C1.11; E1.11 - Rates and average speed | Cambridge IGCSE International Mathematics

If you travel one part of a journey at 20 km/h and another at 30 km/h, must the average be 25 km/h? Rates compare quantities with different units, and the time spent at each speed matters. Use C1.11 and E1.11 to calculate pay, flow, fuel consumption and average speeds for complete journeys.

Common rates

A rate tells you how much of one quantity corresponds to a unit of another. The word “per” means “for each”: $14 per hour means$ 14 for each hour worked at that rate.

Rate

A rate compares two quantities, usually with different units. A rate per one unit is found by dividing the total quantity by the number of those units.

The units tell you which division to perform. Litres per minute means litres divided by minutes; minutes per litre is the reverse rate. If the rate is constant, multiplying by the number of units gives the total quantity.

Worked example 1: hourly pay

Question. A worker earns $63 for$ 4.5$ hours at a constant hourly rate. Find the hourly rate and the pay for 6 hours.

Plan. Find the pay for one hour, then scale up to 6 hours.

\begin{aligned} \text{hourly rate}&=63\div4.5=14\text{ dollars per hour}\\ \text{pay for 6 hours}&=14\times6=84\text{ dollars} \end{aligned}

Answer. $14 per hour;$ 84 for 6 hours.

Check. $14\times4.5=63$ , recovering the given pay.

The next example changes the quantities to water and time, but uses the same unitary method.

Worked example 2: a flow rate

Question. A pump delivers 270 litres in $4.5$ minutes at a constant rate. Find its rate in litres per minute and the volume delivered in 6 minutes.

Plan. Divide by the time to find one minute's volume, then multiply by 6.

\begin{aligned} \text{flow rate}&=270\div4.5=60\,\mathrm{L/min}\\ \text{volume in 6 minutes}&=60\times6=360\,\mathrm{L} \end{aligned}

Answer. $60\,\mathrm{L/min}$ ; 360 litres.

Check. $60\times4.5=270$ litres. The longer time should produce more than 270 litres.

Fuel consumption is often quoted per 100 km rather than per 1 km. A car using $6.4\,\mathrm{L}$ per $100\,\mathrm{km}$ uses $0.064\,\mathrm{L}$ per km under the same conditions. Alternatively, $20\,\mathrm{km/L}$ means each litre takes the car 20 km. A larger km/L value means better fuel economy; a smaller L per 100 km value means better fuel economy.

Worked example 3: fuel per hundred kilometres

Question. A car uses $6.4\,\mathrm{L}$ of fuel per $100\,\mathrm{km}$ . Estimate the fuel needed for $375\,\mathrm{km}$ , assuming the same consumption rate.

Plan. Find how many lots of 100 km are travelled, then scale the fuel.

\begin{aligned} \text{number of 100 km lots}&=375\div100=3.75\\ \text{fuel}&=3.75\times6.4=24\,\mathrm{L} \end{aligned}

Answer. 24 litres.

Check. $24\div6.4=3.75$ lots of 100 km, which is 375 km. Actual fuel use can change with driving conditions, so this is a constant-rate estimate.

Read the entire unit before calculating.

Watch out

At $6.4\,\mathrm{L}$ per 100 km, do not calculate $375\times6.4$ : that treats the rate as litres per kilometre. Divide the distance by 100 first.

Apply the rate to a new journey.

Average speed

Average speed is another rate: total distance for each unit of time. It describes a whole journey even when the speed changes during it.

Speed, distance and time

For a journey of distance $d$ , taking time $t>0$ , with average speed $v$ ,

v=\frac{d}{t},\qquad d=vt,\qquad t=\frac{d}{v}\quad(v>0)

Match units: km and hours give km/h; metres and seconds give m/s.

Not in the supplied formula list: learn it.

The first formula follows from “distance per unit time”. The other two undo division or multiplication. For example, travelling at 5 km/h for 2 hours gives $5\times2=10$ km.

Remember that 1 hour 36 minutes is $1+36/60=1.6$ hours. It is not $1.36$ hours.

Worked example 4: calculate an average speed

Question. A coach travels $120\,\mathrm{km}$ in 1 hour 36 minutes. Calculate its average speed in km/h.

