What is a rate?
A rate is a measure of how quickly one measurement changes with respect to another. Some commonly used rates in our everyday lives are speed, which measures distance per time, and the price of food, which is often measured in dollars per kilogram.
When we combine two different units into a single compound unit we call this a unit rate. We can write these compound units using a slash (/) between the different units, so "metres per second" becomes "m/s" and "dollars per kilogram" becomes "$/kg". This compound unit represents the division of one measurement by another to get a rate.
Let's have a look at an example.
Exploration
Consider an Olympic sprinter who runs $100$100 metres in $10$10 seconds. How fast does he run?
We can find how far the sprinter runs in $1$1 second by dividing the $100$100 metres evenly between the $10$10 seconds.
This calculation tells us that the sprinter runs $10$10 metres in one second.
We can write this a rate for the sprinter's speed in metres per second using the compound unit m/s to give us:
Sprinter's speed $=$= $10$10 m/s
Now let's try a more direct method to finding the sprinter's speed.
Since the sprinter runs $100$100 metres in $10$10 seconds we can say that he runs at a rate of $100$100 metres per $10$10 seconds. Writing this as a fraction gives us:
Sprinter's speed $=$= $100$100m/$10$10s $=$= $\frac{100}{10}$10010 m/s $=$= $10$10 m/s
After some simplifying we find that the speed of the sprinter matches that from the previous method. Notice that we were able to separate the numbers and the units into separate fractions, this is the core concept we use for turning fractions of measurements into rates.
Not all compound units are written using a slash and instead use the letter "p" to represent "per".
For example, "beats per minute" uses the compound unit bpm and "frames per second" uses fps.
Simplifying rates
Rates have two components, the numeric value and the compound unit. The compound unit tells us which units are being measured and the numeric value tells us how quickly the numerator unit changes with respect to the denominator unit.
When constructing a rate we usually start with just a fraction of measurements.
For example, let's find the speed of a car that travels $180$180 kilometres in $3$3 hours.
We start by setting up the fraction as distance per time, written:
Speed $=$= $180$180km/$3$3hr
Then we can separate the fraction into its numeric value and its compound unit. This gives us:
| Numeric value | $=$= | $\frac{180}{3}$1803 | $=$= | $60$60 |
|---|---|---|---|---|
| Compound unit | $=$= | km/hr |
We can then combine them again to get the rate which is:
Speed $=$= $60$60 km/hr
Whenever we can, simplify the fraction to get a whole number value for the rate. This is much nicer to work with as we can now say that the car travels $60$60 kilometres in $1$1 hour, rather than $180$180 kilometres in $3$3 hours.
Practice question
Question 1
Write the following as a unit rate:
$91$91 people per $7$7 buses
Converting units
When applying rates it's important to make sure that we are applying the right one.
Consider the car from before that travelled at a speed of $60$60 km/hr. How many kilometres will the car travel in $7$7 minutes?
Before we use one of the methods we learned for applying rates we should first notice that the units in the question don't quite match up with our rate. Specifically, the question is asking for minutes as the units for time instead of hours.
We can fix this by converting hours into minutes for our rate. Using the fact that $1$1 hour $=$= $60$60 minutes we can convert our speed from km/hr to km/min like so:
Speed $=$= $60$60km/hr $=$= $60$60km/$60$60min $=$= $\frac{60}{60}$6060 km/min $=$= $1$1 km/min
Now that we have a speed with the appropriate units we can apply the rate to the question to find how far the car will travel in $7$7 minutes:
Distance $=$= $7$7min $\times$× $1$1 km/min $=$= $\left(7\times1\right)$(7×1) km$\times$×min/min $=$= $7$7km
Now that we have some experience with this type of question you can try one yourself.
$\text{1 metre}=\text{100 centimetres}$1 metre=100 centimetres
$\text{1 metre}=\text{1000 millimetres}$1 metre=1000 millimetres
$\text{1 kilometre}=\text{1000 metres}$1 kilometre=1000 metres
$\text{1 litre}=\text{1000 millilitres}$1 litre=1000 millilitres
$\text{1 hour}=\text{60 minutes}$1 hour=60 minutes
$\text{1 minute}=\text{60 seconds}$1 minute=60 seconds
Practice question
Question 2
Consider the following rate:
$192$192 metres per $240$240 seconds
Express this is as a unit rate in terms of metres and seconds.
Express this is as a unit rate in terms of metres and minutes.
Negative rates
If a rate represents a loss or a decrease, the rate should be negative.
So if a person earns $\$180$$180 each day in their job, we would write this is a positive rate to represent an earning: $\$180$$180/day.
But if they have to pay $\$6$$6 in tolls each day to get to and from work, we would represent this as a negative rate to represent a loss: $-\$6$−$6/day.
Worked Example
example 1
Sasha earns $\$90$$90 per shift at her job as a waitress, and receives on average $\$14$$14 in tips per shift. However she must also pay $\$4$$4 in tolls to get to her work and $\$4$$4 to get back each shift.
a) Write her total earnings as a positive rate, in dollars per shift.
Think: We need to add her earnings to find the rate.
Do: $90+14=104$90+14=104. So she earns$\$104$$104/shift.
b) Write her expenses as a negative rate, in dollars per shift.
Think: Her expenses are just the cost of the tolls, but she pays it twice.
Do: $2\times4=8$2×4=8. So her expenses are given by the negative rate $-\$8$−$8/shift.
c) If Sasha works $5$5 shifts this week, how much money will she actually have kept?
Think: For each shift, she earns $\$104$$104 but loses $\$8$$8. So we need to find the difference of these amounts and multiply that by $5$5 for $5$5 shifts.
Do: $104-8=96$104−8=96, and $5\times96=480$5×96=480. So she has $\$480$$480.