Rate and Average Rate

What a rate is and how it differs from a ratio; finding a rate and using it to find an amount or a time; comparing rates; and the average rate when the rate does not stay the same.

Before you start

A ratio compares quantities of the same kind, in the same unit. A class of 15 boys and 20 girls has boys : girls $= 15 : 20 = 3 : 4$ , and the ratio has no unit.

The unitary method: find the value of one unit first, then of as many units as you need. If 4 pens cost $6, then 1 pen costs $6 \div 4 = 1.5$ , which is $1.50, and 7 pens cost $1.5 \times 7 = 10.5$ , which is $10.50.

Times are changed to one unit before you divide or multiply: 1 h 12 min $= 60 + 12 = 72$ min, and 30 min $= \frac{30}{60}$ h $= \frac{1}{2}$ h.

What a Rate Is

A ratio compares two quantities of the same kind: 6 kg of flour to 4 kg of sugar is the ratio $6 : 4 = 3 : 2$ . Both are masses, the units cancel, and the ratio has no unit.

A rate compares two quantities of different kinds. A tap that lets out 45 litres of water in 3 minutes compares a volume with a time. Dividing, $45 \div 3 = 15$ , tells us that 15 litres come out for each minute. The rate is 15 litres per minute, written 15 L/min. A rate always has a unit, and the unit is made of the units of both quantities.

Definition: rate

A rate compares two quantities of different kinds. It tells us how much of the first quantity there is for one unit of the second quantity.

Rate = first quantity ÷ second quantity. Its unit is "unit of the first quantity per unit of the second quantity", such as L/min (litres per minute) or $ per hour.

Rate	What it compares	Example
L/min (litres per minute)	volume with time	A tap lets out 15 L of water each minute.
$ per hour	money with time	A part-time worker is paid $12.50 for each hour.
$/kg (dollars per kilogram)	money with mass	Prawns cost $9 for each kilogram.
words per minute	a number of things with time	A student types 60 words each minute.
km/L (kilometres per litre)	distance with volume of petrol	A car travels 15 km on each litre of petrol.
people per km²	a number of people with area	A town has 8000 people living on each square kilometre.
km/h (kilometres per hour)	distance with time	A bus travels 40 km each hour. This rate is a speed, the subject of the next lessons.

Example 1: Pay per hour

Mr Tan is paid $100 for 8 hours of work at a hawker stall. Find his rate of pay.

Solution

The rate is money for each hour, so divide the money by the number of hours.

Rate of pay $= 100 \div 8 = 12.5$ dollars per hour.

Check: $12.50 \times 8 = 100$ , so 8 hours at $12.50 per hour is $100.

Mr Tan's rate of pay is $12.50 per hour.

Example 2: A rate in other units

A 400 g packet of rice crackers costs $3.20. Find the cost (a) in cents per gram, (b) in cents per 100 g.

Solution

(a) The question asks for cents, so first write the price in cents: $3.20 = 320 cents.

Cost per gram $= 320 \div 400 = 0.8$ cents per gram.

(b) 100 g is 100 times as much as 1 g, so it costs 100 times as much: $0.8 \times 100 = 80$ cents.

Or: 400 g is 4 lots of 100 g, so 100 g costs $320 \div 4 = 80$ cents.

(a) The crackers cost 0.8 cents per gram. (b) They cost 80 cents per 100 g.

Using a Rate to Find an Amount or a Time

When a rate stays the same, it tells you how much is done in every unit of time. A tap at 12 L/min lets out 12 litres in the first minute, 12 more in the second, and so on. So in 5 minutes it lets out $12 \times 5 = 60$ litres, and to let out 300 litres it needs as many minutes as there are 12s in 300.

A double number line shows this: time on one line, the amount on the other, and marks one above the other belong together. Whatever the time is multiplied by, the amount is multiplied by the same number.

Rule: using a rate that stays the same

Amount = rate × time. A rate of 12 L/min for 5 minutes gives $12 \times 5 = 60$ litres.

Time = amount ÷ rate. To let out 300 L at 12 L/min takes $300 \div 12 = 25$ minutes.

The time must be in the unit of the rate: minutes for a rate per minute, hours for a rate per hour.

Example 3: Laying bricks

Ravi lays bricks at a steady rate of 3 bricks every 5 minutes. How many bricks does he lay in 1 hour?

Solution

Work in minutes, the unit of the rate: 1 hour = 60 minutes.

