3.05 Unit rates

What is unit 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 pound.

Notice how both of these examples combines two different units into a single compound unit. We can write these compound units using a slash ( / ) between the different units, so "meters per second" becomes "m/s" and "dollars per pound" becomes "$/lb". This compound unit represents the division of one measurement by another to get a rate.  When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates. 

Let's have a look at an example.

 

Exploration

Consider an Olympic sprinter who runs $100$100 meters 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 meters evenly between the $10$10 seconds.
This calculation tells us that the sprinter runs $10$10 meters in one second.
We can write this as a unit rate for the sprinter's speed in meters 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 meters in $10$10 seconds we can say that he runs at a rate of $100$100 meters 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.

 

 

Finding unit 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 miles in $3$3 hours.

We start by setting up the fraction as distance per time, written:

Speed $=$= $180$180mi/$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	$=$=	mi/hr

 

We can then combine them again to get the unit rate which is:

Speed $=$= $60$60 mi/hr

Whenever we can, simplify the fraction to get the unit rate. This is much nicer to work with as we can now say that the car travels $60$60 miles per $1$1 hour, rather than $180$180 miles per $3$3 hours.

 

 

Applying unit rate

Now that we know how to make unit rates, it's time to use them. Rates are very similar to ratios in that we can use them to calculate how much one measurement changes based on the change in another.

 

Worked examples

question 1

Returning to our sprinter, we found that they could run at a speed of $10$10 m/s. Assuming that they can maintain this speed, how far will the sprinter run in $15$15 seconds?

Think:  One way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by $15$15. This will give us:

Do:  Speed $=$= $10$10 m/s $\times$×$\frac{15}{15}$1515​ $=$= $150$150m/$15$15s

By turning the rate back into a fraction we can see that the sprinter will run $150$150 meters in $15$15 seconds.

Reflect:  Another way to solve this problem is to apply the rate directly to the question. We can do this by multiplying the time by the rate. This gives us:

Distance $=$= $15$15s $\times$× $10$10m/s $=$= $\left(15\times10\right)$(15×10) m$\times$×s/s $=$= $150$150m

Notice that the units for seconds from the time canceled with the units for second in the compound unit leaving only meters as the unit for distance.

 

question 2

We can also ask the similar question, how long will it take for the sprinter to run $220$220 meters?

Think:  Again, one way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by $22$22. This gives us:

Do:  Speed $=$= $10$10 m/s $\times$×$\frac{22}{22}$2222​ $=$= $220$220m/$22$22s

By turning the rate back into a fraction we can see that the sprinter will take $22$22 seconds to run $220$220 meters.

Reflect:  Another way to solve this problem is to apply the rate directly to the question. We can do this by dividing the distance by the rate. This gives us:

Time $=$= $220$220m $\div$÷​ $10$10 m/s $=$= $\frac{220}{10}$22010​ s$\times$×m/m = $22$22s

This time we divided by the rate so that the compound unit would be flipped and the meters units would cancel out to leave only seconds as the unit for time.

 

 

 

Converting units

When applying rates it's important to make sure that we are applying the right one.

Worked example

question 3

Consider the car from before that traveled at a speed of $60$60 mi/hr.
How many miles will the car travel in $7$7 minutes?

Think:  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 unit 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 unit rate. Using the fact that $1$1 hour $=$= $60$60 minutes we can convert our speed from mi/hr to mi/min like so:

Do:   Speed $=$= $60$60mi/hr $=$= $60$60mi/$60$60min $=$= $\frac{60}{60}$6060​ mi/min $=$= $1$1 mi/min

Now that we have a speed with the appropriate units we can apply the unit rate to the question to find how far the car will travel in $7$7 minutes:

Distance $=$= $7$7min $\times$× $1$1 mi/min $=$= $\left(7\times1\right)$(7×1) mi$\times$×min/min $=$= $7$7mi

Now that we have some experience with this type of question you can try one yourself.

 

Practice question

question 4

question 5

question 6