Rates, unit conversions and the unitary method for HSC Maths Standard 2

Syllabus dot point

“Use rates and unit conversions to solve practical problems including fuel consumption, dosage, power consumption and energy efficiency”

HSCMaths Standard 2Year 12: Measurement12 min readUpdated 20 Jun 2026

Quick answer

A rate is a quantity per single unit; carry the units as algebra so the unwanted unit cancels, use the unitary method (down to one, then up to many) to scale, square or cube the factor when converting area or volume units, and sanity-check answers against the real world.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA wants you to compute rates, use the unitary method to solve worded rate problems, convert between metric units, and apply rates to everyday Australian contexts including fuel consumption, energy use and medication dosages.

The skill being tested is not really the arithmetic; it is the care you take with units. Most rate questions are one or two multiplications. Marks are lost by multiplying when you should divide, by leaving the time in minutes when the rate is per hour, or by forgetting that area and volume conversions need a squared or cubed factor. Write the units on every line and set up each step so the unwanted unit cancels. Do that and a wordy problem turns into one confident calculation.

The answer

What a rate is

A rate compares two quantities with different units, expressed per single unit of the denominator.

$80$ km/h: a speed.
$8.5$ L per $100$ km: fuel consumption.
$1.92/L: unit cost.
$400$ mg per kg per day: a dosage rate.

A rate is a fraction with units on both the top and the bottom, and that fraction drives every calculation here. The units of the rate tell you what to do with it. For example, litres per kilometre, $\frac{\text{L}}{\text{km}}$ , multiplied by a distance in kilometres leaves litres, because the kilometres cancel.

The unitary method

To solve "if $a$ produces $b$ , how much does $c$ produce", scale by the ratio $\frac{c}{a}$ :

\text{answer} = b \times \frac{c}{a}.

A safer way under exam pressure is to find the value for $1$ unit first, then multiply by $c$ . This two-step pattern (down to one, then up to many) works for almost every rate question: fuel, wages, recipe scaling, currency. So when a problem says "if so much gives so much, how much does this much give", find the value for one unit first.

Example: if $5$ kg of meat costs $84, then $1$ kg costs $\frac{84}{5} = 16.80$ , i.e. $16.80, and $8$ kg costs $16.80 \times 8 = 134.40$ , i.e. $134.40.

Treating units as algebra: the cancellation chain

The reliable way through every rate problem is to carry the units along with the numbers and cancel them like the letters in algebra (a "symbol" you cancel from top and bottom). Set up each calculation so the unwanted units cancel and the unit you want is left. This turns a wordy problem into a single line of arithmetic. It also shows up at once if you have multiplied when you should have divided.

Read the chain left to right. A distance in kilometres times a rate in litres per kilometre gives litres (kilometres cancel); litres times a price in dollars per litre gives dollars (litres cancel). At no point do you have to guess whether to multiply or divide: you choose the operation that cancels the unit you want gone.

Metric conversions

From	To	Multiply by
km	m	$1000$
m	cm	$100$
cm	mm	$10$
kg	g	$1000$
L	mL	$1000$
h	min	$60$
min	s	$60$

Going from a larger unit to a smaller one, you multiply and the number gets bigger; going from smaller to larger, you divide and it gets smaller. The metric staircase below is the picture: step down towards smaller units and multiply, step up towards larger units and divide.

Squared and cubed conversion factors

The trap that catches the most students is converting area and volume units. Length conversions use the plain factor. Area uses that factor squared and volume uses it cubed. This is because area is length times length, and volume is length times length times length. So $1$ metre is $100$ centimetres, but $1$ square metre is $100^2 = 10000$ square centimetres, and $1$ cubic metre is $100^3 = 1\,000\,000$ cubic centimetres (equivalently $1000$ litres). Whenever a question involves $\text{m}^2$ or $\text{m}^3$ , stop and square or cube the conversion factor before you use it.

Fuel consumption

Standard Australian unit: litres per $100$ km. Smaller is better, and a hybrid or efficient small car sits around $4$ to $6$ L per $100$ km while a large SUV may be $10$ or more.

To find fuel used for a trip of $d$ km at consumption $c$ L per $100$ km:

\text{fuel} = \frac{c}{100} \times d.

