Lesson Video: Unit Rate | Nagwa Lesson Video: Unit Rate | Nagwa

Lesson Video: Unit Rate Mathematics • 6th Grade

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


Video Transcript

In this video, we will learn how to find a unit rate and how to use that unit rate to solve problems. But before we think about unit rates, we need to remember what ratios are.

We use ratios to compare two numbers or quantities. For this triangle, we might create a ratio of its length to its width, which would be seven to four. You can also write ratios as a fraction, seven over four. Both of these are ratios of length to width for the given rectangle. Rates are a bit different. Rates are ratios that compare two quantities of a different nature or two quantities that are expressed in different units.

For example, we can compare the distance of a run with how long that run took. If we consider that Sophia ran six kilometers in 40 minutes, we can represent that with this double-lined diagram, in pink, the time in minutes and, in yellow, the distance in kilometers. The rate here will be comparing the distance to time. And while you could write six kilometers to 40 minutes like this, rates are typically given in this quotient form, where we have six kilometers as the numerator and 40 minutes as the denominator. We have found a rate for Sophia. Looking back on our diagram, we see that the time and the distance are being vertically aligned.

If Sophia runs at a constant rate or speed, it means she runs the same distance for every given period of time. And that means when Sophia has been running for half the time, for 20 minutes, she will’ve gone half the distance, three kilometers. And in half of that time, in 10 minutes, she would’ve gone one and a half kilometers. She would’ve gone half the distance. But now, we don’t just want the rate. We also want to consider the unit rate. And that will be the distance that Sophia covers in one time unit, which here is one minute. To calculate the time in one minute, we would divide the total time, 40 minutes, by 40.

And because these rates are proportional, if we divide the time by 40, we need to divide the distance by 40. Six divided by 40 is 0.15. And so we found that Sophia ran 0.15 kilometers per minute. We’ve gone from the rate she was running to the unit rate. This unit rate is useful because now we can easily work out any distance she runs as long as we know the number of minutes she’s running. And we assume that she’s still running at the same speed. To calculate any new distance, we would simply take this unit rate and multiply it by the number of minutes she was running.

For instance, we could try to answer the question, how far could Sophia run in 30 minutes? She runs 0.15 kilometers every minute. If we multiply that by 30 minutes, we get 4.5 kilometers as the distance Sophia ran. 0.15 times 30 is 4.5. We also might answer a question like this: use the information to find Sophia’s speed in kilometers per hour. If Sophia ran 0.15 kilometers in one minute and we know that 60 minutes equals one hour, we can multiply our unit rate by 60 minutes per hour. We multiply the numerator 0.15 times 60. And we’re left with nine kilometers in our numerator and one hour in our denominator. Sophia is running at a speed of nine kilometers per hour.

We could describe a formal definition of the unit rate like this. The unit rate is the coefficient of proportionality between two quantities that are directly proportional. It is expressed in a compound unit and gives the first quantity per unit of the second quantity. With this, we’re ready to look at some examples.

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

The word “average speed” here tells us that we’re going to be looking at the unit rate that this sound wave is traveling. We know that it travels 895 meters every 2.5 seconds, which we can write as the fraction 895 meters over 2.5 seconds. To find this average speed or unit rate, we’ll need to find out how far the sound wave travels per second. We’ll need a proportional fraction to this one.

To go from 2.5 to one second, we’ll need to divide the denominator by 2.5. But to keep the values in the numerator and the denominator proportional, we’ll also divide the numerator by 2.5, which gives us 358 meters. If the sound wave travels 358 meters every one second, then we say the average speed is 358. And the units will be meters per second. We don’t include this denominator of one because 358 divided by one is just 358. But we do keep those unit rate compound units as meters per second, which makes our final answer an average speed of 358 meters per second.

Before we move on to another example, we need to note something about unit rate. We’ve been talking about unit rate as the coefficient of proportionality that links two quantities that are proportional. However, it depends on how we see the proportional relationship. There is not necessarily only one way to consider this proportional relationship. For example, in a situation where we have three pizzas and 12 people, one form of the unit rate could be the number of people per pizza.

Our initial ratio might look like this, 12 people over three pizzas. In this case, the unit would be one pizza. And there will be four people per pizza. And we have a unit rate of four, four people per pizza. But we also have another option for our unit rate. We could also have a unit rate of pizza per person. In this case, we start out with a fraction of three pizzas divided into 12 people. Our single unit would be one person. 12 divided by 12 is one, and three pizzas divided by 12 will be one-fourth a pizza. And the unit rate of pizza per person is one-fourth. We’ll write that like this: one-fourth a pizza per person.

The proportional relationship between the number of people and the number of pizzas could be interpreted in both directions. And most times, the context of our situation will determine which relationship is more helpful. In both cases, we have a compound unit and a coefficient of that compound unit.

Now, we’re ready to consider a few more examples.

William counted his strides while walking. He found that he walked at a constant speed as the number of strides were proportional to the walking time. The table shows how many strides he had taken at different times. What is the constant of proportionality between the walking time and the number of strides, i.e., his speed?

When we think about this constant of proportionality, specifically, the speed with which William walked, we should be considering the unit rate for which he was walking. And that means we’ll have a decision to make. Are we looking for the number of seconds per stride? Or are we looking for the number of strides per second? One simple way to think about speed is the measured distance traveled per unit of time. And so we can choose strides per second as the unit rate. We know that it takes four seconds to walk six strides for William. We can write that ratio six strides out of four seconds.

