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hscscience Maths Std · Y11
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Module 2 · L11 of 22 50–55 min ⚡ +95 XP available

Rates

A rate compares two quantities with different units. Master the unitary method — find the rate per one unit — and you can solve any rate problem by simple multiplication.

Today's hook — Supermarket A sells 2 kg of flour for $3.80. Supermarket B sells 5 kg for $8.75. Which is better value? How would you decide?
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall — your gut answer first
+5 XP warm-up

Supermarket A sells 2 kg of flour for $3.80. Supermarket B sells 5 kg for $8.75. Which is better value? How would you decide? What calculation would you do?

Without calculating — make a prediction and explain your reasoning.

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02
The formulas you need to own
+5 XP to read

Rate problems hinge on two ideas: expressing everything per one unit (the unitary method), and the distance–speed–time triangle. Get these right and every rate question becomes routine.

A rate compares two quantities with different units — e.g. 60 km/h, $4.50/kg, 15 L/min. The unitary method converts any rate to "per one unit", making different rates directly comparable. For speed: $D = S \times T$ — cover the unknown to read the formula.

D S T × COVER UNKNOWN D = S × T S = D ÷ T T = D ÷ S Fuel (L) = c L/100km × distance ÷ 100
$D = S \times T$ — cover the unknown to reveal the formula
Rate — a comparison of two units
Examples: 60 km/h, $4.50/kg, 12 L/min. Always write units — they tell you the operation.
Unitary method
Divide total value by number of units to get rate per 1 unit. Then scale up by multiplying.
Fuel consumption
Measured in L/100 km. Litres used = consumption × distance ÷ 100. Lower number = more economical.
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What you'll master
Know

Key facts

  • A rate compares two different types of quantities
  • The unitary method finds the rate per one unit
  • $D = ST$ for speed problems; rearranges to $S = D/T$ and $T = D/S$
  • Fuel consumption expressed as L/100 km
Understand

Concepts

  • Why expressing rates per one unit makes comparison easy
  • How unit labels guide the calculation method
  • The difference between average speed and instantaneous speed
Can do

Skills

  • Apply the unitary method to find best value
  • Solve speed/distance/time problems
  • Calculate fuel consumption and cost
  • Compare unit rates in practical contexts
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Key terms
RateA comparison of two quantities with different units, e.g. 60 km/h, $4.50/kg, 12 L/min.
Unitary methodFinding the value of one unit first, then multiplying or dividing to find the required amount.
Average speedTotal distance divided by total time for a journey; does not reflect varying speed during the trip.
Fuel consumptionThe number of litres of fuel used per 100 km travelled; a lower number means more economical.
ScaleA ratio that compares measurements on a map or diagram to real-world measurements.
05
Rate per one unit — the unitary method
core concept

The unitary method converts any rate into "per one unit", making different rates directly comparable. Divide the total quantity by the number of units.

For best-value comparisons, find the cost per gram or per unit — then the cheapest rate wins.

$$\text{Rate per 1 unit} = \frac{\text{total value}}{\text{total number of units}}$$
Method. Rate per 1 unit = (total value) ÷ (total number of units). Then scale up or down by multiplying.

Average speed warning: Average speed = total distance ÷ total time. It is NOT the arithmetic mean of two speeds unless the time spent at each speed is equal.

What to write in your book
  • Rate: a comparison of two different quantities, expressed per unit — always write the units.
  • Unitary method: Rate per 1 = total ÷ number of units. Then multiply to scale.
  • $D = S \times T$ · $S = D \div T$ · $T = D \div S$ — cover the unknown to read the formula.
  • Fuel: Litres used = (L/100 km) × distance ÷ 100; Cost = litres × price per litre.
  • Always check units before substituting — convert minutes to hours by dividing by 60.

Did you get this? True or false: to find average speed for a whole journey, you add the two speeds and divide by 2.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BEST VALUE — UNITARY METHOD

Store A sells orange juice: 1.25 L for $2.80. Store B sells 2 L for $4.20. Which is better value?