Then cost is fuel times price per litre. The two multiplications are exactly the cancellation chain shown above.

Energy use (kWh)

Household electricity is billed in kilowatt-hours (kWh). One kWh is the energy used by a $1000$ watt appliance running for $1$ hour.

For an appliance rated $P$ watts running $t$ hours:

E = \frac{P \times t}{1000} \text{ kWh}.

Cost is energy times price per kWh. The division by $1000$ converts watts to kilowatts; if the appliance is already rated in kilowatts, skip it and just multiply kW by hours.

Dosage calculations

For weight-based dosing: total dose = (mg per kg) $\times$ patient weight (kg). The number of tablets or doses is the total dose divided by the strength per tablet.

Always check the answer against the real world. If a patient ends up taking $100$ tablets a day, you have almost certainly made a unit error. Common slips are reading a microgram (a thousandth of a milligram) as a milligram, or reading a "once per dose" rate as a "once per day" rate. In the exam, work the maths as stated, but add a one-line note if the result is clearly impossible. Markers reward both the correct working and the fact that you noticed.

Key fact

A rate is a quantity per single unit of something else; treat its units as algebra and set up each step so the unwanted unit cancels, leaving the unit you want. Length conversions use the plain factor, area the factor squared, volume the factor cubed. Larger unit to smaller unit: multiply (number grows); smaller to larger: divide (number shrinks).

How exam questions ask about this dot point

"... uses fuel at $c$ L per $100$ km ... cost of a trip of $d$ km." Fuel $= \frac{c}{100} \times d$ , then cost $=$ fuel $\times$ price per litre.
"An appliance rated $P$ W (or kW) runs for $t$ hours ... cost." Energy $= \frac{P \times t}{1000}$ kWh (or kW $\times$ hours), then $\times$ price per kWh.
"A patient weighing ... is prescribed ... mg per kg ... how many tablets / mL?" Total dose $=$ rate $\times$ weight, then $\div$ strength per tablet (or dose).
"Which is better value?" Reduce both options to a common per-unit cost (cost per $100$ g, per litre) and compare.
"Convert ... to ... / express ... in ..." A pure unit conversion: multiply or divide by the factor, and square or cube it for area or volume.
"A car travels at ... km/h for ... minutes." Convert the time to hours first (the rate is per hour), then multiply.

Worked example

Fuel cost for a round trip

A Toyota Camry hybrid uses $4.8$ L per $100$ km. The driver does the Sydney-to-Canberra round trip ( $580$ km). Petrol is $1.98/L. Find the fuel cost.

Calculate the fuel used: The consumption rate is $4.8$ L per $100$ km, so for every kilometre the car uses $\frac{4.8}{100}$ L. Multiply by the total distance, and the kilometres cancel to leave litres:

\text{fuel} = \frac{4.8}{100} \times 580 = 27.84 \text{ L}

Multiply fuel by price per litre: Litres times dollars-per-litre leaves dollars:

\text{cost} = 27.84 \text{ L} \times 1.98 \text{ dollars/L} \approx 55.12 \text{ dollars}

Answer: the fuel cost for the round trip is approximately $55.12.

Running cost of an air-conditioner

A $2.5$ kW air-conditioner runs $6$ hours a day. Electricity costs $0.32/kWh (typical NSW residential rate, 2025). Find the daily and seasonal cost.

Find the energy used: The appliance is already in kilowatts, so energy is just kilowatts times hours:

E = 2.5 \times 6 = 15 \text{ kWh per day}

Find the daily and seasonal cost:

\text{cost per day} = 15 \times 0.32 = 4.80 \text{ dollars}

Over a $90$ -day summer season:

4.80 \times 90 = 432 \text{ dollars}

Answer: running the air-conditioner $6$ hours daily costs $4.80 per day, or $432 over a $90$ -day summer.

Distance from speed and time

A train travels at $130$ km/h for $45$ minutes. How far does it travel?

Convert the time to hours: The speed is in kilometres per hour, so the time must be in hours before multiplying:

45 \text{ min} = \frac{45}{60} = 0.75 \text{ h}

Multiply speed by time:

\text{distance} = 130 \text{ km/h} \times 0.75 \text{ h} = 97.5 \text{ km}

If a question instead wanted the speed in metres per second, multiply by $\frac{1000 \text{ m}}{3600 \text{ s}}$ so the original units cancel: $130 \times \frac{1000}{3600} \approx 36.1$ m/s.