Method 1, in lots of 5 minutes. 60 minutes is $60 \div 5 = 12$ lots of 5 minutes. In each lot he lays 3 bricks, so in 60 minutes he lays $12 \times 3 = 36$ bricks.

Method 2, the unitary method. In 5 minutes he lays 3 bricks, so in 1 minute he lays $\frac{3}{5}$ of a brick. In 60 minutes he lays $60 \times \frac{3}{5} = 36$ bricks.

His rate is $\frac{3}{5} = 0.6$ bricks per minute. A rate may be a fraction or a decimal even when the things counted are whole.

Ravi lays 36 bricks in 1 hour.

Example 4: Filling a tank

A tap fills an empty tank with water at 12 L/min. The tank holds 300 L.
(a) How long does the tap take to fill the tank?
(b) After 18 minutes, how many more litres are needed to fill the tank?

Solution

(a) Each minute the tap adds 12 L. The time is the number of 12-litre lots in 300 L.

Time $= 300 \div 12 = 25$ minutes.

(b) In 18 minutes the tap adds $12 \times 18 = 216$ litres.

Still needed: $300 - 216 = 84$ litres.

Check: the tap needs $84 \div 12 = 7$ more minutes, and $18 + 7 = 25$ minutes, as in (a).

(a) The tap takes 25 minutes to fill the tank. (b) 84 more litres are needed.

Example 5: A rate with hours and minutes

The school printer prints worksheets at 25 pages per minute.
(a) How many pages does it print in 1 hour 12 minutes?
(b) How long does it take to print 1000 pages?

Solution

(a) The rate is per minute, so write the time in minutes: 1 h 12 min $= 60 + 12 = 72$ min. (It is not 1.12 hours: 12 minutes is $\frac{12}{60} = 0.2$ of an hour.)

Pages $= 25 \times 72 = 1800$ .

(b) Time $= 1000 \div 25 = 40$ minutes.

(a) The printer prints 1800 pages in 1 hour 12 minutes. (b) It takes 40 minutes to print 1000 pages.

A Rate Either Way Round

Two quantities give two rates, one each way round. A car that travels 450 km on 40 L of petrol can be described by the distance it goes on each litre (km/L) or by the petrol it uses for each 100 km (L per 100 km). Both are correct; they answer different questions.

Which is better depends on the rate. For km/L, a larger number is better: the car goes further on each litre. For L per 100 km, a smaller number is better: the car uses less petrol for the same distance. In the same way a lower price per kilogram is better for a buyer, and more grams per dollar is better.

Example 6: Petrol, both ways round

A car travelled 450 km on 40 L of petrol. Find (a) how far it travels on each litre, in km/L, (b) how much petrol it uses for each 100 km, in L per 100 km.

Solution

(a) km/L is kilometres for each litre, so divide the distance by the volume: $450 \div 40 = 11.25$ km/L.

(b) Litres for each kilometre: 450 km → 40 L, so 1 km → $\frac{40}{450}$ L.

100 km → $\frac{40}{450} \times 100 = \frac{4000}{450} = \frac{80}{9} = 8\frac{8}{9}$ L.

$8\frac{8}{9} = 8.888\ldots \approx 8.89$ to 3 significant figures.

(a) The car travels 11.25 km/L. (b) It uses about 8.89 L per 100 km, correct to 3 significant figures.

Example 7: Which car is more economical?

Car A travels 15 km/L. Car B uses 6 L of petrol per 100 km. Which car is more economical with petrol?

Solution

The two rates are the opposite way round, so they cannot be compared as they stand. Write Car B's rate in km/L.

Car B: 6 L → 100 km, so 1 L → $100 \div 6 = 16\frac{2}{3}$ km. That is about 16.7 km/L.

Car B goes $16\frac{2}{3}$ km on a litre and Car A goes 15 km, so Car B goes further on each litre.

Check the other way round: for 100 km Car A needs $100 \div 15 = 6\frac{2}{3}$ L, which is more than Car B's 6 L.

Car B is more economical with petrol.

Comparing by Rates

Two prices, or two workers, are hard to compare when the amounts differ: $8.70 for 3 litres or $13.50 for 5 litres? Find each rate for one unit, in the same units, and compare those.

Example 8: The better buy

Brand A detergent costs $8.70 for 3 litres. Brand B costs $13.50 for 5 litres. Which is the better buy?

Solution

Find the price of 1 litre of each.

