Lesson Explainer: Unit Rate Mathematics • 6th Grade

In this explainer, we will learn how to find a unit rate and how to use it to solve problems in the real world.

Ratios are used to compare two numbers or quantities. Rates are ratios that compare two quantities of different nature (expressed in different units). For instance, we can compare the distance run with the running time; let us say that Dalia ran the other day 6 km in 40 minutes. The rate is then the ratio distance to time, expressed as a quotient: 640kmmin.

We can represent this rate on a double-line diagram.

In the diagram, 40 minutes and 6 kilometres are associated by being aligned vertically. If Dalia runs at a constant rate (or speed), it means that she runs the same distance for every given period of time: the distance she runs is directly proportional to the time she runs. So, for instance, in 20 minutes, she runs half of 6 km, that is, 3 km. In 10 minutes (a quarter of 40 minutes), she runs a quarter of 6 km, that is, 1.5 km. For each pair of values, the time and distance are then vertically aligned on the double-line diagram.

It is very useful to find the unit rate of the proportional relationship between the distance run and the running time, that is, the distance she covers in one time unit, which here is one minute.

We can find it on the double-line diagram, by working out what 6 divided by 40 is, that is, 640. We find that she runs 0.15 km in one minute.

It is very useful to work out the unit rate because now we can easily work out any distance she runs for a given number of minutes (assuming she still runs at a steady speed) by multiplying the unit rate by this number of minutes.

For instance, we often measure speeds in kilometres per hour (km/h). As one hour is 60 minutes, we need to find the distance she covers in 60 minutes to find her speed in kilometres per hour (it is the distance run in one hour). We multiply the unit rate (the distance she covers in one minute) by 60 minutes: we find 9 km. Her speed is thus 9 km/h.

Definition: Unit Rate

The unit rate is the coefficient of proportionality between two quantities that are directly proportional. It is expressed in a compound unit (e.g., kilometres per hour) and gives the amount of the first quantity per (which means “for one”) unit of the second quantity.

Let us look at the first example to check our understanding.

Example 1: Identifying a Unit Rate from a Description

What is the average speed of a sound wave that travels 895 metres in 2.5 seconds?

Answer

The speed of sound depends on the substance in which it is traveling. Here, we know that, in 2.5 seconds, the sound wave traveled 895 metres. We want to find the unit rate (here the speed), that is, how many meters the sound wave traveled in one second.

This can be represented on a double-line diagram.

The distance traveled in one second is given by 8952.5=358.m

Hence, the speed is 358 m/s.

In the next examples, we are going to look at proportional relationships given in different forms, such as tables and graphs and, more importantly, we are going to focus on the units used for both quantities that are proportional and the unit of the constant of proportionality, or unit rate.

As we have seen, the unit rate is the coefficient of proportionality linking two quantities that are proportional. However, it depends on how we see the proportional relationship, that is, which of both quantities depends on the other. There is not necessarily only one way to consider a proportional relationship. For instance, let us imagine that a pizza serves four people. The number of people served per pizza (the unit rate) links the number of people served and a given number of pizzas.

But we can also consider this relationship the other way around. Indeed, we can work out the number of pizzas necessary to serve a given number of people.

The proportional relationship between the number of people and the number of pizzas can be interpreted in both directions. Often, the context will determine in which way we are considering the relationship between the quantities. Mathematically, it is just a matter of rearranging equations (but this is beyond the scope of the present lesson). In both cases, we observe that the constant of proportionality, or unit rate, is expressed with a compound unit: it is a quantity per unit of the other quantity.

We notice as well that both unit rates found above are multiplicative inverses of each other: 4×=1.peopleservedperpizzaaquarterofapizzaperperson

Let us now find a unit rate from the values of two proportional quantities given in a table.

Example 2: Identifying a Unit Rate from a Table

Bassem counted his strides while walking. He found that he walked at a constant speed, as the number of strides was proportional to the walking time. The table shows how many strides he had taken at different times.

Walking Time (s) 4 10 30 60
Number of Strides6154590

What is the constant of proportionality between the number of strides and the walking time (i.e., his speed)?

Answer

The constant of proportionality, or unit rate, which is the speed here, is the number of strides per unit of time (here seconds).

