'.
Units:
Dollars.
Worked sample
A laptop worth $\$3000$ depreciates at $20\%$ p.a. Find its value after $3$ years.
$S = 3000(1 - 0.20)^3 = 3000(0.8)^3 = 3000 \times 0.512 = \$1536$.
Your turn:
A car is purchased for $\$40\,000$ and depreciates at $15\%$ p.a. Find its value after $4$ years (nearest dollar).
$S = 40000(0.85)^4 = 40000 \times 0.5220 \approx \$20\,880$.
Compare with compound interest: for depreciation you multiply by $(1-r)$ each period; for growth you multiply by $(1+r)$. The structures are mirror images.
What each symbol means
$FV$ — future value of the annuity; $M$ — regular payment amount per period; $r$ — interest rate per period (decimal); $n$ — total number of payments.
When to use it
Equal deposits or payments are made at the end of each period and earn compound interest (e.g. regular superannuation contributions, savings plans). Use when the question asks for the accumulated value of a series of payments.
Units:
Dollars.
Worked sample
$\$500$ is deposited at the end of each quarter into an account earning $4\%$ p.a. compounding quarterly. Find the future value after $3$ years.
$r = 0.04/4 = 0.01$, $n = 12$. $FV = 500 \times \dfrac{(1.01)^{12}-1}{0.01} = 500 \times \dfrac{0.1268}{0.01} = 500 \times 12.68 \approx \$6341$.
Your turn:
$\$200$/month into an account at $6\%$ p.a. compounding monthly for $5$ years. Find $FV$ (nearest dollar).
$r = 0.005$, $n = 60$. $FV = 200 \times \dfrac{(1.005)^{60}-1}{0.005} = 200 \times 69.77 \approx \$13\,954$.
Break the formula into steps: compute $(1+r)^n$ first, subtract 1, divide by $r$, then multiply by $M$. Don't try to enter it all in one calculator step.
What each symbol means
$PV$ — lump sum now that is equivalent to the annuity stream; $M$ — regular payment per period; $r$ — interest rate per period (decimal); $n$ — total number of payments.
When to use it
Loans, mortgages, and 'how much is this stream of payments worth today?' questions. The opposite of the future-value annuity formula.
Units:
Dollars.
Worked sample
A loan is repaid with monthly payments of $\$800$ at $6\%$ p.a. compounding monthly over $3$ years. Find the loan principal.
$r = 0.005$, $n = 36$. $PV = 800 \times \dfrac{1-(1.005)^{-36}}{0.005} = 800 \times 32.87 \approx \$26\,296$.
Your turn:
An annuity pays $\$1000$/quarter for $4$ years. Interest rate is $8\%$ p.a. compounding quarterly. Find the PV (nearest dollar).
$r = 0.02$, $n = 16$. $PV = 1000 \times \dfrac{1-(1.02)^{-16}}{0.02} \approx 1000 \times 13.58 \approx \$13\,578$.
For a loan, the PV is the amount borrowed. For an investment, it is the lump sum that gives the same return. Check which direction the question is asking.
Statistical Analysis
7What each symbol means
$\bar{x}$ — sample mean; $\sum x$ — sum of all data values; $n$ — number of values.
When to use it
Finding the central value of a data set. Use on ungrouped data, or as a weighted average when frequencies are given.
Units:
Same units as the data.
Worked sample
Find the mean of $3, 7, 8, 12, 5$.
$\bar{x} = \dfrac{3+7+8+12+5}{5} = \dfrac{35}{5} = 7$.
Your turn:
Data: $15, 22, 8, 30, 10, 19$. Find the mean.
$\bar{x} = \dfrac{104}{6} \approx 17.3$.
For a frequency table, multiply each value by its frequency, sum the products, then divide by the total frequency: $\bar{x} = \dfrac{\sum fx}{\sum f}$.
What each symbol means
$\sigma$ — population standard deviation; $x$ — each data value; $\bar{x}$ — mean; $n$ — number of values.
When to use it
Measuring the spread of a data set around the mean. A larger $\sigma$ means data is more spread out. In HSC Standard 2, use the calculator's $\sigma_n$ (population SD) unless told otherwise.
Units:
Same units as the data.
Worked sample
For the data set $2, 4, 4, 4, 5, 5, 7, 9$, find the population standard deviation (given $\bar{x}=5$).
$\sigma = \sqrt{\dfrac{(2-5)^2+(4-5)^2+\cdots+(9-5)^2}{8}} = \sqrt{\dfrac{32}{8}} = \sqrt{4} = 2$.