Answer: the train travels $97.5$ km in $45$ minutes.

Which jar is better value

A $375$ g jar of Vegemite costs $8.50. A $560$ g jar costs $12.40. Which is better value?

Reduce to a common unit cost: Reduce both to a common basis, the cost per $100$ g:

\text{small}: \frac{8.50}{375} \times 100 \approx 2.27 \text{ dollars per } 100 \text{ g}

\text{large}: \frac{12.40}{560} \times 100 \approx 2.21 \text{ dollars per } 100 \text{ g}

Compare and conclude: The large jar costs about $6$ cents less per $100$ g.

Answer: the larger jar is slightly better value at about $2.21 per $100$ g, compared with $2.27 per $100$ g for the smaller jar.

Weight-based dosage with a sanity check

A child weighing $20$ kg is prescribed paracetamol at $15$ mg per kg per dose. The syrup contains $120$ mg per $5$ mL. How many millilitres per dose?

Compute the volume needed: Total dose $= 15 \text{ mg/kg} \times 20 \text{ kg} = 300$ mg. The syrup is $120$ mg per $5$ mL, so per millilitre it is $\frac{120}{5} = 24$ mg/mL. The volume needed is the dose divided by that concentration:

\frac{300 \text{ mg}}{24 \text{ mg/mL}} = 12.5 \text{ mL}

A $12.5$ mL dose of children's paracetamol is realistic, so the answer passes the sanity check.

Answer: $12.5$ mL per dose.

Common traps

Confusing the rate with its reciprocal: Fuel consumption $8.5$ L/ $100$ km is L on top, $100$ km on the bottom. Use it the way it is written.
Wrong unit conversion direction: $1$ km = $1000$ m, so converting $5$ km to m gives $5000$ , not $0.005$ . From larger to smaller the number grows; from smaller to larger it shrinks.
Area and volume factors: $1$ m $^2$ = $10000$ cm $^2$ (not $100$ ) and $1$ m $^3$ = $1\,000\,000$ cm $^3$ . Squared units need squared factors; cubed units need cubed factors.
Mismatched units in a rate: A speed in km/h needs the time in hours, not minutes. Convert first.
Forgetting unit cancellation: Treat units like algebraic symbols; multiplying L by $/L gives $. If the leftover unit is wrong, the setup is wrong.
Not sanity-checking dosages: An answer of dozens of tablets a day signals a unit error. Flag it.

Exam technique

Write each calculation with its units attached and arrange every step so the unwanted unit cancels and the unit you want survives; this single habit removes the "do I multiply or divide?" doubt and earns the method marks. Convert times, masses and lengths to the unit the rate uses before multiplying. For area or volume, square or cube the conversion factor and show that step. Round money to the nearest cent with a dollar sign, and finish every numerical answer with its unit. Where a context answer is wildly unrealistic (a $5000 tank of petrol, $100$ tablets a day), add a brief note that the figure looks implausible; it shows the marker you understand the rate, not just the arithmetic.

Note

A rate is just a way of saying how much of one thing you get for one of another. Think "how many kilometres per hour" when you drive, "how many dollars per litre" at the petrol bowser, or "how many litres of fuel per 100 km" for a car. Once you know the amount for one, you can find the amount for any number by multiplying. A unit conversion is simply swapping the name of a measurement, like turning kilometres into metres, without changing how much there really is. The maths here is mostly one or two times or divide steps, so the real skill is keeping track of the labels and checking the answer makes sense.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 HSC-style3 marks

A car uses fuel at a rate of

8.5

L per

100

km. Petrol costs $1.92 per litre. Find the fuel cost for a trip of

620

km.

Show worked answer →

Fuel used: $\frac{8.5}{100} \times 620 = 0.085 \times 620 = 52.7$ L.

Cost: $52.7 \times 1.92 = 101.18$ , i.e. $101.18 (round to cents).

Markers reward the fuel-used calculation, the cost calculation, and the answer rounded to cents with the dollar sign.

2021 HSC-style3 marks

A patient is prescribed

400

mg of medication per kg of body weight per day. The patient weighs

65

kg. The medication comes in

250

mg tablets. How many tablets per day?

