3.04 Rates in context

Let's review how to identify and convert rates, as well as using rates to make useful comparisons. Recall the following definition and distinction between a rate and a ratio:

• A ratio is a comparison of the relative size of two or more quantities. For example, if a bag of marbles contains $3$3 blue marbles and $2$2 red marbles, we can describe the comparison between the number of red and blue marbles as a ratio like red:blue  = $2:3$2:3.
• A rate is a ratio between two quantities that are measured in different units. For example, the rate a tap leaks may be $30$30 mL every $5$5 minutes. Rates are often expressed as unitary rates (or a unit rate) where the second quantity in the rate has a measure of $1$1. The unitary rate for the leaking tap would be $6$6 mL for every $1$1 minute. To abbreviate this we drop the $1$1 and use the word per or the notation '/' to separate the two units, thus we could write $6$6 mL/min. This compound unit, mL/min, represents the division of one measurement by another to obtain the rate.

Common examples of rates include:

• Speeds, for example $60$60 km/h, $20$20 m/s, $400$400 m every $5$5 minutes
• Growth rates, for example $20$20 cm/year, $15$15 mm/month, $2$2 kg per week
• Cost of groceries, for example $\$5$$5 per kg, $\$1.20$$1.20 per $100$100 g, $\$10$$10 per box

 

Simplifying and converting rates

A rate is usually considered simplified when it is represented as a unit rate. Remember that a unit rate is a rate where the second quantity is just $1$1 of the unit prescribed. To calculate a rate, divide one quantity by another. This will give us two components, the numeric value and the compound unit. The numeric component can often be further simplified just as we simplify fractions.

Converting rates allows us to compare rates given in different units and to also obtain a rate in units suitable for a particular application. Common applications are comparing unit prices to find the best deal and comparing speeds of different objects.

 

Worked example

Example 1

Holly runs a $42$42 kilometre marathon in $3$3 hours and $30$30 minutes.

a) Find Holly's simplified running rate (speed) in kilometres per hour.

Think: To obtain a rate in km/h, we need to divide the distance in kilometres by the time in hours. ($3$3 hours and $30$30 minutes$=3.5$=3.5 hours)

Do:

$\text{Speed}$Speed	$=$=	$\frac{42\text{ km}}{3.5\text{ h}}$42 km3.5 h​	Divide distance in kilometres by time in hours
	$=$=	$12\text{ km/h}$12 km/h	Simplify the fraction and don't forget units

 

b) Convert her speed to metres per second.

Think: Both the distance unit and time unit are being converted. Let's first change the distance unit. In this case, we are converting distance from kilometres to metres. The number of metres travelled in a given amount of time is $1000$1000 times greater than the number of kilometres, so we want to multiply the rate by $1000$1000. 

Then we want to convert the new rate in m/h to m/s. There are $60$60 seconds in a minute, and then $60$60 minutes in an hour, so the number of metres travelled in a second will be the rate in m/h divided by $60^2$602.

Do:

Speed	$=$=	$12$12 km/h	Write the given rate including units
	$=$=	$12\times1000$12×1000 m/h	Convert the kilometres to metres
	$=$=	$12000$12000 m/h	Simplify the rate
	$=$=	$\frac{12000}{60^2}$12000602​ m/s	Convert the rate from per hours to per second
	$=$=	$3\frac{1}{3}$313​ m/s	Simplify the rate

 

Sometimes a rate can be measured in several different units. For example, speed could be measured in km/h, m/s, mm/s and the units we use will depend on what is reasonable for the speed we are measuring.

A car travelling down the freeway is going to cover many kilometres each hour, so the reasonable units to use would be km/h.

A $100$100-metre sprinter will cover many metres over a smaller period of time, so m/s is a more reasonable unit to describe the speed. (Usain Bolt's $100$100 m world record was achieved at a speed of $12.2$12.2 m/s)

A snail travels a very small distance over a small period of time, so mm/s is a reasonable unit to use. (Speed of a common snail is $1$1 mm/s).

But if we want to compare the speed of a car travelling $40$40 km/h and a sprinter who runs at $12.2$12.2 m/s, we would need to convert one of these rates to the units of the other.

Converting rates allows us to compare rates given in different units and to also obtain a rate in unit suitable for a particular application.

To convert a rate, we want to multiply or divide the rate by the appropriate constant. So to convert a rate in m/min to m/s, we want to divide the rate by $60$60. This is because a rate given in m/min, tells us the number of metres per minute, which is the number of metres per $60$60 seconds.

 

Worked examples

Example 2

Convert the rate of $20$20 L/h to a rate in L/min.

Think: We first need to identify which unit we are changing. Is it the first unit, the second unit or are we converting both?  In this case, we are only changing time from hours to minutes.  We also need to identify what the conversion factor between these units is. In our case, $1$1 hour is equivalent to $60$60 minutes.