Your turn:
Data: $10, 12, 14, 16, 18$. Find $\sigma$ (2 dp).
$\bar{x}=14$. $\sigma = \sqrt{\dfrac{16+4+0+4+16}{5}} = \sqrt{8} \approx 2.83$.
Use your calculator's statistics mode (STAT or SD) to compute $\sigma_n$ directly — do not calculate it by hand in an exam.
What each symbol means
$x$ — score; $\mu$ — mean; $\sigma$ — standard deviation.
When to use it
Comparing scores from different distributions, or finding how many standard deviations a value lies from the mean. A $z$-score of $0$ means the value equals the mean; $z=1$ is one SD above.
Units:
Dimensionless (no units).
Worked sample
A class has $\bar{x} = 65$ and $\sigma = 8$. Find the $z$-score for a mark of $73$.
$z = \dfrac{73 - 65}{8} = \dfrac{8}{8} = 1.0$.
Your turn:
Two tests: Test A has $\bar{x}=50, \sigma=10$; Test B has $\bar{x}=60, \sigma=15$. Student scored $65$ on A and $78$ on B. Which was the better relative performance?
Test A: $z = (65-50)/10 = 1.5$. Test B: $z = (78-60)/15 = 1.2$. Test A was the better relative performance.
Negative $z$-scores are fine — they mean the raw score is below the mean. The sign matters when comparing performance.
What each symbol means
For a normal (bell-shaped) distribution: approximately $68\%$ of data falls within $1\sigma$ of the mean, $95\%$ within $2\sigma$, and $99.7\%$ within $3\sigma$.
When to use it
Any question about normal distributions that asks what proportion of data or people fall within certain score ranges.
Units:
Percentages / proportions.
Worked sample
IQ scores are normally distributed with $\bar{x}=100$ and $\sigma=15$. What percentage of people have IQ between $85$ and $115$?
$85 = 100 - 1\sigma$ and $115 = 100 + 1\sigma$, so $68\%$ of people fall in this range.
Your turn:
Heights are $N(170, 6^2)$. What percentage have heights between $158$ cm and $182$ cm?
$158 = 170 - 2\sigma$ and $182 = 170 + 2\sigma$, so approximately $95\%$.
The rule gives both-sided intervals. If the question asks for one side only (e.g. above the mean), halve the percentage: $34\%$ lie between the mean and $+1\sigma$.
What each symbol means
$\hat{y}$ — predicted value of the response variable; $x$ — explanatory variable; $b$ — gradient of the regression line; $a$ — $y$-intercept.
When to use it
Modelling a linear relationship between two numerical variables and making predictions. Use when correlation is strong (Pearson's $r$ close to $\pm 1$).
Units:
Units of $y$ per unit of $x$ for the gradient.
Worked sample
The regression line for study hours ($x$) vs test mark ($y$) is $\hat{y} = 40 + 5x$. Predict the mark for $6$ hours of study.
$\hat{y} = 40 + 5(6) = 40 + 30 = 70$.
Your turn:
The regression line is $\hat{y} = 12 + 0.8x$. Predict $\hat{y}$ when $x = 25$.
$\hat{y} = 12 + 0.8 \times 25 = 12 + 20 = 32$.
Only use the line to predict within the range of the data (interpolation). Extrapolating far outside the data range is unreliable and the HSC often tests whether you recognise this.
What each symbol means
$r$ — Pearson's correlation coefficient; $r = 1$ perfect positive linear correlation; $r = -1$ perfect negative; $r = 0$ no linear correlation.
When to use it
Describing the strength and direction of a linear relationship between two variables. Calculated by the spreadsheet or the given table in the HSC — you interpret rather than compute it by hand.
Units:
Dimensionless.
Worked sample
Scatter plot data gives $r = -0.85$. Describe the correlation.
Strong negative linear correlation — as $x$ increases, $y$ tends to decrease.
Your turn:
Is $r = 0.3$ considered strong, moderate, or weak? In which direction?
Weak positive linear correlation.
Correlation does not imply causation — a classic HSC extended-response prompt. Identify any possible confounding variables.
What each symbol means
$Q_1$ — lower quartile; $Q_3$ — upper quartile; $\text{IQR} = Q_3 - Q_1$ — interquartile range; $x$ — the data value being tested.
