Rates and Unit Rates
A rate compares quantities measured in different units. Find a unit rate to compare fairly, then use it to solve speed, price and planning problems.
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A cyclist travels 36 km in 2 hours at a steady average speed. How far would the cyclist travel in 1 hour? Write the distance and time units with your answer.
A ratio compares quantities. A rate is a ratio of quantities measured in different units, such as kilometres and hours or dollars and kilograms. Units tell you what the comparison means: 60 km/h is a distance travelled per hour.
A train travels 240 km in 3 h. Its average rate is $240\text{ km} \div 3\text{ h}=80\text{ km/h}$. This is a unit rate: it tells us the distance for 1 hour. The units do not disappear; keep “km per h” in the answer.
Text description
- Two columns show distance and time: 240 km and 3 h.
- A vertical coral arrow in the distance column is labelled divide by 3; below it is 80 km.
- A separate vertical plum arrow in the time column is labelled divide by 3; below it is 1 h.
- The resulting unit rate is 80 km per hour.
Know
- A rate compares quantities with different units
- A unit rate is expressed for one unit
- Units are part of the answer
Understand
- Why comparisons need the same unit basis
- How rate depends on both quantities
- How a constant rate scales both quantities
Can Do
- Find speed, price per unit and other unit rates
- Compare two offers fairly
- Use a constant rate to find a missing quantity
Wrong: “Shop B costs more dollars, so it must be worse value.” The packs contain different amounts, so total prices alone do not compare their value.
Compare per 100 g: Shop A: $4.50 ÷ 500 × 100 = $0.90 per 100 g. Shop B: $6.30 ÷ 750 × 100 = $0.84 per 100 g. Shop B is cheaper by $0.06 per 100 g.
Wrong: 90 beats in 1 minute is 90 beats per second.
Convert the time: 1 minute = 60 seconds, so 90 ÷ 60 = 1.5 beats per second. Check that the denominator unit matches what the question asks for.
Divide the first quantity by the number of second-units. A car travels 240 km in 3 h: $240÷3=80$ km/h. A printer makes 60 pages in 4 min: $60÷4=15$ pages/min. If the second quantity is not already in the unit you need, convert it first.
Convert both offers to the same unit rate. For 3 kg at $8.40 and 4 kg at $11.20: $8.40 ÷ 3 = $2.80/kg and $11.20 ÷ 4 = $2.80/kg. The packs give the same value, even though their sizes and total prices differ.
For cereal, $4.50 for 500 g is $0.90 per 100 g; $6.30 for 750 g is $0.84 per 100 g. Shop B saves $0.06 per 100 g. Lower unit price means better value when the products are otherwise comparable.
If a rate stays constant, multiplying the time or quantity by a factor multiplies the matching amount by the same factor. At 72 km/h for 2.5 h, a vehicle travels $72×2.5=180$ km. Use the units to choose the operation: distance = speed × time; time = distance ÷ speed.
First find the rate: $90÷6=15$ pages per minute. Then $15×10=150$ pages. The constant-rate assumption is stated, so the rate can be scaled.
A train travels 420 km in 3 hours 30 minutes. Find its average speed in km/h.
Convert 3 h 30 min to 3.5 h. Then $420÷3.5=120$. Average speed = 120 km/h.
Compare 600 g for $7.20 with 800 g for $9.60, using dollars per kilogram.
First offer: $7.20 ÷ 0.6 = $12/kg. Second offer: $9.60 ÷ 0.8 = $12/kg. They have the same unit price.
A recipe makes 4 muffins with 300 g of flour and 200 mL of milk. How much of each is needed for 10 muffins?
Scale factor $=10÷4=2.5$. Flour: $300×2.5=750$ g. Milk: $200×2.5=500$ mL. Both amounts scale by the same factor, so the recipe proportions stay unchanged.
Do not compare different amounts by total price alone. Convert both offers to dollars per the same mass or volume.
Do not mix time units. Convert minutes to hours before calculating km/h.
Sanity check: 420 km in 3.5 h is 120 km/h because $3.5×120=420$.
How are you completing this lesson?
Brain Trainer · 4 problems
Find each rate. Write the unit with every answer, then reveal to check.
1 A car travels 240 km in 3 hours. Find its speed in km/h.
$240÷3=80$. The speed is 80 km/h.2 A 750 g pack costs $6. Find the price per 100 g.
$6 ÷ 750 × 100 = $0.80. The price is $0.80 per 100 g.3 A printer makes 60 pages in 4 minutes. Find pages per minute.
$60÷4=15$. The rate is 15 pages per minute.4 A train covers 315 km in 3.5 hours. Find average speed.
$315÷3.5=90$. Average speed is 90 km/h.
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. (a) A train travels 420 km in 3 hours 30 minutes. Find its average speed in km/h. (2 marks) (b) A printer makes 60 pages in 4 minutes. Find its rate in pages per minute. (2 marks)
Q7. Compare the value of 500 g of cereal for $4.50 and 750 g for $6.30. Calculate each price per 100 g, say which costs less on that basis, and find the difference. Show working.
Q8. A coach travels at a constant average speed of 72 km/h for 2.5 hours. How far does it travel? Explain why multiplying speed by time gives a distance, and include the answer unit.
Quick Check
1. B $240÷3=80$ km/h.
2. C $84÷2=42$ km/h.
3. A $60÷4=15$ pages per minute.
4. D $2.40 ÷ 3 = $0.80/kg.
5. B Both prices are $2.80/kg, so they are equal value.
Show Your Working Model Answers
Q6 (4 marks): (a) $3$ h $30$ min $=3.5$ h [0.5]; $420÷3.5=120$ km/h [1.5]. (b) $60÷4=15$ pages per minute [2].
Q7 (4 marks): Shop A: $4.50 ÷ 500 × 100 = $0.90 per 100 g [1]. Shop B: $6.30 ÷ 750 × 100 = $0.84 per 100 g [1]. Shop B costs less per 100 g [1]. Difference: $0.90 − $0.84 = $0.06 per 100 g [1].
Q8 (3 marks): Distance = speed × time [1]. $72$ km/h × $2.5$ h = $180$ km [1.5]. The hours cancel, leaving kilometres, so the answer unit is km [0.5].
Class Trip: Compare Two Cost Rates
Thirty students are planning a class trip. Company A charges $2.50 per kilometre for the bus, plus a $50 booking fee. Company B charges a flat $250 for the bus. The venue costs $8 per student for either company. (a) At 80 km, calculate the total and per-student cost for each. (b) At 100 km, which company is cheaper and by how much per student? (c) Find the distance where the bus costs are equal. Show why the venue fee does not affect that break-even distance.
Reveal solution
At 80 km, A costs $2.50 × 80 + $50 = $250 for the bus; B also costs $250. The venue adds $8 × 30 = $240, so each trip totals $490, or $490 ÷ 30 = $16.33 per student (to the nearest cent). At 100 km, A costs $250 + $50 + $240 = $540, or $18 per student; B costs $250 + $240 = $490, or $16.33 per student. B saves $50 ÷ 30 = $1.67 per student (to the nearest cent). Break-even: $2.50d + 50 = 250$, so $d=200÷2.50=80$ km. The same venue charge is added to both options, so it cancels when comparing bus costs.
Rate
Different units, for example km/h
Unit rate
Amount for one unit
Speed
Distance ÷ time
Unit price
Price ÷ quantity
Fair comparison
Use the same unit basis
Constant rate
Scale matching quantities together
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