Lesson Video: Comparing Ratios Mathematics • 6th Grade

In this video, we will learn how to compare ratios.

17:48

Video Transcript

In this video, we will look at comparing ratios using unit rates in real-life problems. We will begin by recapping how we can simplify ratios. If we consider the ratios four to 10 and six to 18, we can simplify them by looking for common factors. Four and 10 have a highest common factor of two, so we can divide both sides of the ratio by two. This tells us that the ratio four to 10 in its simplest form is two to five.

We can repeat this process for the ratio six to 18. The highest common factor this time is six. As six divided by six is equal to one and 18 divided by six is equal to three, the ratio in its simplest form is one to three. This is all well and good if we want a ratio in its simplest form. But what if we wanted to compare the two ratios? How can we compare two to five and one to three? In order to compare two or more ratios, we need to write them in the form one to 𝑛. This is known as the unit ratio or unit rate.

In terms of our example, the second ratio, one to three, is already written in this form. This means that, for every one unit of the first part, we get three units of the second part. In order to write the ratio two to five as a unit ratio, we need to divide both sides by two. Two divided by two is equal to one. Five divided by two can be written as the fraction five-halves or five over two. When comparing ratios, it is useful to turn this fraction into a decimal. The ratio four to 10 or two to five written as a unit ratio is one to 2.5. For every one unit of the first part, we get two and a half or 2.5 units of the second part. We’re now in a position to compare the ratios as required in the question.

An alternative method to compare ratios is to consider the first part as a fraction of the whole. In our ratio four to 10, the first part four is four out of 14 parts altogether. In the same way, the first part of the second ratio, six to 18, is six parts out of 24 in total. Both of these fractions can be simplified by dividing the numerator and denominator by two and six, respectively.

We still have a problem when trying to compare two-sevenths and one-quarter. The easiest way to do so would be to find the lowest common multiple of four and seven. This is 28. Two-sevenths is equivalent to eight twenty-eighths, whereas one-quarter is equivalent to seven twenty-eighths. As the denominators are now equal, we can compare the fractions by looking at the numerators. Whilst the second method is sometimes useful, for the majority of this video, we will compare ratios using unit ratios or unit rates.

Scarlett uses two tablespoons of sugar for every three glasses of lemonade, while Natalie uses three tablespoons of sugar for every six glasses of lemonade. Who makes the sweeter lemonade?

We can begin this question by writing down a ratio of sugar to lemonade for both of the girls. Scarlett used two tablespoons for every three glasses of lemonade. Therefore, her ratio is two to three. Natalie used three tablespoons of sugar for every six glasses of lemonade. So, her ratio is three to six. One way of comparing two or more ratios is to write them in the form one to 𝑛. This is known as the unit ratio. When finding an equivalent ratio, we must divide or multiply both sides by the same value. Two divided by two is equal to one. And three divided by two is equal to 1.5. This means that, for every one tablespoon of sugar that Scarlett uses, she will fill 1.5 glasses of lemonade.

Repeating this process for Natalie, we divide both sides of her ratio by three. The ratio three to six simplifies to one to two. This means that, for every one tablespoon of sugar that Natalie uses, she is able to fill two glasses of lemonade. We’re asked to work out who makes the sweeter lemonade. This will be the person who has less glasses per tablespoon of sugar. As Scarlett is only able to make one and a half glasses of lemonade for one tablespoon of sugar, she makes the sweeter lemonade.

Had we noticed that Natalie had double the number of glasses originally than Scarlett, we could’ve used an alternative method for this question. Multiplying both sides of Scarlett’s ratio by two gives us a new ratio, four to six. As Scarlett used four tablespoons of sugar for six glasses of lemonade whereas Natalie only used three tablespoons, once again, we have proven that Scarlett made the sweeter lemonade.

Michael wants to sign up for his school’s sports competition. In order to be accepted, he has to be able to run 400 meters in one minute. Michael took 20 seconds to run 100 meters. If he could run at the same rate, would he qualify to take part?

We’re told in the question that Michael needs to be able to run 400 meters in one minute. If we write this as a ratio of time in minutes to distance in meters, this would be one to 400. We’re also told that Michael can run 100 meters in 20 seconds. There are 60 seconds in one minute. And 20 multiplied by three is equal to 60. Multiplying 100 by three gives us 300. So, if Michael runs at the same rate, he will cover 300 meters in 60 seconds. This ratio of time in minutes to distance in meters is one to 300.

As Michael needed to cover a distance of 400 meters in one minute, the correct answer is no. He would not qualify to take part. Writing any ratio in this form, one to 𝑛, is known as the unit ratio or unit rate. For every one unit, or minute in this case, of time, we can see the distance in meters that Michael covered.

The next question we look at, we’ll look at unit rates to compare three different ratios.

Mason, Liam, and James are biking. Mason can bike two miles in 20 minutes, Liam can bike three miles in 25 minutes, and James can bike six miles in 66 minutes. Who cycles at the fastest rate?