To find the constant of proportionality, we divide the number of strides given in the table by the corresponding time in seconds. With 15 strides in 10 seconds, for instance, we find that the speed is 1510=1.5.stridessecondsstridespersecond

In the next example, we have the description of a proportional relationship.

Example 3: Identifying a Unit Rate from a Description

A car uses 10 litres of fuel to travel 50 km. What is the car’s fuel consumption rate in litres per kilometre?

Answer

We can find the car’s average fuel consumption rate using the pair of values we are given, namely, 10 litres for a distance of 50 km. Be careful that fuel consumption can be expressed in two different ways, either as the amount in liters of fuel to travel 1 km (like here) or 100 km or as the distance that can be traveled with a certain volume of fuel.

Here, we want to find the number of litres used to travel 1 km. As 1 km is 50 times smaller than 50 km, the quantity of fuel used to travel 1 km is 50 times smaller than 10 litres. We see that the fuel consumption rate is given by 1050=0.2/.LkmLkm

[The units have been written in the calculation for learning purposes. You may not have to write them in your workings.]

Different units are usually proportional (the famous exception is temperature units). Let us see with the next example how to use what we know about proportional relationships to convert between units.

Example 4: Finding a Unit Rate and Converting between Units

Given that a water tap leaks 7‎ ‎800 litres of water in 5 hours, find the leakage rate per minute.

Answer

In this question, we have two quantities: the amount of water that is leaking, in liters, during a given time, in hours. We want to find the leakage rate per minute.

The leakage rate per hour is found by dividing the volume of water leaked in five hours by five: 78005=1560/.LhLh

Because one hour is 60 minutes, the volume of water that leaked in one minute is then one-sixtieth of 1‎ ‎560 L/h; that is, 1560/60/=26/.LhminhLmin

Note that the units have been written in the above calculation to highlight the fact that units need to be consistent. The whole calculation could have been done in one step. See how multiplying five hours by 60 minutes per hour here gives the number of minutes in five hours: 78005×60=26/.LhLminminh

You may not be supposed to write the units in your workings, but, in any case, you need to write your final result with its unit, as otherwise it has no meaning.

This last example involves fractions and will allow us to explore, reasoning with diagrams, the meaning of both unit rates when considering the given proportional relationship in both directions.

Example 5: Finding a Unit Rate from a Pair of Values Involving Fractions

Shady sweeps two-ninths of the hall floor in 15 minutes, or one-quarter of an hour.

  • What fraction of the floor does he sweep in one hour?
  • How long does it take for him to sweep the whole floor?

Answer

We know that Shady sweeps two-ninths of the hall in one-quarter of an hour. To find what fraction of the floor he sweeps in one hour, we need to multiply two-ninths by four, as the diagram suggests.

Hence, we find that 29×4=89.oftheoorperhour

Note that what we have done is actually dividing by one-quarter. Writing the units as well, we have oftheoorhoftheoorhoftheoorh=29×41=89/.

In the second part, we are asked to find the unit rate taken in the other direction, that is, the time it takes him to sweep the whole floor. For this, we can use the fact that he sweeps two-ninths of the floor in one-quarter of hour.

As the double-line diagram suggests, we can split the two-ninths in two shares, which gives one-ninth, and then multiply by nine to get one (which is the whole floor). This is equivalent to multiplying by 92.

Hence, we find that the time it takes him to sweep the whole floor is 14×92=98.h

Therefore, the unit rate here is 98 h for the whole floor.

Again, our calculation is equivalent to dividing the time (in hours) by the fraction of floor swept in that time, which gives the number of hours per unit of floor, which here is the whole floor: hoftheoorhoftheoorhoor=14×92=98/.

Note that both unit rates are multiplicative inverses of each other: 89×98=1.oorhhoor

Key Points

  • The unit rate is the coefficient of proportionality between two quantities that are directly proportional. It is expressed in a compound unit (e.g., kilometres per hour) and gives the amount of the first quantity per unit of the other quantity.
  • A double-line diagram can help us visualize how to find a unit rate when we know a pair of values of two directly proportional quantities.
    In the example shown below, the unit rate is 0.15 km/min.
  • Using the units when working out the ratio of one quantity to the other also shows what unit rate we are calculating. In the previous example, we have 640=0.15/.kmminkmmin Kilometers divided by minutes give a unit rate in kilometres per minute.
  • A proportional relationship can often be understood in both ways; for instance,

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