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$\text{Store A: } \$2.80 \div 1.25 = \$2.24 \text{ per litre}$
Find cost per litre for Store A using the unitary method.
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PROBLEM 2 · SPEED–DISTANCE–TIME

A car travels 270 km in 3 hours and 15 minutes. Find the average speed in km/h.

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$T = 3 + \dfrac{15}{60} = 3.25 \text{ h}$
Convert time to hours: 15 min = 15/60 h = 0.25 h.
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PROBLEM 3 · FUEL CONSUMPTION

A car has fuel consumption of 8.5 L/100 km. Petrol costs $2.05 per litre. Find the cost of driving 420 km.

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$\text{Fuel} = 8.5 \times \dfrac{420}{100} = 8.5 \times 4.2 = 35.7 \text{ L}$
Multiply consumption rate by number of 100 km units.
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What to write in your book
  • Best-value method: divide each price by its quantity to get cost per unit. Lowest wins.
  • Speed: always convert time to hours before using $S = D \div T$. 45 min = 0.75 h.
  • Fuel: Litres = (L/100 km) × distance ÷ 100. Then Cost = litres × price/L.
  • Unitary method and DST triangle are the two core tools — master both.

Quick check: A car travels at 80 km/h for 2.5 hours. What distance does it cover?

D = S × T FORMULA TRIANGLE D S T × Cover the unknown → read the formula COVER THE UNKNOWN D D = S × T cover D → multiply S and T S S = D ÷ T cover S → D on top, T on bottom T T = D ÷ S cover T → D on top, S on bottom

Common errors · the traps that cost marks

Trap 01
Average speed ≠ mean of two speeds
Wrong: Average speed = (speed1 + speed2) ÷ 2. Right: Average speed = total distance ÷ total time. The arithmetic mean only works when equal time is spent at each speed.
Trap 02
Forgetting to convert minutes to hours
If speed is in km/h, time must be in hours. Divide minutes by 60 before substituting. 45 min = 0.75 h, not 45 h.
Trap 03
Comparing prices without finding per-unit rate
A larger pack is not automatically cheaper per unit. Always divide price by quantity to get the rate per gram, litre, or unit before comparing.
What to write in your book
  • Always check units are consistent before substituting into D = ST.
  • Average speed = total distance ÷ total time (not mean of speeds).
  • Best value: divide each price by quantity; lowest unit rate wins.
  • Fuel cost = (L/100 km) × (distance ÷ 100) × (price per litre).

Fill the gap: A car uses 9 L/100 km. For a 350 km trip: litres used = 9 × (350 ÷ ) = L.

Quick-fire practice · 12 rate calculations

1

A printer produces 420 pages in 7 minutes. Find the rate in pages per minute.

2

A tap fills a 180 L bath in 12 minutes. Find the flow rate in L/min.

3

Pack A: 400 g of coffee for $7.60. Pack B: 600 g for $11.10. Which is better value?

4

A worker earns $336 for 8 hours. Another earns $270 for 6 hours. Who has the better hourly rate?

5

A cyclist travels at 24 km/h for 2.5 hours. Find the distance.

6

A train travels 480 km at an average speed of 120 km/h. Find the time in hours and minutes.

7

A runner completes a 10 km race in 52 minutes and 30 seconds. Find the average speed in km/h (to 2 d.p.).

8

A car travels 140 km in 1 h 45 min. Find the average speed.

9

A car uses 9.2 L/100 km. How much fuel is needed for a 350 km journey?

10

Using Q9 above, if petrol costs $2.15/L, find the total fuel cost.

11

A tap drips at 0.3 L/min. How many litres are wasted in 24 hours?

12

A pump drains a 2400 L tank in 40 minutes. Find the drainage rate in L/min. How long to drain a 3600 L tank at the same rate?