In order to compare the three speeds, we will write the ratio of distance in miles to time in minutes. For Mason, this is a ratio of two to 20. For Liam, the ratio is three to 25. And finally, for James, the ratio is six to 66. In order to compare the three ratios, we need to calculate the unit rate or unit ratio. This is written in the form one to 𝑛. In this question, this will calculate the time it would take each boy to cycle a distance of one mile. When simplifying or finding equivalent ratios, we need to multiply or divide both sides by the same number. For mason, we need to divide both sides by two. This means that the ratio two to 20 is equivalent to one to 10. It takes Mason 10 minutes to bike one mile.

To make the left-hand side of Liam’s ratio equal to one, we need to divide both sides by three. 25 divided by three is equal to eight and one-third or 8.3 recurring, written with a dot or bar above the three. It takes Liam eight and a third minutes to cycle one mile. Dividing both sides of James’s ratio by six gives us the new ratio one to 11. It takes James 11 minutes to cycle one mile. As all three ratios are now written in terms of their unit rate, we can compare them. The person who cycles at the fastest rate will be the person who takes the least amount of time to cycle one mile. In this question, this is Liam.

The next question involves comparing ratios using tables.

Use the table to determine which runners ran at the same rate.

In order to answer this question, we will look at the ratio of each runner of their time in hours to distance in miles. For Liam, this is two to 10. He jogged 10 miles in two hours. James’s ratio was three to 18 as he ran 18 miles in three hours. The corresponding ratios for David and Michael were four to 20 and three to 12, respectively. In order to compare two or more ratios, we need to write them in the form one to 𝑛. This is the unit rate or unit ratio. In this question, it will be the distance that each runner jogged in one hour.

For Liam, we will divide both sides of the ratio by two. This means that Liam ran a distance of five miles per hour. We divide the two parts of James’s ratio by three. This gives us a ratio of one to six. So, James ran six miles in one hour. Repeating this for David and Michael tells us that David ran five miles per hour and Michael ran four miles per hour. We’ve been asked to identify which runners ran at the same rate. As both Liam and David had the same unit ratio of one to five, we can conclude that they ran at the same rate.

Our final question will involve using a calculator to calculate density to compare populations.

Daniel and Charlotte both have a keen interest in gardening and are concerned by the number of slugs that they keep finding in their respective vegetable patches. They want to compare the number of slugs in each of their gardens. But due to the differences in size of their vegetable patches, they decide to compare the number of slugs per square foot. Daniel’s vegetable patch is a rectangle with dimensions five foot by three foot. And Charlotte’s is a circular patch with a radius of three foot. One Saturday morning, Daniel counts 21 slugs in his entire vegetable patch and Charlotte counts 36. There are three parts to this question. Work out the density of slugs in Daniel’s vegetable patch. Work out the density of slugs in Charlotte’s vegetable patch. Who has the more severe slug problem?

We’re told that Daniel’s patch is rectangular and measures five foot by three foot, whereas Charlotte’s is circular with a radius of three foot. There were 21 slugs in Daniel’s vegetable patch and 36 in Charlotte’s. We will now clear some space to calculate the number of slugs they had per square foot. Let’s consider Daniel’s vegetable patch first. His patch was rectangular with dimensions three foot and five foot, and he found 21 slugs in his patch. We can calculate the area of any rectangle by multiplying the length by the width. In this case, we need to multiply five by three. This is equal to 15. Therefore, Daniel’s patch has an area of 15 square foot.

In order to calculate the density per square foot, we can, firstly, write the ratio of the area to the number of slugs. This is equal to 15 to 21. To calculate the density of slugs in Daniel’s patch, we need to calculate the unit ratio, how many slugs there are per square foot. This is written in the form one to 𝑛. We divide both sides of the ratio by 15, giving us the ratio one to 1.4. The density of slugs in Daniel’s vegetable patch is therefore 1.4 slugs per square foot.

We can now repeat this process for Charlotte. Charlotte’s vegetable patch was circular and had a ratio [radius] of three foot. She found 36 slugs in her vegetable patch. The area of any circle can be calculated by multiplying 𝜋 by the radius squared. In this question, this is equal to 𝜋 multiplied by three squared. This is equal to 28.2743 and so on. This means that the area of Charlotte’s vegetable patch is 28.27 square feet. For the purposes of this question, we will keep this as nine 𝜋. The ratio of area to slugs for Charlotte is therefore nine 𝜋 to 36. To find the unit ratio or density, we can divide both sides by nine 𝜋. 36 divided by nine 𝜋 is equal to 1.273 and so on. Rounding this to one decimal place gives us 1.3 slugs per square foot.

The three correct answers are 1.4, 1.3, and Daniel. As 1.4 is greater than 1.3, Daniel has the more severe slug problem.

We will finish this video by summarizing the key points. In order to compare two or more ratios, we need to calculate the unit rate or unit ratio. This is written in the form one to 𝑛. To simplify a ratio or find an equivalent ratio, we must multiply or divide all parts of the ratio by the same number. For example, the ratio four to 12 can be written as a unit ratio by dividing both parts by four. The ratio four to 12 is equivalent to the unit ratio one to three. For every one unit of the first part, we have three units of the second part.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.