R11 - Speed, density and other compound units | AQA GCSE Mathematics

Use speed, rates of pay, unit prices, density and pressure to connect two measurements. You will choose a calculation from the meaning of “per”, keep units consistent and compare quantities fairly.

Average speed, distance and time

Travel half a journey at $20\,\mathrm{km/h}$ and the other half at $60\,\mathrm{km/h}$ . Is your average speed $40\,\mathrm{km/h}$ ? The slower half takes longer, so averaging the two displayed speeds will not work.

You already know

There are $60$ minutes in an hour, so $45$ minutes is $\frac{45}{60}=0.75$ hours. R1 covered converting compound units. Convert before combining measurements in a formula.

A compound unit combines units. In $\mathrm{km/h}$ , “per hour” means “for each hour”. A steady $50\,\mathrm{km/h}$ covers $50\,\mathrm{km}$ in one hour, or $25\,\mathrm{km}$ in half an hour. Average speed spreads the total distance evenly over the total time, even if the actual speed changes.

Average speed

\begin{aligned} \text{average speed}&=\frac{\text{total distance}}{\text{total time}}\\ \text{distance}&=\text{speed}\times\text{time}\\ \text{time}&=\frac{\text{distance}}{\text{speed}} \end{aligned}

Use matching units. For average speed over several stages, include all distances and all elapsed times, including stops.

Not on the formulae sheet: learn it.

A written total-time calculation makes your choice clear, especially when a journey includes a stop.

Worked example 1: a journey without a stop

Question. A cyclist travels $18\,\mathrm{km}$ in $45$ minutes. Work out the average speed in $\mathrm{km/h}$ . Non-calculator.

Plan. Convert the time to hours, then divide distance by time.

\begin{aligned} 45\div60&=\frac34 &&\text{time in hours}\\ 18\div\frac34&=18\times\frac43 &&\text{distance divided by time}\\ &=24 &&\text{evaluate} \end{aligned}

Why convert first? Dividing kilometres by minutes gives kilometres per minute, not kilometres per hour.

Answer. $24\,\mathrm{km/h}$ .

Check. $24\times\frac34=18\,\mathrm{km}$ .

Add only a $15$ -minute stop to the same ride. The distance stays the same, while the total elapsed time increases.

Worked example 2: including a stop

Question. A cyclist rides $18\,\mathrm{km}$ in $45$ minutes, then rests for $15$ minutes. Work out the average speed over the whole hour, including the rest. Non-calculator.

Plan. Add riding and resting times, then use total distance divided by total time.

\begin{aligned} 45+15&=60 &&\text{elapsed minutes}\\ 60\div60&=1 &&\text{elapsed hours}\\ 18\div1&=18 &&\text{average speed} \end{aligned}

Why include the stop? The question asks about the whole hour. No extra distance is covered during those fifteen minutes.

Answer. $18\,\mathrm{km/h}$ .

Check. A stop must reduce the whole-journey average below the riding average of $24\,\mathrm{km/h}$ .

Now settle the opening puzzle with a $60\,\mathrm{km}$ journey. The first $30\,\mathrm{km}$ takes $30\div20=1.5$ hours; the second takes $30\div60=0.5$ hours. The average is $60\div2=30\,\mathrm{km/h}$ , not $40$ . Surprisingly, equal distances do not give equal importance to the two speeds: three quarters of the time is spent at the slower one.

Watch out

Wrong: $(20+60)\div2=40$ for equal-distance stages. Right: divide total distance by total time. The simple mean of two speeds works for two stages of equal duration, not generally for equal distances.

The next ride changes the distance and the rest time. Include both pieces of time.

Rates of pay and unit prices

A pay rate tells you the money for one unit of time; a unit price tells you the money for one unit of a product.

At $\pounds 12$ per hour, three hours earns $\pounds 36$ . A $\pounds 6$ pack containing $2\,\mathrm{kg}$ costs $\pounds 3$ per kilogram. “Per” tells you which quantity belongs underneath the division.

Pay rates and unit prices

For a fixed hourly rate and no other payments or deductions,

\begin{aligned} \text{pay}&=\text{hourly rate}\times\text{hours}\\ \text{hourly rate}&=\frac{\text{pay}}{\text{hours}}\\ \text{unit price}&=\frac{\text{total price}}{\text{quantity}} \end{aligned}

Match time units for pay and quantity units for price comparisons.

Not on the formulae sheet: learn it.

Compare rates for the same amount. A larger total payment can simply mean more hours of work.

Worked example 3: comparing hourly pay

Question. Job A pays $\pounds 52$ for $4$ hours. Job B pays $\pounds 67.50$ for $5$ hours. Which has the higher hourly rate, and by how much? Non-calculator.

Plan. Divide each payment by its own number of hours, then compare.

\begin{aligned} 52\div4&=13 &&\text{A: pounds per hour}\\ 67.50\div5&=13.50 &&\text{B: pounds per hour}\\ 13.50-13&=0.50 &&\text{difference per hour} \end{aligned}

Why not compare $52$ with $67.50$ directly? They pay for different amounts of time.

Answer. Job B pays $\pounds 0.50$ more per hour.

Check. At A's rate, five hours pays $5\times13=65$ , which is $\pounds 2.50$ less than B over five hours.

Change the measured quantity from hours to kilograms. The same divide-to-one-unit method compares prices, but now the lower rate is better value.

Worked example 4: comparing prices per kilogram

Question. Rice A costs $\pounds 2.40$ for $0.8\,\mathrm{kg}$ . Rice B costs $\pounds 3.75$ for $1.5\,\mathrm{kg}$ . Which has the lower price per kilogram, and by how much? Non-calculator.

Plan. Divide each price by its mass in kilograms.

\begin{aligned} 2.40\div0.8&=24\div8=3 &&\text{A: pounds per kilogram}\\ 3.75\div1.5&=7.50\div3=2.50 &&\text{B: pounds per kilogram}\\ 3-2.50&=0.50 &&\text{difference per kilogram} \end{aligned}

Why use kilograms for both? The rates must refer to the same amount before their numbers can be compared.

Answer. Rice B costs $\pounds 0.50$ less per kilogram.

Check. At A's rate, $1.5\,\mathrm{kg}$ would cost $1.5\times3=\pounds 4.50$ , more than B's $\pounds 3.75$ .

Rates can be multiplied back by a new amount: $3.5$ hours at $\pounds 13$ per hour earns $\pounds 45.50$ . Keep hours and minutes distinct; $3$ hours $30$ minutes is $3.5$ hours, not $3.30$ .

Watch out

Wrong: compare $\pounds 2.40$ per $800\,\mathrm{g}$ with $\pounds 3.75$ per $1.5\,\mathrm{kg}$ by dividing by $800$ and $1.5$ and comparing the results. One rate would be pounds per gram and the other pounds per kilogram. Right: convert the quantities to matching units first.

A new pair of packs uses grams for one quantity. R1's conversion is the first step.

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R11 - Speed, density and other compound units

Average speed, distance and time

Rates of pay and unit prices

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