Odd one out: Three of these are correct statements about rates. Which one is wrong?

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Revisit your thinking

Earlier you predicted which supermarket had better value flour. Let's check:

Supermarket A: $3.80 ÷ 2 = $1.90/kg

Supermarket B: $8.75 ÷ 5 = $1.75/kg

Supermarket B is cheaper per kilogram despite having the higher total price. The unitary method — find the rate per one unit — removes all ambiguity from comparison problems.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 43 marks

SA 4. A family drives from Sydney to Melbourne, a distance of 880 km. They drive at an average speed of 100 km/h for the first 400 km, then stop for 45 minutes, then continue at 90 km/h for the rest. (a) Find the time for the first 400 km. (1 mark) (b) Find the time for the remaining distance. (1 mark) (c) Find the total travel time including the rest stop, in hours and minutes. (1 mark)

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ApplyBand 43 marks

SA 5. A supermarket sells three sizes of olive oil: 375 mL for $6.45, 750 mL for $11.40, and 1.5 L for $24.00. (a) Find the cost per 100 mL for each size. (2 marks) (b) State which size offers the best value. (1 mark)

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AnalyseBand 54 marks

SA 6. Riley drives a car with fuel consumption of 7.8 L/100 km. Petrol costs $2.10/L. (a) How many litres are needed for a 520 km trip? (1 mark) (b) What is the fuel cost for the trip? (1 mark) (c) Riley's tank holds 60 L and starts full. After the trip, how many litres remain? (1 mark) (d) At the same consumption rate, how far could Riley travel on a full tank? Give your answer to the nearest km. (1 mark)

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📖 Comprehensive answers (click to reveal)

Drill 1: $420 \div 7 = 60$ pages/min  ·  2: $180 \div 12 = 15$ L/min  ·  3: A: $7.60 \div 400 = \$0.019$/g; B: $11.10 \div 600 = \$0.0185$/g → Pack B  ·  4: W1: $336 \div 8 = \$42$/h; W2: $270 \div 6 = \$45$/h → Worker 2  ·  5: $D = 24 \times 2.5 = 60$ km  ·  6: $T = 480 \div 120 = 4$ h  ·  7: $T = 52.5/60$ h; $S = 10 \div (52.5/60) \approx 11.43$ km/h  ·  8: $T = 1.75$ h; $S = 140 \div 1.75 = 80$ km/h  ·  9: $9.2 \times 3.5 = 32.2$ L  ·  10: $32.2 \times 2.15 = \$69.23$  ·  11: $0.3 \times 60 \times 24 = 432$ L  ·  12: Rate $= 2400 \div 40 = 60$ L/min; Time $= 3600 \div 60 = 60$ min

SA 4 (3 marks): (a) $T_1 = 400 \div 100 = 4$ h [1]. (b) $T_2 = 480 \div 90 = 5\frac{1}{3}$ h $= 5$ h 20 min [1]. (c) Total $= 4 + 5$ h 20 min $+ 45$ min $= 10$ h 5 min [1].

SA 5 (3 marks): (a) 375 mL: $6.45/3.75 = \$1.72$/100 mL; 750 mL: $11.40/7.5 = \$1.52$/100 mL; 1.5 L: $24.00/15 = \$1.60$/100 mL [2]. (b) 750 mL at $1.52/100 mL is best value [1].

SA 6 (4 marks): (a) $7.8 \times 5.2 = 40.56$ L [1]. (b) $40.56 \times 2.10 = \$85.18$ [1]. (c) $60 - 40.56 = 19.44$ L [1]. (d) $D = 60 \div 7.8 \times 100 \approx 769$ km [1].

01
Boss battle · The Rate Ruler
earn bronze · silver · gold

Five timed questions on rates and speed. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms using your knowledge of rates and unit conversions. Pool: lessons 1–11.

Mark lesson as complete

Tick when you've finished the practice and review.

Module overview · Year 11 · Maths Standard