Since the unit rate needs to be one second, we divide four by one to get one. And if we divide by four in the denominator, we need to divide by four in the numerator. Six divided by four is one and a half. And so we can say that William walked one and a half strides in one second.

It might be worth checking a few of the other data points to make sure that this is the constant rate he was walking. From the table, we know that, in 30 seconds, William walked 45 strides. If we take the unit rate and multiply it by 30 seconds, we should get 45 strides. When we multiply the numerators, 1.5 times 30 does equal 45. We could also multiply 1.5 by 60 which would give us 90, 90 strides in 60 seconds. And 1.5 times 10 seconds also equals 15 strides. We found his speed, his rate, to be 1.5 strides for every one second. And we can write that constant of proportionality as the number 1.5 with the units strides per second.

Here’s another example.

A car uses 10 liters of fuel to travel 50 kilometers. What is the car’s fuel consumption rate in liters per kilometer?

We’re looking for a fuel-consumption rate in liters per kilometer, which is a unit rate for how many liters the car consumes per kilometer, where the unit is a kilometer. We’ve been given the rate, 10 liters, for every 50 kilometers. If we write that as fraction, we have 10 liters over 50 kilometers. Our unit rate will be how many liters it takes to go one kilometer. To go from 50 to one, we can divide the denominator by 50. And because we want to keep these values proportional, we’ll also have to divide the numerator by 50. 10 divided by 50 is 0.2. That’s gonna be two-tenths of a liter per kilometer. And when we write that out as a unit rate, we write the numerator 0.2. And then we list the units as liters per kilometer.

In this example, we’ll have to consider a unit rate where we need to change the units we’re operating in.

Given that a water tap leaks 7800 liters of water in five hours, find the leakage rate per minute.

This leakage rate per minute will be a unit rate for the leak in liters per minute. We can imagine the information we’re given with this double-line diagram. The first two data points tell us that 7800 liters leak out every five hours. Before we can find the leakage rate per minute, we’ll need to find out how much is leaking in one hour, instead of in five hours. To find that out, we can divide the five hours by five to get one hour. And because these two lines are proportional, these two values are proportional, we’ll also need to divide the 7800 by five, which will give us 1560 liters.

And so we can say that the leakage rate is 1560 liters per hour. But we haven’t found the leakage rate in liters per minute. Now, we know that one hour is equal to 60 minutes. And so we can say that the leakage rate is 1560 liters for every 60 minutes. But how will we find that one minute value? How will we know how many liters leak per minute? Well, to go from 60 minutes to one minute, we need to divide by 60. And that means to find out how many liters are leaking in one minute, we need to divide the 1560 liters in 60 minutes by 60. Since we’re dividing the numerator and the denominator by 60, this rate is staying proportional. And that proportion will be 26 liters every one minute. The leakage rate is 26 liters per minute. And written as a unit rate, we have the whole number, 26, and then the units liters per minute.

For reference, a kitchen tap has a flow rate of anywhere from four to six liters per minute. A garden hose might have a pressure of 90 liters per minute, while certain fire hoses can have flow rates upwards of 360 liters per minute. With our leakage rate of 26 liters per minute, we could may be guess that this is a garden hose that hasn’t been turned all the way off.

In our final example, we’ll consider what happens when both of the values in our rate are fractional.

Matthew 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? And how long does it take for him to sweep the whole floor?

We’ve been given this really helpful diagram. Our bottom line is the time and our top line is how much of the floor has been swept in that same amount of time. In one-fourth of an hour, two-ninths of the hall was swept. And that means every fourth of an hour, two-ninths of the floor can be swept. Two-ninths plus two-ninths will be four-ninths. In half an hour, four-ninths of the floor can be swept. And then by that measure, in three-fourths of an hour, six-ninths of the floor can be swept. And then, in one hour, eight-ninths of the floor can be swept.

Our first question asked, what fraction of the floor does he sweep in one hour? And we found that to be eight-ninths. The second question might not feel as straightforward. How long does it take for him to sweep the whole floor? If we continue out our line a little bit, there is some point where nine-ninths of the floor is swept. And that value is going to be a bit longer than an hour. Now, nine-ninths is one-ninth more than eight-ninths. So one thing that would be helpful for us to find out is how long does it take for Matthew to sweep one-ninth of the floor.

Because these values are proportional, we know that one-ninth is half of two-ninths, and that means it will take half of one-fourth to sweep one-ninth of the floor. Half of one-fourth is one-eighth. What we’re saying here is one-ninth of the floor is swept every one-eighth hour. If we wanted to get nine-ninths of the floor swept, we would multiply the numerator one-ninth by nine. And because these values are proportional, we’ll need to multiply the one-eighth of an hour by nine as well, which gives us nine-eighths of an hour for the whole floor swept. Nine-eighths of an hour is one and one-eighth. And so we can say it would take Matthew one and one-eighth of an hour to sweep the whole floor, which is 67 and a half minutes.

For a quick look at our key points, the unit rate is the coefficient of proportionality between two quantities that are directly proportional. Unit rates have compound units and gives the amount of the first quantity per unit of the other quantity. And finally, a proportional relationship can often be understood in both ways, for example, four people per pizza or one-fourth a pizza per person. And we let the context of the situation determine which ways we’re using this proportional relationship.

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