C1.12; E1.12 - Rates and average speed | Cambridge IGCSE Mathematics

A rate tells you how much of one quantity corresponds to one unit of another. You will use pay, flow and fuel rates, calculate average speed and apply supplied formulas for density, pressure and population density. C1.12 and E1.12 share these methods, with units helping you choose each calculation.

Rates per unit

A cyclist travels $30$ km at $10$ km/h and another $30$ km at $30$ km/h. Is the average speed $20$ km/h? The two speeds do not last for equal times. We will find the answer by returning to the meaning of “per hour”.

Rate

A rate compares two quantities with different units. “Per” means “for each one”: $\$ 9 $per hour gives$ $9 $for each hour, and$ 3,\mathrm{L/min} $means$ 3$ litres each minute.

To find a rate, divide the total amount by the number of units. To find the total amount at a constant rate, multiply rate by number of units. To find how many units, divide total by rate. These are three directions through the same relationship.

Worked example 1: hourly pay

Question. A worker earns $\$ 9 $per hour. Calculate the pay for$ 3 $hours$ 30$ minutes.

Plan. Express all the time in hours, then multiply by the hourly rate.

\begin{aligned} 3+\frac{30}{60}&=3.5 && \text{time in hours}\\ 9\times3.5&=31.5 && \text{dollars per hour times hours} \end{aligned}

Answer. The pay is $\$ 31.50$.

Check. Three hours earns $\$ 27 $, and half an hour earns$ $4.50 $. Their sum is$ $31.50$.

Increase only the working time by one hour. The pay should increase by exactly one hourly rate.

Worked example 2: an extra hour

Question. At the same rate, calculate the pay for $4$ hours $30$ minutes.

Plan. Convert the duration to hours and multiply by $9$ .

\begin{aligned} 4+\frac{30}{60}&=4.5 && \text{time in hours}\\ 9\times4.5&=40.5 && \text{apply the constant hourly rate} \end{aligned}

Answer. The pay is $\$ 40.50$.

Check. $40.50-31.50=9$ , the pay for the extra hour.

The units change in other rate problems, but the reasoning stays the same. A tap delivering $3\,\mathrm{L/min}$ fills $18$ litres in $18\div3=6$ minutes. A pump delivering $24$ litres in $8$ minutes has flow rate $24\div8=3\,\mathrm{L/min}$ . These calculations assume a constant flow and no lost water.

Fuel consumption needs especially careful reading. If a car travels $90$ km using $6$ litres, its rate in kilometres per litre is

90\div6=15\,\mathrm{km/L}.

At that rate, $150$ km requires $150\div15=10$ litres. A different quote, $6$ litres per $100$ km, measures fuel used instead of distance travelled: a $250$ km journey uses $6\times2.5=15$ litres. A larger km/L value means less fuel for a fixed journey, while a larger L per $100$ km value means more fuel.

Watch out

“ $15\,\mathrm{km/L}$ ” does not mean $15$ litres per kilometre. Write what one unit produces. For $150$ km, multiplying by $15$ would not give a fuel amount consistent with this rate.

Use the rate's units to choose the operation.

Average speed

Average speed is the total distance travelled divided by the total elapsed time. It is the constant speed that would cover the same distance in the same time. Actual speed can change throughout a journey.

Speed, distance and time

\text{average speed}=\frac{\text{total distance}}{\text{total time}},\qquad \text{distance}=\text{speed}\times\text{time},\qquad \text{time}=\frac{\text{distance}}{\text{speed}}

Use consistent units. In the distance and time relationships, speed means a constant speed or the overall average speed for that same journey. Time must be positive when calculating speed; speed must be positive when dividing to calculate journey time.

Not in the List of formulas: learn these relationships.

For km/h, use kilometres and hours. For m/s, use metres and seconds. You know that $1$ km is $1000$ m and $1$ hour is $3600$ s, so $72$ km/h is $72\times1000\div3600=20$ m/s. Converting only the distance or only the time gives the wrong rate.

Worked example 3: an exact number of hours

Question. A bus travels $90$ km in $2$ hours. Calculate its average speed.

Plan. Divide the total distance by the total time in hours.

\begin{aligned} \text{average speed}&=90\div2 && \text{kilometres divided by hours}\\ &=45 && \text{distance travelled per hour on average} \end{aligned}

Answer. $45\,\mathrm{km/h}$ .

Check. $45\times2=90$ km, reproducing the distance.

Keep the distance but increase the journey time by fifteen minutes. The average speed must fall.

Worked example 4: hours and minutes

Question. A bus travels $90$ km in $2$ hours $15$ minutes. Calculate its average speed.

