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.

50–55 min MS-M2 — LOW-MEDIUM 3 MC 3 SA Lesson 11 of 22 Free
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Think First

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?

Type your initial response below — you will revisit this at the end of the lesson.

Write your initial response in your book. You will revisit it at the end of the lesson.

Write your initial thinking in your book
Saved

Come back to this at the end of the lesson.

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Key Rate Formulas — This Lesson

$\text{Rate} = \dfrac{\text{quantity}}{\text{time (or other unit)}}$
Definition of a rate — expressed as "per unit", e.g. km/h, $/kg, L/min Always write the units — they tell you what operation to perform
$D = S \times T$
Speed–distance–time — $D$ = distance, $S$ = speed, $T$ = time Rearranges: $S = D/T$ and $T = D/S$
$\text{Fuel use} = \dfrac{\text{litres}}{100 \text{ km}} \times \dfrac{\text{distance}}{100}$
Fuel consumption — if consumption is $c$ L/100 km and distance is $d$ km: litres used $= c \times d \div 100$
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

🧠 Know

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

💡 Understand

  • 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

  • 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

Rate A comparison of two quantities with different units, e.g. 60 km/h, $4.50/kg, 12 L/min
Unitary method Finding the value of one unit first, then multiplying or dividing to find the required amount
Average speed Total distance divided by total time for a journey; does not reflect varying speed during the trip
Fuel consumption The number of litres of fuel used per 100 km travelled; a lower number means more economical

Misconceptions to Fix

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Wrong: Average speed = (speed1 + speed2) / 2.

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

Unitary Method and Comparing Rates

Key Point

Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.

Key Terms
RateA comparison of two different quantities, often expressed per unit of time (e.g. km/h, $/L).
Unitary MethodFinding the value of one unit first, then scaling to find the value of any number of units.
Average SpeedTotal distance travelled divided by total time taken: speed = distance ÷ time.
Fuel ConsumptionThe rate at which a vehicle uses fuel, usually measured in litres per 100 km (L/100km).
ScaleA ratio that compares measurements on a map or diagram to real-world measurements.

Rate per One Unit

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.

Method: Rate per 1 unit = (total value) ÷ (total number of units). Then scale up or down by multiplying.
Worked Example 1 Best Value

Problem

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

Solution

1 Store A: $2.80 \div 1.25 = \$2.24 \text{ per litre}$ Find cost per litre for Store A

Speed, Distance, Time

The DST Triangle

Distance, speed, and time are linked by $D = S \times T$. Cover the unknown with your thumb — what remains shows the operation.

Worked Example 2 Speed–Distance–Time

Problem

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

Solution

1 $T = 3 + \dfrac{15}{60} = 3.25 \text{ h}$ Convert time to hours: 15 min = 15/60 h = 0.25 h

Fuel Consumption

Calculating Fuel Use and Cost

Fuel consumption is measured in litres per 100 km (L/100 km). To find litres used for a trip, multiply the consumption rate by the number of hundreds of kilometres.

Steps:

  1. Litres used $= \text{consumption (L/100 km)} \times \dfrac{\text{distance (km)}}{100}$
  2. Cost $= \text{litres used} \times \text{price per litre}$
Worked Example 3 Fuel Consumption

Problem

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

Solution

1 $\text{Fuel} = 8.5 \times \dfrac{420}{100} = 8.5 \times 4.2 = 35.7 \text{ L}$ Multiply consumption rate by number of 100 km units
Practice

Practice Questions

Show all working. Include units in every answer.

Section A — Unitary Method and Unit Rates

  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?

Section B — Speed, Distance, Time

  1. A cyclist travels at 24 km/h for 2.5 hours. Find the distance.
  2. A train travels 480 km at an average speed of 120 km/h. Find the time in hours and minutes.
  3. A runner completes a 10 km race in 52 minutes and 30 seconds. Find the average speed in km/h (to 2 d.p.).
  4. A car travels 140 km in 1 h 45 min. Find the average speed.

Section C — Fuel and Other Rates

  1. A car uses 9.2 L/100 km. How much fuel is needed for a 350 km journey?
  2. Using the information in Q9, if petrol costs $2.15/L, find the total fuel cost.
  3. A tap drips at 0.3 L/min. How many litres are wasted in 24 hours?
  4. A pump drains a 2400 L tank in 40 minutes. Find the drainage rate in L/min. How long would it take to drain a 3600 L tank at the same rate?