Show worked answer →

Daily dose: $400 \text{ mg/kg} \times 65 \text{ kg} = 26000$ mg/day.

Convert to tablets: $\frac{26000}{250} = 104$ tablets per day.

That is implausibly many, so the question likely intended $4$ mg per kg, giving $260$ mg per day and roughly $1$ tablet per day. In the exam, work the maths as stated and add a brief flag if the result is implausible. Markers reward correct unit-by-unit computation and an answer to the right precision.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation1 mark

A walking track is signposted as

2.4

km long. Convert this length to metres.

Show worked solution →

Choose the direction and factor. Kilometres are larger than metres, so going from km to m you multiply, and the number grows. The factor is $1000$ because $1$ km $= 1000$ m.

Multiply by the factor.

2.4 \times 1000 = 2400

Answer: the track is $2400 \text{ m}$ long. Check the direction: a larger unit converting to a smaller one should give a bigger number, and $2400$ is bigger than $2.4$ , so this is right.

foundation1 mark

A cyclist rides

24

km in

1.5

hours at a steady pace. Find the average speed in kilometres per hour.

Show worked solution →

Set up the rate. Average speed is distance divided by time, so the kilometres stay on top and the hours on the bottom to give km/h directly:

\text{speed} = \frac{24 \text{ km}}{1.5 \text{ h}}

Divide.

\frac{24}{1.5} = 16

Answer: the average speed is $16 \text{ km/h}$ . Check the size: $16$ km/h for $1.5$ hours covers $16 \times 1.5 = 24$ km, which matches the distance given.

foundation2 marks

During a fitness test, a student counts

19

heartbeats in

15

seconds. Express the heart rate in beats per minute.

Show worked solution →

Read the rate from the data. The count is $19$ beats per $15$ seconds, so the rate is $\dfrac{19 \text{ beats}}{15 \text{ s}}$ .

Scale the time to one minute. A heart rate in beats per minute needs the time as $60$ s, so multiply the rate by $60$ ; the seconds cancel and beats survive:

\frac{19}{15} \times 60 = 19 \times 4 = 76

Answer with units. The heart rate is $76$ beats per minute. Check the size: a resting rate near $76$ bpm is realistic, and $19$ beats in a quarter of a minute should give about four times as many in a full minute, which it does.

foundation2 marks

At a supermarket a

1

kg bag of rice costs $3.60 and a

5

kg bag of the same rice costs $16.00. Which bag is better value?

Show worked solution →

Reduce both to a common unit cost. Compare cost per $100$ g. For the $1$ kg ( $1000$ g) bag:

\frac{3.60}{1000} \times 100 = 0.36 \text{ dollars per } 100 \text{ g}

For the $5$ kg ( $5000$ g) bag:

\frac{16.00}{5000} \times 100 = 0.32 \text{ dollars per } 100 \text{ g}

Compare and conclude. The large bag costs $0.32 per $100$ g against $0.36 for the small bag, so it is $4$ cents cheaper per $100$ g.

Answer. The $5$ kg bag is better value. Check the logic: the bigger pack giving the lower unit price is the usual pattern, so the result is sensible.

core3 marks

(a) Convert a speed of

54

km/h to metres per second. (b) Convert a speed of

12.5

m/s to kilometres per hour.

Show worked solution →

Part (a) km/h to m/s, so turn km into m and h into s. Multiply by $1000$ to change kilometres to metres and divide by $3600$ to change hours to seconds:

54 \text{ km/h} = \frac{54 \times 1000}{3600} = \frac{54\,000}{3600} = 15 \text{ m/s}

so $54$ km/h $= 15$ m/s. The quick rule is to divide by $3.6$ : $54 \div 3.6 = 15$ .

Part (b) m/s to km/h, so multiply by $3.6$ . Going the other way reverses the factor:

12.5 \times 3.6 = 45

so $12.5$ m/s $= 45$ km/h. The factor is $3.6$ because $1$ km/h $= \dfrac{1000 \text{ m}}{3600 \text{ s}} = \dfrac{1}{3.6}$ m/s, and m/s is always the smaller of the two numbers, which is the check on direction.

core3 marks

A large SUV uses fuel at a rate of

9.2

L per

100

km. The driver completes a trip of

540

km. Petrol costs $2.05 per litre. Find the cost of the fuel for the trip.