Brand A: $8.70 \div 3 = 2.90$ , so $2.90 per litre.

Brand B: $13.50 \div 5 = 2.70$ , so $2.70 per litre.

Brand B costs less for each litre.

Brand B is the better buy, at $2.70 per litre against $2.90 per litre.

Example 9: The faster worker

At a hotel, Mrs Goh cleans 17 rooms in 5 hours and Mrs Lim cleans 14 rooms in 4 hours. Who works at the faster rate?

Solution

Find each rate in rooms per hour.

Mrs Goh: $17 \div 5 = 3.4$ rooms per hour.

Mrs Lim: $14 \div 4 = 3.5$ rooms per hour.

3.5 is more than 3.4, so Mrs Lim cleans more rooms in each hour.

The other way round, in minutes per room: Mrs Goh takes $300 \div 17 \approx 17.6$ minutes per room and Mrs Lim $240 \div 14 \approx 17.1$ minutes. A shorter time for each room is faster, so the answer is the same.

Mrs Lim works at the faster rate, 3.5 rooms per hour against 3.4.

Average Rate

Mei Ling types 300 words in 5 minutes. Dividing, $300 \div 5 = 60$ , gives 60 words per minute. But she does not type exactly 60 words in every minute: she slows down to think and speeds up on easy sentences. The table shows what she typed in each minute, and what a typist at a constant 60 words per minute would type.

Minute	1st	2nd	3rd	4th	5th	Total
Mei Ling	52	63	66	58	61	300
A constant 60 words per minute	60	60	60	60	60	300

Both type 300 words in 5 minutes. 60 words per minute is Mei Ling's average rate: the steady rate that would do the same total in the same time. Most rates in real life change from moment to moment, so a rate found from a total and a time is an average rate. A train that covers 500 km in 4 hours has an average rate of $500 \div 4 = 125$ km/h, though it does not go at 125 km/h the whole way.

Definition: constant rate and average rate

A rate is constant when the same amount is done in every interval of time (or of the other quantity) of the same length, however short. Then the rate in every interval is the same.

Average rate = total amount ÷ total time (or ÷ the total of the other quantity). It is the constant rate that would give the same total in the same time.

To find an average rate over several parts, add the amounts and add the times first, then divide. Do not take the average of the rates of the parts.

Example 10: An average rate in two parts

A tank is being filled. In the first 4 minutes 40 L of water flows in. The tap is then turned up, and in the next 6 minutes 90 L flows in. Find the average rate of flow over the 10 minutes.

Solution

Total amount $= 40 + 90 = 130$ L.

Total time $= 4 + 6 = 10$ min.

Average rate $= 130 \div 10 = 13$ L/min.

The two parts had rates $40 \div 4 = 10$ L/min and $90 \div 6 = 15$ L/min. The average of these, $(10 + 15) \div 2 = 12.5$ , is not the average rate: the tap ran for more minutes at 15 L/min than at 10 L/min, so the average rate is nearer 15.

Check: at a constant 13 L/min, 10 minutes gives $13 \times 10 = 130$ L, the same total.

The average rate of flow is 13 L/min.

Example 11: Using an average rate

In one week (7 days) Siti used 10.5 GB of data on her phone. Her plan gives her 40 GB for a month of 30 days.
(a) Find her average rate of data use, in GB per day.
(b) If she keeps to this average rate, will 40 GB last the 30 days?

Solution

(a) Average rate $= 10.5 \div 7 = 1.5$ GB per day.

(b) In 30 days at 1.5 GB per day she would use $1.5 \times 30 = 45$ GB.

45 GB is more than 40 GB, so the data would run out.

It lasts $40 \div 1.5 = 26\frac{2}{3}$ days, so it runs out during the 27th day.

(a) Her average rate is 1.5 GB per day. (b) No: at this rate she would need 45 GB, and her 40 GB runs out during the 27th day.