Do: The number of litres in an hour must be $60$60 times more than the number of litres in a minute. So we want to divide the given rate by $60$60.

Rate	$=$=	$20$20 L/h	Write the rate as a division including units.
	$=$=	$\frac{20}{60}$2060​ L/min	Convert hours to minutes by dividing by $60$60.
	$=$=	$\frac{1}{3}$13​ L/min	Simplify the fraction if possible.

Example 3

How fast is Usain Bolt's world record speed of $12.2$12.2 m/s in km/h?

Think: Both the distance unit and time unit are being converted. Let's first change the distance unit. In this case, we are changing distance from metres to kilometres. The number of metres travelled in a given second is $1000$1000 times greater than the number of kilometres, so we want to divide the rate by $1000$1000. 

Then we want to convert the new rate in km/s to km/h. There are $60$60 seconds in a minute, and then $60$60 minutes in an hour, so the number of kilometres travelled in an hour will be the rate in km/s multiplied by $60^2$602.

Do:

Rate	$=$=	$12.2$12.2 m/s	Write the rate as a fraction including units.
	$=$=	$12.2/1000$12.2/1000 km/s	Convert metres to kilometres by dividing by $1000$1000.
	$=$=	$0.0122$0.0122 km/s	Simplify the rate.
	$=$=	$0.0122\times60^2$0.0122×602 km/h	Convert seconds to hours by multiplying by $60^2$602
	$=$=	$43.92$43.92 km/h	Simplify the rate.

Reflect: Alternatively we can convert km/s to km/h by converting in stages. First we can find the rate in km/m, by multiplying by $60$60 and then we can find the rate in km/h by multiplying by $60$60 again.

 

 

Practice questions

Question 1

Question 2

 

Rates in context

Best buys

Everyone loves a great deal or a sale when they're shopping! However, most products today come in different varieties or quantities and are sold at more than one store. To determine which item is better value, we need to find a common amount to compare. Often it's easiest to find the unit price of each item, where we find the cost per unit of measurement, for example per litre, per kilogram or per item. We can also compare the amount per dollar, where we work out how much of something we would get for one dollar.

 

Amount per unit

Are we better off paying $\$10.50$$10.50 for $3$3 kg of apples or $\$6.20$$6.20 for $2$2 kg of apples? An easy way to compare the two options is to find the price per kilogram for each option.

So we're better off paying $\$6.20$$6.20 for $2$2 kg of apples because it's a cheaper price per kilogram.

 

Amount per dollar

If David's deli sells $620$620 grams of salmon for $\$18$$18 and Fred's fish market sells $460$460 grams of salmon for $\$13$$13, and if the fish is of similar quality, which is better value? Let's see how much salmon we would get for one dollar at each shop.

David's:

Amount per dollar	$=$=	$\frac{620}{18}$62018​ grams per dollar	Expressing the rate as the amount of grams per dollar
	$=$=	$34.\overline{4}$34.4 grams per dollar	Expressing the rate as a decimal

So at David's, we'd get approximately $34.4$34.4 grams of salmon for a dollar.

Fred's:

Amount per dollar	$=$=	$\frac{460}{13}$46013​ grams per dollar	Expressing the rate as the amount of grams per dollar
	$\approx$≈	$35.4$35.4 grams per dollar	Expressing the rate as a decimal

We get approximately $35.4$35.4 grams of salmon at Fred's, which is slightly better value than at David's.

 

Worked example

example 4

At a growers' market, durians are sold at stand A for $80c$80c per kilogram. At stand B, each $4.5$4.5 kg durian is sold for $\$4.00$$4.00.

a) Calculate the price of $4.5$4.5 kg of durian from store A. Round your answer to two decimal places.

Think: How do we change the price per kilogram to work out the price for $4.5$4.5 kg?

Do: $4.5\times0.8=\$3.60$4.5×0.8=$3.60

b) Which store is the best buy?

Think: In other words, at which store would you pay less for $4.5$4.5 kg of durian?

Do: Choose Store A, as you pay less for a $4.5$4.5 kg in this store.

c) What should the owner of store C charge for his $4.5$4.5 kg durians if he wants to beat the best buy by $6%$6%? Round your answer to two decimal places.

Think: This means he wants to charge $6%$6% less that the best buy or  $94%$94% of the lowest price.

Do:

$0.94\times3.60$0.94×3.60	$=$=	$3.384$3.384	Finding $94%$94% of the price of $4.5$4.5 kg of durian
	$=$=	$\$3.38$$3.38 (to $2$2 d.p.)

 

So the owner of store C should charge $\$3.38$$3.38 for his $4.5$4.5 kg durians.

 

Practice questions

Question 3

question 4

question 5