Rates - Year 8 Mathematics

What you will learn

distinguish a ratio (same unit) from a rate (different units),
write a rate as a unit rate (e.g. km per hour),
use the speed-distance-time triangle,
compare prices, fuel consumption, exchange rates, and pay rates,
solve multi-step problems that combine rates and percentages.

Worked example 0 Real-world example: road trip fuel cost

Your family is driving Melbourne to Adelaide ( $730$ km). The car uses $7.2$ L per $100$ km and petrol costs $1.85/L. What will the fuel cost?

Fuel needed: $\dfrac{730}{100} \times 7.2 = 7.3 \times 7.2 = 52.56$ L.
Cost: $52.56 \times 1.85 = 97.24$ dollars.
Budget about $100 for fuel each way.

Key idea: two rates interact here — fuel consumption (L/100 km) and fuel price ($/L). Multiplying them in the right order gives dollars per distance.

1. Rate vs ratio

A ratio compares two quantities of the same kind (both times, both amounts of money, both lengths).

A rate compares two quantities of different kinds and keeps the units: km/h, $/kg, L/min.

Unit rate

\text{unit rate} = \dfrac{\text{amount of A}}{\text{amount of B, per 1 unit}}.

Worked example 1 Speed as a unit rate

A car covers $240$ km in $3$ hours. Its average speed is

\dfrac{240 \text{ km}}{3 \text{ h}} = 80 \text{ km/h}.

2. Speed, distance, time

Speed-distance-time

Three forms

\text{speed} = \dfrac{\text{distance}}{\text{time}}, \quad \text{distance} = \text{speed} \times \text{time}, \quad \text{time} = \dfrac{\text{distance}}{\text{speed}}.

Units must match

If speed is in km/h, time must be in hours and distance in km. Convert first if the units do not match.

Worked example 2 Distance from speed and time

A cyclist rides at $18$ km/h for $2$ h $30$ min. How far?

Time in hours: $2.5$ .
Distance: $18 \times 2.5 = 45$ km.

Worked example 3 Time from speed and distance

A train travels $350$ km at $80$ km/h. How long does it take?

t = \dfrac{350}{80} = 4.375 \text{ h} = 4 \text{ h } 22 \text{ min } 30 \text{ s}.

3. Other common rates

Density: $\dfrac{\text{mass}}{\text{volume}}$ (g/cm $^3$ or kg/m $^3$ ).
Pay rate: $/hour.
Fuel consumption: L/100 km.
Exchange rate: units of currency A per 1 unit of currency B.
Interest rate (simple): $ per $100 per year, i.e. a percentage.

Worked example 4 Fuel consumption

A car uses $7$ L per $100$ km. On a $450$ km trip, how much fuel is used?

\dfrac{7}{100} \times 450 = 31.5 \text{ L}.

Worked example 5 Currency conversion

The exchange rate is $1$ AUD = $0.65$ USD. Convert $240 AUD to USD, then $600 USD back to AUD.

AUD → USD: $240 \times 0.65 = 156$ USD.
USD → AUD: $\dfrac{600}{0.65} \approx 923.08$ AUD.

Practice: Year 8 core

Fluency

Unit rates

A car goes $240$ km in $4$ h. Find the average speed.
A tap fills a tank at $600$ L in $20$ min. Find the rate in L/min.
A worker earns $540 for $20$ hours. Find the hourly rate.
A mass of $180$ g has volume $20$ cm $^3$ . Find the density.
A printer prints $60$ pages in $4$ minutes. Find pages per minute.

Fluency

Speed, distance, time

Distance from $65$ km/h × $3$ h.
Time to cover $300$ km at $50$ km/h.
Speed of $200$ m in $25$ seconds (in m/s).
Convert $72$ km/h to m/s.
How long to cover $120$ km at $80$ km/h?
A train covers $400$ km in $2$ h $30$ min. Find its speed.

Fluency

Fuel, pay, exchange

A car uses $8$ L/ $100$ km. Fuel for $350$ km?
A worker earns $22/h. Find pay for $36.5$ hours.
Exchange rate $1$ AUD $= 0.85$ NZD. Convert $150 AUD to NZD.
$1$ USD $= 1.50$ AUD. Convert $200 USD to AUD.
Simple interest: $1000 at $4\%$ for $3$ years. How much interest?