Q1

$420 \div 7 = \mathbf{60 \text{ pages/min}}$

Q2

$180 \div 12 = \mathbf{15 \text{ L/min}}$

Q3

A: $7.60 \div 400 = \$0.019$/g; B: $11.10 \div 600 = \$0.0185$/g → Pack B is better value

Q4

Worker 1: $336 \div 8 = \$42$/h; Worker 2: $270 \div 6 = \$45$/h → Worker 2 has better rate

Q5

$D = 24 \times 2.5 = \mathbf{60 \text{ km}}$

Q6

$T = 480 \div 120 = 4 \text{ h} = \mathbf{4 \text{ hours 0 minutes}}$

Q7

$T = 52.5 \text{ min} = 52.5/60 \text{ h}$; $S = 10 \div (52.5/60) = 10 \times 60/52.5 \approx \mathbf{11.43 \text{ km/h}}$

Q8

$T = 1.75 \text{ h}$; $S = 140 \div 1.75 = \mathbf{80 \text{ km/h}}$

Q9

$9.2 \times 3.5 = \mathbf{32.2 \text{ L}}$

Q10

$32.2 \times 2.15 = \mathbf{\$69.23}$

Q11

$0.3 \times 60 \times 24 = \mathbf{432 \text{ L}}$

Q12

Rate $= 2400 \div 40 = 60$ L/min; Time $= 3600 \div 60 = \mathbf{60 \text{ min}}$

Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Multiple Choice

1 A car travels at 90 km/h for 2 hours 20 minutes. The distance travelled is:

A   180 km
B   200 km
C   210 km
D   225 km

? Regarding this topic, 1 A car travels at 90 km/h for 2 hours 20 minutes. The distance travelled is:

A     180 km
B     200 km
C     210 km
D     225 km
C - Correct!
C — $T = 2 + 20/60 = 7/3$ h; $D = 90 \times 7/3 = 210$ km.

2 Which represents the best value?

A   250 mL for $1.40
B   600 mL for $3.00
C   1 L for $4.60
D   1.5 L for $6.60

? Regarding this topic, 2 Which represents the best value?

A     250 mL for $1.40
B     600 mL for $3.00
C     1 L for $4.60
D     1.5 L for $6.60
B - Correct!
B — A: $5.60/L; B: $5.00/L; C: $4.60/L; D: $4.40/L. Wait — D is cheapest at $4.40/L. Correction: D at $4.40/L is best value.

3 A car uses 10.5 L/100 km. Petrol costs $1.98/L. The fuel cost for a 600 km trip is closest to:

A   $62.37
B   $74.69
C   $124.74
D   $149.37

? Regarding this topic, 3 A car uses 10.5 L/100 km. Petrol costs $1.98/L. The fuel cost for a 600 km trip is closest to:

A     $62.37
B     $74.69
C     $124.74
D     $149.37
C - Correct!
C — Fuel: $10.5 \times 6 = 63$ L; Cost: $63 \times 1.98 = \$124.74$.

Short Answer

01

SA 4 3 marks 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)

Work in your book
Saved

(a)

$T_1 = 400 \div 100 = \mathbf{4 \text{ h}}$

(b)

$T_2 = 480 \div 90 = 5.\overline{3} \text{ h} = \mathbf{5 \text{ h } 20 \text{ min}}$

(c)

Total $= 4 \text{ h} + 5 \text{ h } 20 \text{ min} + 45 \text{ min} = \mathbf{10 \text{ h } 5 \text{ min}}$

02

SA 5 3 marks 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)

Work in your book
Saved

(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

(b)

750 mL at $1.52 per 100 mL is the best value

03

SA 6 4 marks 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)

Work in your book
Saved

(a)

$7.8 \times 5.2 = \mathbf{40.56 \text{ L}}$

(b)

$40.56 \times 2.10 = \mathbf{\$85.18}$

(c)

$60 - 40.56 = \mathbf{19.44 \text{ L}}$

(d)

$D = 60 \div 7.8 \times 100 = 6000 \div 7.8 \approx \mathbf{769 \text{ km}}$

Interactive

D = S × T — Cover the Unknown

Click a letter to cover it and reveal the formula for that variable.

D
S
×
T
Click D, S, or T above

D = Distance  |  S = Speed  |  T = Time

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Science Jump

Jump Through Rates!

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