Common Mistakes

Dividing the wrong way round. 450 km on 40 L: $40 \div 450$ gives litres per kilometre, not km/L. The quantity before "per" is the one you divide: km/L is km ÷ L, so $450 \div 40 = 11.25$ .
Leaving out the unit. "11.25" alone does not say whether it is km/L or L per 100 km. Every rate has a unit.
Comparing rates that are not in the same form. 15 km/L and 6 L per 100 km cannot be compared as they stand; nor can a price per kilogram and a price per 100 g. Change one so both are the same way round and in the same units.
Thinking a bigger rate is always better. More km/L is better, but fewer L per 100 km is better, and a lower price per kilogram is better for the buyer.
Taking the average of the rates. 10 L/min for 4 minutes and then 15 L/min for 6 minutes is an average rate of $130 \div 10 = 13$ L/min, not $(10 + 15) \div 2 = 12.5$ L/min. Always use total amount ÷ total time.
Writing hours and minutes as a decimal. 1 h 12 min is 72 minutes, or 1.2 hours. It is not 1.12 hours.
Using a rounded rate for a final answer. Keep the exact rate (such as $\frac{240}{7}$ pages per hour, not 34.3) in the working and round only the answer.

Practice

Work each question on paper before you open its solution. Write the unit of every rate, and check that you are dividing the right way round.

Question 1

Say whether each is a ratio or a rate, and write it in its simplest form (a ratio) or as an amount per one unit (a rate).
(a) 8 kg of flour to 6 kg of sugar
(b) $42 for 3 kg of prawns
(c) 56 L of water in 4 minutes
(d) 18 boys to 24 girls
(e) 15 000 people living on 2.5 km² of land

ShowHide solution

(a) Both are masses in kg: a ratio. $8 : 6 = 4 : 3$ , dividing both by 2.

(b) Money and mass: a rate. $42 \div 3 = 14$ , so $14 per kg.

(d) Both are numbers of children: a ratio. $18 : 24 = 3 : 4$ , dividing both by 6.

(e) People and area: a rate. $15\,000 \div 2.5 = 6000$ , so 6000 people per km².

(a) ratio 4 : 3 (b) rate, $14 per kg (c) rate, 14 L/min (d) ratio 3 : 4 (e) rate, 6000 people per km²

Question 2

A 750 g box of cereal costs $5.25. Find the cost (a) in cents per gram, (b) per 100 g.

ShowHide solution

(a) $5.25 = 525 cents. Cost per gram $= 525 \div 750 = 0.7$ cents.

(b) 750 g is $750 \div 100 = 7.5$ lots of 100 g, so 100 g costs $5.25 \div 7.5 = 0.70$ dollars.

Check with (a): $0.7 \times 100 = 70$ cents, which is $0.70.

(a) 0.7 cents per gram (b) $0.70 per 100 g

Question 3

A machine in a factory fills 5040 bottles in a week. It runs for 8 hours on each of the 7 days. Find its rate of filling (a) in bottles per day, (b) in bottles per hour, (c) in bottles per minute.

ShowHide solution

(a) $5040 \div 7 = 720$ bottles per day.

(b) Each day is 8 hours of running: $720 \div 8 = 90$ bottles per hour.

Check: $1.5 \times 60 \times 8 \times 7 = 5040$ .

(a) 720 bottles per day (b) 90 bottles per hour (c) 1.5 bottles per minute

Question 4

A photocopier copies 40 pages per minute.
(a) How many pages does it copy in 2 minutes 45 seconds?
(b) How long does it take to copy 1000 pages?

ShowHide solution

(a) 45 seconds is $\frac{45}{60} = \frac{3}{4}$ of a minute, so the time is $2\frac{3}{4}$ minutes.

Pages $= 40 \times 2\frac{3}{4} = 80 + 30 = 110$ .

(b) Time $= 1000 \div 40 = 25$ minutes.

(a) It copies 110 pages. (b) It takes 25 minutes.

Question 5

A pump fills an empty swimming pool with 51 000 L of water at a constant rate of 150 L/min. How long does it take? Give your answer in hours and minutes.

ShowHide solution

Time $= 51\,000 \div 150 = 340$ minutes.

$340 = 5 \times 60 + 40$ , so 340 minutes is 5 h 40 min.

Check: $150 \times 340 = 51\,000$ .

The pump takes 5 hours 40 minutes.

Question 6

Mr Tan's van used 18 L of diesel to travel 240 km. Find (a) how far it travels on each litre, in km/L, correct to 3 significant figures, (b) how much diesel it uses for each 100 km, in L per 100 km.

ShowHide solution

(a) $240 \div 18 = 13\frac{1}{3} = 13.333\ldots \approx 13.3$ km/L.

(b) 240 km → 18 L, so 1 km → $\frac{18}{240}$ L and 100 km → $\frac{18}{240} \times 100 = 7.5$ L.

(a) about 13.3 km/L (b) 7.5 L per 100 km

Question 7

A supermarket sells the same rice in three bags: 2 kg for $3.98, 5 kg for $9.60 and 10 kg for $18.60. Which bag is the best value?