Reasoning

Explain and spot the mistake

Sam writes “the speed is $50$ km in $1$ hour, so $50$ per hour, so $50$ km”. What units are missing? What is the correct way to report speed?
Two cars: A does $60$ km in $1$ hour; B does $30$ km in $30$ minutes. Are they the same speed? Show working.
Explain why a “rate” and a “unit rate” are slightly different ideas. Give an example of each.
Without calculating, compare: a pool fills at $10$ L/min for $20$ min, or at $4$ L/min for $60$ min - which delivers more water?

Problem solving

Real contexts

A road trip is $1600$ km. If the driver averages $100$ km/h (including breaks in planned driving time), how long does the trip take?
A swimming pool holds $120\,000$ L. A hose delivers $120$ L/min. How long to fill (hours)?
A box of $24$ pens costs $7.20. A single pen costs $0.40. Which is better value, and by how much per pen?
Two phone plans: A is $25/month + $0.10/min; B is $35/month with unlimited calls. For what usage does B beat A?
An alloy uses $5$ kg of copper for every $3$ kg of tin. For a $40$ kg alloy, how much of each?

Challenge

Reasoning

Harder problems

A car averages $80$ km/h for $2$ hours and then $60$ km/h for $3$ hours. What is its average speed for the whole trip?
A shopkeeper buys tea at $12/kg and sells it at $15/kg. What percentage profit is this?
A tap fills a tank at $8$ L/min while a drain removes water at $3$ L/min. The tank holds $600$ L; it starts empty. How long to fill?
Currency arbitrage: $1 AUD $=$ $0.65 USD; $1 USD $=$ $0.80 EUR; $1 EUR $=$ $1.60 AUD. Is there a profit in converting $100 AUD → USD → EUR → AUD? If so, how much?

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 8 core - answers

Fluency

Unit rates

$60$ km/h
$30$ L/min
$27/h
$9$ g/cm $^3$
$15$ pages/min

Fluency

Speed, distance, time

$195$ km
$6$ h
$8$ m/s
$20$ m/s. Method: $72 \times \tfrac{1000}{3600}$ .
$1$ h $30$ min
$160$ km/h. Method: $400 \div 2.5$ .

Fluency

Fuel, pay, exchange

$28$ L. Method: $\tfrac{8}{100} \times 350$ .
$803. Method: $22 \times 36.5$ .
$127.50 NZD. Method: $150 \times 0.85$ .
$300 AUD. Method: $200 \times 1.50$ .
$120 interest. Method: $\tfrac{4}{100} \times 1000 \times 3$ .

Reasoning

Explain and spot the mistake

Units km/hour (or km per hour). Sam dropped the “per hour” and collapsed the rate to a distance, losing the time dimension. Correct: $50$ km/h.
Same speed. B does $30$ km in $0.5$ h $= 60$ km/h.
A rate is any comparison of two different quantities (e.g. $60$ km in $1.5$ h). A unit rate is the per-one-unit form (e.g. $40$ km/h). A rate can be simplified to a unit rate by division.
$10 \times 20 = 200$ L vs $4 \times 60 = 240$ L. The second delivers more.

Problem solving

Real contexts

$16$ h. Method: $1600 \div 100$ .
$1000$ min $\approx 16$ h $40$ min.
Single pen is dearer by $10$ c each. Box price $= 30$ c/pen; individual $40$ c/pen; difference $10$ c.
B beats A when $25 + 0.10 m > 35$ , i.e. $m > 100$ minutes.
Copper $25$ kg, tin $15$ kg. Method: $5 + 3 = 8$ parts; each part $= 5$ kg.

Challenge - answers

Reasoning

Harder problems

$68$ km/h. Method: total distance $= 160 + 180 = 340$ km; total time $= 5$ h; $340/5$ .
$25\%$ . Method: profit $= 3$ dollars/kg; $3/12$ .
$120$ min. Method: net $5$ L/min; $600/5$ .
Yes, small profit. $100 AUD → $65 USD → $52 EUR → $83.20 AUD. Wait: that’s a loss. Let me redo - actually the chain gives $100 \times 0.65 \times 0.80 \times 1.60 = 83.20$ AUD, so a loss, not a profit. No arbitrage exists; the student has lost $16.80 on fees-free conversion.

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