Plan. Convert the whole duration to hours before dividing.

\begin{aligned} 2+\frac{15}{60}&=2.25 && \text{time in hours}\\ 90\div2.25&=9000\div225 && \text{equal scaling for division}\\ &=40 && \text{average speed in kilometres per hour} \end{aligned}

Answer. $40\,\mathrm{km/h}$ .

Check. Two hours at $40$ km/h gives $80$ km, and a quarter hour gives $10$ km. Total $90$ km; the speed is below $45$ km/h as expected.

For a journey with several stages, add the distances and add the times first. Include stopping time if the question asks for the average over the whole elapsed journey.

The opening cyclist takes $30\div10=3$ hours for the first stage and $30\div30=1$ hour for the second. The overall average is $60\div4=15$ km/h. The slower stage lasts three times as long, so averaging $10$ and $30$ gives the wrong result.

Watch out

$2$ hours $15$ minutes is $2.25$ hours, not $2.15$ hours. For equal-duration stages the arithmetic mean of the stage speeds gives the overall average; otherwise use total distance divided by total time.

Try a new distance with a fractional-hour duration.

Density and other supplied rate formulas

Some rates describe how closely an amount is packed into a space or how a force is spread. Read the supplied formula before substituting. The units tell you what is being measured per unit of what.

Density is mass per unit volume. In $\mathrm{g/cm^3}$ , it tells you the grams in each cubic centimetre. Pressure is force per unit area; $\mathrm{N/cm^2}$ means newtons per square centimetre. Population density is population per unit area, often people per square kilometre. These are averages: a district's population density does not mean every square kilometre contains exactly that many people.

For these rate questions the necessary formula is supplied. You must still put the quantities into matching units. All divisor volumes and areas in these contexts must be positive.

Worked example 5: density in matching units

Question. A solid has mass $600$ g and volume $200\,\mathrm{cm^3}$ . Calculate its density in $\mathrm{g/cm^3}$ .

$\text{Density}=\text{mass}\div\text{volume}$ .

Plan. The units already match, so divide the mass by the volume.

\begin{aligned} \text{density}&=600\div200 && \text{use the supplied formula}\\ &=3 && \text{grams per cubic centimetre} \end{aligned}

Answer. $3\,\mathrm{g/cm^3}$ .

Check. $3\times200=600$ g, recovering the mass.

Change only the way the mass is stated. The physical solid and its density stay the same.

Worked example 6: converting mass first

Question. A solid has mass $0.6$ kg and volume $200\,\mathrm{cm^3}$ . Calculate its density in $\mathrm{g/cm^3}$ .

$\text{Density}=\text{mass}\div\text{volume}$ .

Plan. Convert kilograms to grams before applying the formula.

\begin{aligned} 0.6\times1000&=600 && \text{mass in grams}\\ 600\div200&=3 && \text{grams divided by cubic centimetres} \end{aligned}

Answer. $3\,\mathrm{g/cm^3}$ .

Check. $3\times200=600$ g, which equals $0.6$ kg. Changing the written unit has not changed the density.

Apply the same division idea to a pressure problem. A $240$ N force acting uniformly over $12\,\mathrm{cm^2}$ , with $\text{pressure}=\text{force}\div\text{area}$ , gives $240\div12=20\,\mathrm{N/cm^2}$ . If the same force acts over $24\,\mathrm{cm^2}$ , pressure halves to $10\,\mathrm{N/cm^2}$ : spreading the force reduces the amount per unit area.

For a district of $48\,\mathrm{km^2}$ with $24\,000$ residents, the supplied formula $\text{population density}=\text{population}\div\text{area}$ gives $24\,000\div48=500$ people per $\mathrm{km^2}$ . To recover population, multiply $500$ by $48$ . To find area instead, divide the population by the population density. These reverse calculations follow from the meaning of “per”.

Watch out

$0.6\div200=0.003$ is a density in $\mathrm{kg/cm^3}$ , not $\mathrm{g/cm^3}$ . A correct numerical division with the wrong unit does not answer the question. Convert before substituting or convert the result consistently.

Use a supplied rate formula in this new setting.

Quick Recap

“Per” means “for each one”. Multiply a rate by the number of units to find a total; divide to reverse that calculation.
Read fuel-consumption units carefully: km/L and litres per $100$ km describe different quantities.
Average speed is total distance divided by total elapsed time, using consistent units.
For density, pressure and population density, use the supplied formula and check the units before calculating.

The opening cyclist averages $15$ km/h, because $60$ km takes $4$ hours. The longer time at the slower speed makes the simple mean of the two speeds unsuitable.

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C1.12; E1.12 - Rates and average speed

Rates per unit

Average speed

Density and other supplied rate formulas

Quick Recap

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