When to use it
Identifying whether a data value is an outlier. Calculate IQR first, then find the lower fence ($Q_1 - 1.5 \times \text{IQR}$) and upper fence ($Q_3 + 1.5 \times \text{IQR}$). Any value outside these fences is an outlier.
Units:
Same units as the data.
Worked sample
A data set has $Q_1 = 20$, $Q_3 = 35$. Is the value $60$ an outlier?
$\text{IQR} = 15$. Upper fence $= 35 + 1.5 \times 15 = 57.5$. Since $60 > 57.5$, yes — $60$ is an outlier.
Your turn:
$Q_1 = 10$, $Q_3 = 22$. Is the value $2$ an outlier?
$\text{IQR} = 12$. Lower fence $= 10 - 18 = -8$. Since $2 > -8$, it is not an outlier.
Calculate both fences even if you only need one — the HSC sometimes asks you to check both ends. IQR fences are on the reference sheet; remember the multiplier is $1.5$, not $2$.
Algebra & Linear/Non-linear Relationships
4What each symbol means
$m$ — gradient (slope); $(x_1, y_1)$ and $(x_2, y_2)$ — two points on the line.
When to use it
Finding the steepness of a straight line given any two points on it. A positive $m$ means rising left to right; negative means falling.
Units:
Change in $y$ per unit change in $x$. Units depend on context.
Worked sample
Find the gradient of the line through $(2, 3)$ and $(6, 11)$.
$m = \dfrac{11-3}{6-2} = \dfrac{8}{4} = 2$.
Your turn:
Find the gradient through $(-1, 4)$ and $(3, -4)$.
$m = \dfrac{-4-4}{3-(-1)} = \dfrac{-8}{4} = -2$.
Be consistent with the order: if you subtract $y_1$ in the numerator, you must subtract $x_1$ in the denominator. Mixing the order gives the wrong sign.
What each symbol means
$m$ — gradient; $b$ — $y$-intercept (value of $y$ when $x = 0$).
When to use it
Graphing a line, reading off its gradient and $y$-intercept, or writing the equation when both $m$ and $b$ are known.
Worked sample
Write the equation of the line with gradient $3$ and $y$-intercept $-2$.
$y = 3x - 2$.
Your turn:
A line has gradient $-\tfrac{1}{2}$ and $y$-intercept $4$. Write its equation and find $y$ when $x = 6$.
$y = -\tfrac{1}{2}x + 4$. When $x=6$: $y = -3 + 4 = 1$.
If the equation is given in general form (e.g. $2x - 3y + 6 = 0$), rearrange to gradient-intercept form to read off $m$ and $b$ directly.
What each symbol means
$m$ — gradient; $(x_1, y_1)$ — a known point on the line.
When to use it
Writing the equation of a line when you know its gradient and one point (but not the $y$-intercept). Faster than finding $b$ separately.
Worked sample
Find the equation of the line through $(3, 5)$ with gradient $2$.
$y - 5 = 2(x - 3) \Rightarrow y = 2x - 1$.
Your turn:
Write the equation of the line through $(1, -2)$ with gradient $-3$.
$y - (-2) = -3(x-1) \Rightarrow y = -3x + 1$.
Always expand and simplify to $y = mx + b$ form for the final answer unless the question asks for a specific form.
What each symbol means
$a$ — initial value (value when $x = 0$); $b$ — growth/decay factor per unit of $x$; $x$ — time or independent variable.
When to use it
Modelling exponential growth (populations, compound interest) when $b > 1$, or exponential decay (radioactive decay, drug concentration) when $0 < b < 1$.
Units:
Depends on context.
Worked sample
A population of $500$ grows by $10\%$ per year. Write an equation and find the population after $5$ years.
$y = 500 \times (1.10)^x$. After $5$ years: $y = 500(1.1)^5 \approx 805$.
Your turn:
A drug concentration starts at $200$ mg and halves every hour. Find the concentration after $3$ hours.
$y = 200 \times (0.5)^3 = 200 \times 0.125 = 25$ mg.
The compound interest formula $FV = PV(1+r)^n$ is a special case of $y = ab^x$ with $a = PV$ and $b = 1+r$.
Networks
4What each symbol means
$V$ — number of vertices; $E$ — number of edges; $F$ — number of faces (regions, including the outer/infinite face).
When to use it
Verifying or finding an unknown count (vertices, edges, or faces) in any connected planar graph.
Units:
Dimensionless counts.
Worked sample
A planar graph has $6$ vertices and $9$ edges. How many faces does it have?
