4.2 Rates, Unit Pricing, Fuel Consumption, and Speed–Distance–Time

Ratios vs. Rates and Unit Pricing

While a ratio compares two quantities of the same kind (such as millilitres to millilitres) and has no units once simplified, a rate compares two quantities measured in different units—for example, dollars per kilogram ($ per $\text{kg}$), words read per minute ($\text{words/min}$), litres per $100\text{ kilometres}$ ($\text{L}/100\text{ km}$), or kilometres per hour ($\text{km/h}$).

A unit rate expresses how much of the first quantity corresponds to one single unit (or one standard benchmark block such as $100\text{ g}$, $100\text{ mL}$, or $100\text{ km}$) of the second quantity. For instance, if a Year 4 student reads $435\text{ words}$ in $3\text{ minutes}$, dividing $435 \div 3$ gives a unit reading fluency rate of $145\text{ words/min}$.

Unit Pricing and "Best Buy" Comparisons

Australian supermarkets display unit prices—typically cost per $100\text{ g}$, per $1\text{ kg}$, or per $1\text{ L}$—so consumers can compare value across different package sizes. School business managers use the same technique when ordering classroom consumables.

To determine the best buy:

1. Identify a convenient common unit (such as $1\text{ exercise book}$, $100\text{ g}$, or $1\text{ L}$).
2. Divide the price of each option by the number of those units in the package.
3. Compare the resulting unit prices; the lowest unit price is the best value for money.

Worked Example: Comparing Bulk Classroom Stationery Orders

A Year 5 coordinator needs to purchase A4 grid books and compares three supplier packages:

• Pack A ($10\text{ books}$ for $14.50): Unit cost is $14.50 \div 10 = 1.45$, or $1.45 per book.
• Pack B ($25\text{ books}$ for $33.00): Unit cost is $33.00 \div 25 = 1.32$, or $1.32 per book.
• Carton C ($60\text{ books}$ for $81.60): Unit cost is $81.60 \div 60 = 1.36$, or $1.36 per book.

Notice that the largest bulk option (Carton C at $1.36 per book) is not automatically the cheapest per unit; Pack B at $1.32 per book is the best buy.

Similarly, when comparing a $350\text{ g}$ tub of spread for $5.60 against a $600\text{ g}$ tub for $9.30, find the cost per $100\text{ g}$:

• $350\text{ g}$ tub contains $3.5\text{ lots of }100\text{ g}$: $5.60 \div 3.5 = 1.60$, or $1.60 per $100\text{ g}$.
• $600\text{ g}$ tub contains $6\text{ lots of }100\text{ g}$: $9.30 \div 6 = 1.55$, or $1.55 per $100\text{ g}$ (cheaper per $100\text{ g}$).

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Australian Fuel Consumption ($\text{L}/100\text{ km}$)

In Australia, automotive fuel efficiency is rated in litres per $100\text{ kilometres}$ ($\text{L}/100\text{ km}$). A lower number indicates a more fuel-efficient vehicle.

$$\text{Fuel Required (L)} = \frac{\text{Distance (km)}}{100} \times \text{Consumption Rate (L/100 km)}$$ $$\text{Maximum Range (km)} = \frac{\text{Fuel Available (L)}}{\text{Consumption Rate (L/100 km)}} \times 100$$

Worked Example 1: Regional School Bus Trip Fuel Cost

A regional school mini-bus travels a $350\text{ km}$ round trip for an athletics carnival. The bus has an average fuel consumption of $11.4\text{ L}/100\text{ km}$, and diesel costs $1.95 per litre (often displayed on service station boards as $195.0\text{ cents/L}$).

1. Count how many $100\text{ km}$ blocks are in the trip: $\frac{350}{100} = 3.5\text{ blocks of }100\text{ km}$.
2. Calculate total fuel required: $3.5 \times 11.4\text{ L} = 39.9\text{ L}$.
3. Calculate total fuel cost: $39.9 \times 1.95 = 77.805$ dollars, which rounds to $77.81.

Worked Example 2: Finding Maximum Driving Range from Tank Capacity

Suppose a school utility vehicle has a $65\text{ L}$ fuel tank and consumes $10.4\text{ L}/100\text{ km}$ on country highways. How far can it travel on one full tank?

$$\text{Maximum Range} = \frac{65}{10.4} \times 100\text{ km} = 6.25 \times 100\text{ km} = 625\text{ km}$$

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Speed, Distance, and Time ($\text{SDT}$)

Speed, distance, and time are linked by three equivalent rearrangements:

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}, \qquad \text{Distance} = \text{Speed} \times \text{Time}, \qquad \text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

Critical Trap 1: Decimal Hours vs. Minutes

Because clocks use base-$60$ ($60\text{ minutes} = 1\text{ hour}$) rather than base-$100$, a calculator result such as $2.35\text{ hours}$ does not mean $2\text{ hours }35\text{ minutes}$. Instead, take the decimal portion ($0.35$) and multiply by $60$: $0.35 \times 60 = 21\text{ minutes}$, giving $2\text{ hours }21\text{ minutes}$.

Conversely, if a teacher drives at $80\text{ km/h}$ for $1\text{ hour }15\text{ minutes}$, never multiply $80 \times 1.15$. Convert $15\text{ minutes}$ to $\frac{15}{60} = 0.25\text{ hours}$ first, then compute $80 \times 1.25 = 100\text{ km}$.

Converting Between $\text{m/s}$ and $\text{km/h}$ in Physical Education

Since $1\text{ km} = 1000\text{ m}$ and $1\text{ hour} = 3600\text{ seconds}$, a speed of $1\text{ m/s}$ covers $3600\text{ m} = 3.6\text{ km}$ in one hour. If a secondary student runs the $100\text{ m}$ sprint in $12.5\text{ seconds}$, their average speed in $\text{m/s}$ is $100 \div 12.5 = 8\text{ m/s}$. Multiplying by $3.6$ converts this to $8 \times 3.6 = 28.8\text{ km/h}$.

Critical Trap 2: Average Speed Across a Multi-Stage Journey

When a journey has two or more legs travelled at different speeds, the overall average speed is never the simple arithmetic mean of the speeds unless the time spent at each speed is identical. Always compute:

$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$

For example, if a teacher drives $60\text{ km}$ through suburban traffic at $40\text{ km/h}$ (taking $\frac{60}{40} = 1.5\text{ hours}$) and then $140\text{ km}$ on a freeway at $70\text{ km/h}$ (taking $\frac{140}{70} = 2.0\text{ hours}$), the total distance is $60 + 140 = 200\text{ km}$ and the total time is $1.5 + 2.0 = 3.5\text{ hours}$. The true average speed is $200 \div 3.5 \approx 57.14\text{ km/h}$, not $\frac{40 + 70}{2} = 55\text{ km/h}$.

Worked Example 3: Excursion Latest Departure Time with a Rest Break

A school camp coach must arrive at Parliament House by $11:15\text{ am}$ after covering $210\text{ km}$ at an average driving speed of $75\text{ km/h}$, plus a mandatory $25\text{-minute}$ driver rest stop. Driving time is $210 \div 75 = 2.8\text{ hours} = 2\text{ hours }48\text{ minutes}$. Adding the $25\text{-minute}$ rest break gives a total journey duration of $3\text{ hours }13\text{ minutes}$. Counting back $3\text{ hours }13\text{ minutes}$ from $11:15\text{ am}$ gives a latest departure time of $8:02\text{ am}$.