ShowHide solution

Find the price of 1 kg in each bag.

2 kg bag: $3.98 \div 2 = 1.99$ , so $1.99 per kg.

5 kg bag: $9.60 \div 5 = 1.92$ , so $1.92 per kg.

10 kg bag: $18.60 \div 10 = 1.86$ , so $1.86 per kg.

The lowest price per kilogram is $1.86.

The 10 kg bag is the best value, at $1.86 per kg.

Question 8

Car P travels 14 km/L. Car Q uses 7.5 L of petrol per 100 km. Which car is more economical with petrol? Show your working.

ShowHide solution

Write Car Q's rate in km/L: 7.5 L → 100 km, so 1 L → $100 \div 7.5 = 13\frac{1}{3}$ km.

Car P goes 14 km on a litre and Car Q only $13\frac{1}{3}$ km, so Car P goes further on each litre.

The other way round: for 100 km Car P needs $100 \div 14 \approx 7.14$ L, less than Car Q's 7.5 L.

Car P is more economical with petrol.

Question 9

Ali packs 84 boxes in 3 hours 30 minutes. Ravi packs 105 boxes in 4 hours 10 minutes. Who packs at the faster rate?

ShowHide solution

Write each time in minutes: 3 h 30 min = 210 min and 4 h 10 min = 250 min.

Ali: $84 \div 210 = 0.4$ boxes per minute, which is $0.4 \times 60 = 24$ boxes per hour.

Ravi: $105 \div 250 = 0.42$ boxes per minute, which is $0.42 \times 60 = 25.2$ boxes per hour.

25.2 is more than 24.

Ravi packs at the faster rate: 25.2 boxes per hour against Ali's 24.

Question 10

Hui Min reads 45 pages of a book in 1 hour 30 minutes and then 75 pages in the next 2 hours.
(a) Find her average rate of reading, in pages per hour, correct to 3 significant figures.
(b) Ali says: "She read at 30 pages per hour and then 37.5 pages per hour, so her average rate is $(30 + 37.5) \div 2 = 33.75$ pages per hour." Explain why he is wrong.

ShowHide solution

(a) Total pages $= 45 + 75 = 120$ .

Total time $= 1\frac{1}{2} + 2 = 3\frac{1}{2}$ hours.

Average rate $= 120 \div 3\frac{1}{2} = 120 \times \frac{2}{7} = \frac{240}{7} = 34.28\ldots \approx 34.3$ pages per hour.

(b) The two rates are right ( $45 \div 1.5 = 30$ and $75 \div 2 = 37.5$ ), but she read for 2 hours at the faster rate and only $1\frac{1}{2}$ hours at the slower one. Averaging the two rates treats the two parts as if they took the same time.

The average rate must use the totals: 120 pages in $3\frac{1}{2}$ hours.

(a) Her average rate is about 34.3 pages per hour. (b) Ali averages the two rates, but she spent longer at the faster rate; the average rate is total pages ÷ total time.

Question 11

An empty 600 L tank was filled in 40 minutes.
(a) Find the average rate of filling.
(b) For the first 25 minutes the water flowed in at 12 L/min. At what constant rate did it flow in for the last 15 minutes?

ShowHide solution

(a) Average rate $= 600 \div 40 = 15$ L/min.

(b) In the first 25 minutes: $12 \times 25 = 300$ L.

Left for the last 15 minutes: $600 - 300 = 300$ L.

Rate $= 300 \div 15 = 20$ L/min.

Check: $300 + 20 \times 15 = 600$ L in $25 + 15 = 40$ minutes.

(a) The average rate is 15 L/min. (b) The water flowed in at 20 L/min for the last 15 minutes.

Key Points

A ratio compares quantities of the same kind and has no unit. A rate compares quantities of different kinds and always has a unit.
Rate = first quantity ÷ second quantity, read as "per one unit" of the second: L/min, $ per hour, words per minute, km/L, people per km².
For a rate that stays the same: amount = rate × time, and time = amount ÷ rate, with the time in the unit of the rate.
A rate can be written either way round (km/L or L per 100 km). For km/L more is better; for L per 100 km and a price per kilogram, less is better.
To compare, find each rate per one unit, the same way round and in the same units.
A rate is constant when the same amount is done in every interval of time of the same length, however short.
Average rate = total amount ÷ total time. Never take the average of the rates of the parts.