$F = 2 - V + E = 2 - 6 + 9 = 5$ faces.
Your turn:
A connected planar graph has $10$ vertices and $7$ faces. How many edges?
$E = V + F - 2 = 10 + 7 - 2 = 15$ edges.
The outer (unbounded) region always counts as one face. Students frequently forget it and get $F$ wrong by 1.
What each symbol means
A spanning tree connects all $V$ vertices using exactly $V-1$ edges with no cycles, minimising total edge weight.
When to use it
Finding the cheapest/shortest way to connect all nodes in a network (e.g. laying cables, building roads). This is an algorithm/procedure, not a formula printed on the reference sheet — you need to know the method. Use Kruskal's (sort all edges by weight, greedily add smallest that does not create a cycle) or Prim's (start from any node, always add the cheapest edge that extends the tree).
Units:
Same units as the edge weights (km, $, hours, etc.).
Worked sample
A network has edges: AB=3, AC=5, BC=2, BD=4, CD=6. Find the minimum spanning tree and its total weight.
Select BC=2 (smallest), then AB=3, then BD=4. Skip CD (creates cycle). Total weight = $2+3+4 = 9$.
Your turn:
Edges: PQ=7, PR=3, QR=5, QS=4, RS=6. Find the minimum spanning tree weight.
PR=3, QS=4, QR=5. Total = 12. (PQ and RS create cycles.)
Minimum spanning tree is an algorithm you apply — it is not a formula on the reference sheet. Always check: a spanning tree on $n$ vertices has exactly $n-1$ edges. If you have more or fewer, you've made an error.
What each symbol means
$d(v)$ — shortest distance to vertex $v$; $d(u)$ — shortest distance to a neighbouring vertex $u$; $w(u,v)$ — weight of edge between $u$ and $v$.
When to use it
Finding the shortest (or quickest, or cheapest) route between two specific nodes in a weighted network. This is an algorithm/procedure, not a formula on the reference sheet — you need to know the method. Use a systematic labelling approach: always extend from the nearest unlabelled vertex.
Units:
Same as edge weights.
Worked sample
Network: A→B=4, A→C=2, B→D=5, C→B=1, C→D=8. Find the shortest path from A to D.
A to C = 2. C to B = 3 (via 2+1). B to D = 8 (via 3+5). A to D via A-C-B-D = 2+1+5 = 8.
Your turn:
Shortest path from S to T: S-A=6, S-B=2, A-T=4, B-A=3, B-T=9. Find minimum distance.
S-B-A-T = 2+3+4 = 9. S-A-T = 6+4 = 10. S-B-T = 2+9 = 11. Minimum = 9 via S-B-A-T.
Dijkstra's shortest-path method is an algorithm you apply — it is not a formula printed on the reference sheet. In an HSC exam a table or diagram is usually provided. Work systematically from the source node — always label the nearest unlabelled vertex next.
What each symbol means
$\text{Float}$ — slack time available for an activity before it delays the project; $\text{LST}$ — latest start time; $\text{EST}$ — earliest start time.
When to use it
Project scheduling. This float calculation is a procedural method, not a formula on the reference sheet — you need to know how to construct and read activity-on-edge network diagrams. Activities on the critical path have float $= 0$ — any delay there delays the whole project. Activities with positive float have some scheduling flexibility.
Units:
Same time units as the activity durations (days, hours, etc.).
Worked sample
Activity D has EST = 8 days and LST = 11 days. Find the float for D.
Float $= 11 - 8 = 3$ days. Activity D can start up to 3 days late without delaying the project.
Your turn:
EST = 5, LST = 5. What is the float, and what does it tell you?
Float $= 0$. The activity is on the critical path — any delay will delay the project.
Critical path float is a procedure you apply — it is not a formula printed on the reference sheet. The critical path is the longest path through the network (maximum total duration). To find it, trace the path(s) where every activity has zero float.
Frequently asked questions
It covers measurement, financial mathematics, statistical analysis, algebra and linear/non-linear relationships, and networks. NESA provides it in every Standard 2 exam.
Yes — it's provided in every HSC Mathematics Standard 2 exam, so you don't need to memorise the listed formulas. You do need to know which one each question calls for.
Some, yes. The sheet provides many, but not all, and it won't tell you which to use. Marks come from selecting and applying the right formula correctly.
Practise with it open so you know where each formula is and the question types it answers. In the exam you'll then find it quickly instead of losing time searching.