Show worked solution →

Calculate the fuel used. For every kilometre the SUV uses $\dfrac{9.2}{100}$ L, so multiply by the distance and the kilometres cancel to leave litres:

\text{fuel} = \frac{9.2}{100} \times 540 = 0.092 \times 540 = 49.68 \text{ L}

Multiply fuel by the price per litre. Litres times dollars per litre leaves dollars:

\text{cost} = 49.68 \times 2.05 = 101.844

Answer rounded to cents. The fuel costs $101.84 (rounding $101.844$ to the nearest cent). Check the leftover unit: L multiplied by $/L gives dollars, so the setup is right.

core3 marks

A sprinter runs

100

m in

12.5

seconds. (a) Find the average speed in metres per second. (b) Convert this speed to kilometres per hour.

Show worked solution →

Part (a) speed in m/s. Speed is distance divided by time, with metres on top and seconds on the bottom to give m/s:

\frac{100 \text{ m}}{12.5 \text{ s}} = 8 \text{ m/s}

Part (b) convert m/s to km/h. Multiply by $1000$ to make metres kilometres and divide by $3600$ to make seconds hours, or use the quick rule of multiplying by $3.6$ :

8 \times 3.6 = 28.8

Answer: the sprinter runs at $8 \text{ m/s}$ , which is $28.8 \text{ km/h}$ . Check the direction: m/s is always the smaller number, and $8$ is less than $28.8$ , so the conversion is the right way round.

core2 marks

An adult is prescribed a

500

mg dose of an antibiotic. The medicine is a suspension containing

250

mg per

5

mL. What volume of the suspension, in millilitres, gives the correct dose?

Show worked solution →

Find the concentration per millilitre. The suspension is $250$ mg per $5$ mL, so per millilitre it is

\frac{250}{5} = 50 \text{ mg/mL}

Divide the dose by the concentration. The volume needed is the dose divided by the milligrams in each millilitre; the milligrams cancel and millilitres survive:

\frac{500 \text{ mg}}{50 \text{ mg/mL}} = 10 \text{ mL}

Answer. The correct dose is $10$ mL. Check the size: $10$ mL is two of the $5$ mL spoonfuls, and two lots of $250$ mg is $500$ mg, which matches the prescription.

exam5 marks

A swimming-pool pump is rated at

1.1

kW and runs for

8

hours each day. Electricity costs $0.30 per kWh. (a) Find the energy the pump uses in one day, in kWh. (b) Find the cost of running the pump for one day. (c) Find the cost of running the pump for a

30

-day month. (d) The owner reduces the daily run time to

5

hours. How much is saved over a

30

-day month?

Show worked solution →

Part (a) energy per day. The pump is rated in kilowatts, so energy is kilowatts times hours:

E = 1.1 \times 8 = 8.8 \text{ kWh per day}

Part (b) cost per day. Multiply the energy by the price per kWh:

8.8 \times 0.30 = 2.64

so the daily cost is $2.64.

Part (c) cost for $30$ days. Multiply the daily cost by $30$ :

2.64 \times 30 = 79.20

so the monthly cost is $79.20.

Part (d) saving from running $5$ hours instead of $8$ . At $5$ hours a day the energy is $1.1 \times 5 = 5.5$ kWh, costing $5.5 \times 0.30 = 1.65$ dollars a day, or $1.65 \times 30 = 49.50$ dollars a month. The saving is

79.20 - 49.50 = 29.70

Answer. The owner saves $29.70 over the month. Check: the run time drops by $3$ of the $8$ hours, so the saving should be $\tfrac{3}{8}$ of $79.20, and $\tfrac{3}{8} \times 79.20 = 29.70$ , which agrees.

exam5 marks

A family drives

500

km from Sydney to Port Macquarie. Their car uses fuel at

7.5

L per

100

km and petrol costs $1.96 per litre. (a) Find the fuel used for the trip, in litres. (b) Find the cost of the fuel, to the nearest cent. (c) If they drive at an average speed of

95

km/h, find the driving time in hours and minutes, to the nearest minute.

