Mathematics • Year 8 • Unit 1 • Lesson 12
Introduction to Rates
Build fluency with rates: comparing two different units and finding the unit rate. One fully-worked example, one guided example with blanks, then eight independent problems from quick reading to comparing two rates.
1. I do — fully worked example
Read every line. Each step has a short reason so you can see why the unit rate is so useful.
Problem. A 2.5 kg bag of rice costs $\$8.75$. Find the price per kg, then use it to find the cost of 7 kg.
Step 1 — Spot the rate.
$\$8.75$ for 2.5 kg compares dollars with kilograms — two DIFFERENT units, so it's a rate.
Reason: a rate compares two unlike things. Here it's $/kg.
Step 2 — Find the UNIT rate (per 1 kg).
$\$8.75 \div 2.5 = \$3.50$ per kg
Reason: to find the cost of 1 kg, divide the total cost by the number of kg.
Step 3 — Use the unit rate to scale UP.
7 kg cost $7 \times \$3.50 = \$24.50$
Reason: once you know the per-1 amount, multiply by however many you need.
Step 4 — Always keep the units on the answer.
$3.50 per kg, and 7 kg = $24.50.
Reason: “3.50” alone is meaningless — per kg makes the rate useful.
Answer: Unit rate = $\$3.50$/kg; 7 kg costs $\$24.50$.
2. We do — fill in the missing steps
Same shape as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. A car travels 240 km on 30 L of fuel. Find the unit rate in km per litre, then use it to find how far it can travel on a full 50 L tank.
Step 1 — What two units are being compared? ________ and ________.
Step 2 — Find the unit rate (km per 1 L):
240 ÷ ______ = ______ km/L
Step 3 — Scale to 50 L:
50 × ______ = ______ km
Step 4 — Put it together with units:
Unit rate = ______ km/L; range on 50 L = ______ km.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (reading rates and computing one unit rate). The middle two are standard (use a unit rate to scale up or down). The last two are extension (compare two rates).
Foundation — read and compute a unit rate
3.1 Which of these is a rate? Tick the rates and put a cross next to the others. (a) 60 km in 1 hour (b) 3 boys to 4 girls (c) $\$22$ per hour (d) 5 oranges. 1 mark
3.2 Lucia earns $\$84$ for 6 hours of work. Find her hourly rate. 1 mark
3.3 A heart beats 75 times in 30 seconds. Find the rate in beats per minute. 1 mark
3.4 A 5 kg bag of apples costs $\$18.50$. Find the price per kg. 1 mark
Standard — use the unit rate to scale
3.5 A car uses 7.2 L per 100 km. How many litres does it need to travel 350 km? 2 marks
3.6 A bike covers 18 km in 45 minutes. Find the speed in km/h. (Hint: 45 minutes = 0.75 hours.) 2 marks
Extension — compare two rates
3.7 Cheese A: 400 g for $\$6$. Cheese B: 250 g for $\$3.50$. (a) Find the unit rate ($/kg) for each. (b) Which is cheaper per kg? 2 marks
3.8 A car drives 150 km in 2.5 hours. A bus drives 180 km in 3 hours. Which has the higher average speed? Show the unit rate (km/h) for each. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (car: 240 km on 30 L)
Step 1: kilometres and litres (different units — so it's a rate).
Step 2: 240 ÷ 30 = 8 km/L.
Step 3: 50 × 8 = 400 km.
Step 4: Unit rate = 8 km/L; range on 50 L = 400 km.
3.1 — Which are rates?
(a) ✓ rate (km per hour), (b) ✗ ratio (same unit on both sides, count of students), (c) ✓ rate ($ per hour), (d) ✗ just a count (no second unit).
3.2 — Lucia's hourly rate
$\$84 \div 6 = \textbf{\$14/h}$.
3.3 — Heart rate
75 beats in 30 seconds → 75 × 2 = 150 beats per minute.
3.4 — Apples
$\$18.50 \div 5 = \textbf{\$3.70/kg}$.
3.5 — Fuel for 350 km
7.2 L per 100 km, so per km it's $7.2 \div 100 = 0.072$ L/km. For 350 km: $350 \times 0.072 = \textbf{25.2 L}$. (Or: $350 \div 100 = 3.5$ “hundreds of km”, so $3.5 \times 7.2 = 25.2$ L.)
3.6 — Bike speed
45 min = 0.75 h. Speed = $18 \div 0.75 = \textbf{24 km/h}$.
3.7 — Cheese A vs B
Cheese A: $\$6 \div 0.4 = \textbf{\$15/kg}$. Cheese B: $\$3.50 \div 0.25 = \textbf{\$14/kg}$. Cheese B is cheaper per kg (by $\$1$/kg).
3.8 — Car vs Bus average speed
Car: $150 \div 2.5 = \textbf{60 km/h}$. Bus: $180 \div 3 = \textbf{60 km/h}$. They have the same average speed.