Plan. Convert the time to hours before dividing distance by time.

\begin{aligned} t&=1+\frac{36}{60}=1.6\,\mathrm{h}\\ v&=\frac{120}{1.6}=75\,\mathrm{km/h} \end{aligned}

Answer. $75\,\mathrm{km/h}$ .

Check. $75\times1.6=120$ km. The units km divided by h give km/h.

Now the unknown is time, so divide distance by speed instead.

Worked example 5: calculate a journey time

Question. A train travels $180\,\mathrm{km}$ at an average speed of $72\,\mathrm{km/h}$ . Find the journey time in hours and minutes.

Plan. Calculate hours using $t=d/v$ , then convert the fractional hour.

\begin{aligned} t&=\frac{180}{72}=2.5\,\mathrm{h}\\ 0.5\times60&=30\,\mathrm{min} \end{aligned}

Answer. 2 hours 30 minutes.

Check. $72\times2.5=180$ km. At 72 km/h, 2 hours covers 144 km and half an hour covers the remaining 36 km.

To convert a speed, convert both units. In one hour an object moving at $1\,\mathrm{m/s}$ travels 3600 m, or 3.6 km. Therefore multiply m/s by $3.6$ for km/h, and divide km/h by $3.6$ for m/s.

Watch out

If distance is in kilometres and time is in minutes, the quotient is km/min, not km/h. Change the time to hours or convert the resulting rate afterwards.

State the units as part of the answer.

Journeys in stages

For a journey in stages, the overall average speed uses the whole distance and the whole time. Each stage contributes its actual duration; the speeds are not usually equally weighted.

Overall average speed

\text{overall average speed}=\frac{\text{total distance}}{\text{total elapsed time}}

If the question includes a stop in the journey, include its time even though it adds no distance. An average speed “while moving” excludes stopped time.

The opening journey shows why simply averaging 20 and 30 can fail.

Worked example 6: equal distances at different speeds

Question. A cyclist travels 30 km at 20 km/h, then 30 km at 30 km/h, without stopping. Calculate the overall average speed.

Plan. Calculate each stage's time, add the times and distances, then divide.

\begin{aligned} t_1&=30\div20=1.5\,\mathrm{h}\\ t_2&=30\div30=1\,\mathrm{h}\\ \text{overall speed}&=\frac{30+30}{1.5+1}=24\,\mathrm{km/h} \end{aligned}

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

Check. $24\times2.5=60$ km. More time is spent at 20 km/h, so the average should be closer to 20 than to 30.

Keep the same moving stages and add a half-hour stop. This isolates the effect of stopped time.

Worked example 7: include a stop

Question. The cyclist travels 30 km at 20 km/h, stops for 30 minutes, then travels 30 km at 30 km/h. Calculate the average speed for the whole journey, including the stop.

Plan. Add the stop to the previous stage times; the total distance stays 60 km.

\begin{aligned} \text{total time}&=1.5+0.5+1=3\,\mathrm{h}\\ \text{overall speed}&=\frac{60}{3}=20\,\mathrm{km/h} \end{aligned}

Answer. $20\,\mathrm{km/h}$ .

Check. The distance is unchanged and the elapsed time is longer, so the result must be below the previous 24 km/h. $20\times3=60$ confirms the calculation.

The simple mean $(20+30)/2=25$ would work if the cyclist spent equal times moving at those speeds. Equal distances do not give equal times.

Problem-solving

Make a stage table with distance, speed and time. Complete missing times before calculating the total-distance divided by total-time average.

For this check, include the stated rest in the elapsed time.

Quick Recap

“Per” determines the division and the units of a rate.
A fuel rate per 100 km applies to the number of 100 km portions travelled.
Use $v=d/t$ , $d=vt$ and $t=d/v$ with consistent units.
For a complete journey, divide total distance by total elapsed time, including any stated stops.

The opening speeds do not force an average of 25 km/h: equal 30 km stages at 20 and 30 km/h give 24 km/h.

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C1.11; E1.11 - Rates and average speed

Common rates

Average speed

Journeys in stages

Quick Recap

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