Show worked solution →

Part (a) fuel used. For every kilometre the car uses $\dfrac{7.5}{100}$ L, so multiply by the distance:

\text{fuel} = \frac{7.5}{100} \times 500 = 0.075 \times 500 = 37.5 \text{ L}

Part (b) cost of the fuel. Litres times dollars per litre:

37.5 \times 1.96 = 73.50

so the fuel costs $73.50.

Part (c) driving time. Time is distance divided by speed, with kilometres cancelling against the km in km/h to leave hours:

T = \frac{500}{95} \approx 5.2632 \text{ h}

Convert the decimal part to minutes by multiplying by $60$ :

0.2632 \times 60 \approx 15.8 \text{ min}

so the time is $5$ hours and $16$ minutes to the nearest minute.

Answer. The trip uses $37.5$ L costing $73.50, and the drive takes about $5$ h $16$ min. Check the time: $5$ h at $95$ km/h covers $475$ km, leaving $25$ km, which at $95$ km/h takes about $16$ minutes, matching.

exam5 marks

A patient weighing

80

kg is prescribed a medication at

5

mg per kg per day, taken once daily as

100

mg tablets. (a) Find the total daily dose, in milligrams. (b) Find how many tablets make up the daily dose. (c) Separately, the patient is put on an IV drip of

1200

mL of fluid to run evenly over

8

hours. Find the flow rate in millilitres per hour. (d) The drip delivers

20

drops per millilitre. Find the rate in drops per minute.

Show worked solution →

Part (a) total daily dose. Multiply the rate by the body weight; the kilograms cancel and milligrams survive:

5 \text{ mg/kg} \times 80 \text{ kg} = 400 \text{ mg/day}

Part (b) number of tablets. Divide the dose by the strength of each tablet:

\frac{400}{100} = 4 \text{ tablets}

Part (c) flow rate in mL/h. Divide the volume by the time:

\frac{1200 \text{ mL}}{8 \text{ h}} = 150 \text{ mL/h}

Part (d) rate in drops per minute. Convert millilitres per hour to drops per minute: multiply by $20$ drops per mL and divide by $60$ minutes per hour, so the millilitres and hours cancel and drops per minute survive:

150 \text{ mL/h} \times \frac{20 \text{ drops/mL}}{60 \text{ min/h}} = \frac{150 \times 20}{60} = 50 \text{ drops/min}

Answer. The daily dose is $400$ mg ( $4$ tablets), the drip runs at $150$ mL/h, and the drop rate is $50$ drops per minute. Check the dose: $400$ mg a day as $4$ tablets is realistic, and $50$ drops per minute is a steady, plausible drip rate.

exam5 marks

A household shower head delivers water at

9

L per minute. (a) Find the water used in one

8

-minute shower, in litres. (b) A family of four each take one such shower a day. Find the water used over a

30

-day month, in kilolitres (

1

= 1000

L). (c) Water costs $2.40 per kilolitre. Find the cost of these showers for the month, to the nearest cent. (d) The family fits a water-saving head delivering

6

L per minute, with shower times unchanged. Find the saving in dollars over the same month.

Show worked solution →

Part (a) water per shower. Flow rate times time, with minutes cancelling to leave litres:

9 \text{ L/min} \times 8 \text{ min} = 72 \text{ L}

Part (b) water for the month. Four showers a day use $72 \times 4 = 288$ L per day, so over $30$ days:

288 \times 30 = 8640 \text{ L} = 8.64 \text{ kL}

dividing by $1000$ to change litres to kilolitres.

Part (c) cost for the month. Kilolitres times dollars per kilolitre leaves dollars:

8.64 \times 2.40 = 20.736

so the cost is $20.74 to the nearest cent.

Part (d) saving with the water-saving head. At $6$ L/min each shower uses $6 \times 8 = 48$ L, so the month uses $48 \times 4 \times 30 = 5760$ L $= 5.76$ kL, costing $5.76 \times 2.40 = 13.824$ , i.e. $13.82. The saving is

20.74 - 13.82 = 6.92

Answer. The new head saves about $6.92 over the month (using the rounded monthly figures). Check: the flow drops by $3$ of every $9$ L/min, which is a third, so the saving should be roughly a third of $20.74, and $\frac{1}{3} \times 20.74 \approx 6.91$ , which agrees.

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA