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.