Video: Ratio Among Three Numbers

In this video, we will learn how to simplify and use the ratio between three numbers to solve problems.

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Video Transcript

In this video, we will learn how to simplify and use the ratio between three numbers to solve problems. We will begin by recalling what we mean by ratio. A ratio shows the relative sizes of two or more values. In this video, we will look at questions with three values. Ratios can be shown in different ways, firstly, using a column to separate example values, secondly, written as fractions to separate one value from the total. Any fraction can be converted to a decimal by dividing the numerator by the denominator, in this case, dividing one value by the total. Finally, we could convert a decimal to a percentage by multiplying the decimal by 100.

As an example, let’s imagine we had one boy and three girls. This can be written as a ratio one to three, where the first number represents the boys and the second the girls. When dealing with ratio, it is important the order is correct. If we wanted the ratio of girls to boys, this would be three to one. As there are four children in total and one is a boy, one-quarter of the children are boys and three-quarters are girls. These can be converted to decimals by dividing one by four and three by four, giving us 0.25 and 0.75. This means that 25 percent of the children are boys and 75 percent are girls. The first question that we’ll look at involves simplifying a ratio.

Express the ratio 72 : 54 : 81 in its simplest form.

In order to simplify any ratio, we must divide each of the numbers by the same value. This means that we’re looking for a common factor of 72, 54, and 81. Whilst it will be quicker if we choose the highest common factor, any common factor will enable us to start the question. 72, 54, and 81 are all in the nine times table. Therefore, they’re all divisible by nine. 72 divided by nine is eight, 54 divided by nine is six, and 81 divided by nine is equal to nine. This means that the ratio simplifies to eight to six to nine. These three numbers have no common factor except one. This means that the ratio is in its simplest form. Whilst eight and six are exactly divisible by two, nine is not. Likewise, six and nine are divisible by three, but eight is not. The ratio 72 : 54 : 81 written in its simplest form is eight : six : nine.

Our next question involves combining two ratios to find the ratio between three numbers.

Determine the ratio between the three numbers π‘Ž, 𝑏, and 𝑐 given that the ratio of π‘Ž to 𝑏 is 10 to one and the ratio of 𝑏 to 𝑐 is two to one.

We’re asked to work out the ratio of π‘Ž to 𝑏 to 𝑐. We’re told in the question that the ratio of π‘Ž to 𝑏 is 10 to one. We’re also told that the ratio of 𝑏 to 𝑐 is two to one. 𝑏 occurs in both of these. So we need to use equivalent ratios to make sure the value of 𝑏 is the same. We can find equivalent ratios by multiplying all of our values by the same number. This means that the ratio 20 to two is equivalent to the ratio 10 to one as we have multiplied both values by two. The ratio of π‘Ž to 𝑏 can therefore be rewritten as 20 to two. As the value for 𝑏 is now the same, the ratio of π‘Ž to 𝑏 to 𝑐 is 20 : two : one.

Our next two questions are ratio problems in context.

The ratio between the heights of three buildings A, B, and C is 10 : four : three. If the height of building A is 60 meters, find the heights of building B and building C.

We’re told in the question that the ratio of the heights is 10 : four : three. We’re also told that the height of building A is 60 meters. We can find equivalent ratios by multiplying each of our values by the same number. 10 multiplied by six is equal to 60. This means that, in order to find an equivalent ratio, we must also multiply four and three by six. Four multiplied by six is 24, and three multiplied by six is 18. This means that the ratio 10 : four : three is equivalent to the ratio 60 : 24 : 18. We can therefore conclude that the height of building B is 24 meters and the height of building C is 18 meters.

The total number of students in the first, second, and third grades at a primary school is 285 in the ratio seven : four : eight. Calculate the number of students in each grade.

We can share a total, in this case, 285 students, in a given ratio by following three steps. Firstly, we find the sum of the ratios. Seven plus four plus eight is equal to 19. Our second step is to divide the total by this answer. 285 divided by 19 is equal to 15. This is equal to one part or one share of the ratio.

Our final step is to multiply the value for one part by each of the ratios. 15 multiplied by seven is 105. 15 multiplied by four is equal to 60. 15 multiplied by eight is equal to 120. As the seven corresponded to the number of first graders, there are 105 students in first grade, 60 students in second grade, and 120 students in third grade. It is always worth checking our answers by adding the values to ensure this makes the total. 105 plus 60 plus 120 is equal to 285. This method works for sharing any total between any number of ratios.

The last two questions we will look at are more complicated problems involving algebra.

If 10π‘₯ is equal to 11𝑦 which is equal to 12𝑧, find the ratio of π‘₯ to 𝑦 to 𝑧.

There are lots of ways of solving this problem. One way would be to find values that solve different parts of the equation first. Let’s begin by considering 10π‘₯ is equal to 11𝑦. Substituting in the values π‘₯ equals 11 and 𝑦 equals 10 would mean that this equation is true. 10 multiplied by 11 and 11 multiplied by 10 are both equal to 110. This means that the ratio of π‘₯ to 𝑦 could be written as 11 to 10. Let’s now consider the fact that 10π‘₯ is also equal to 12𝑧. In this equation, π‘₯ equals 12 and 𝑧 equals 10 is a solution. 10 multiplied by 12 and 12 multiplied by 10 are both equal to 120. This means that the ratio of π‘₯ to 𝑧 is 12 to 10.

We now have two ratios, a ratio of π‘₯ to 𝑦 and a ratio of π‘₯ to 𝑧. In order to combine these ratios, we need to use equivalent ratios to ensure that the value for π‘₯ is the same. The lowest common multiple of 11 and 12 is 132. We can therefore multiply the top ratio by 12 and the bottom ratio by 11. The ratio 11 to 10 is equivalent to the ratio 132 to 120. Likewise, the ratio of π‘₯ to 𝑧 of 12 to 10 is equivalent to 132 to 110. As our value for π‘₯ is the same, we can now combine the ratios. The ratio of π‘₯ to 𝑦 to 𝑧 is 132 : 120 : 110.

This ratio can be simplified as all of our values are even and are therefore divisible by two. 132 divided by two is equal to 66. 120 divided by two is equal to 60. And 110 divided by two is equal to 55. The ratio of π‘₯ to 𝑦 to 𝑧 in its simplest form is 66 : 60 : 55 as these three numbers have no common factor apart from one.

If the ratio of π‘₯ to 𝑦 to 𝑧 is three : four : eight, then what is the value of π‘₯ squared plus 𝑦 squared plus 𝑧 squared divided by 𝑦 multiplied by π‘₯ plus 𝑧.

We can see from the ratio given that our value of π‘₯ is equal to three, our value of 𝑦 is equal to four, and our value of 𝑧 is equal to eight. We can substitute these values directly into our expression. Three squared is equal to nine, four squared is 16, and eight squared is 64. So the numerator simplifies to nine plus 16 plus 64. We can simplify the denominator by using our order of operations known as PEMDAS or BIDMAS. We do the parentheses or brackets first, leaving us with four multiplied by 11. Nine plus 16 plus 64 is equal to 89. Four multiplied by 11 is 44.

The value of the expression π‘₯ squared plus 𝑦 squared plus 𝑧 squared divided by 𝑦 multiplied by π‘₯ plus 𝑧 is 89 over 44. This fraction can not be simplified as 89 and 44 have no common factor apart from one.

We will now summarize the key points from this video. We recalled at the start of the video that a ratio shows the relative sizes of two or more values. In this video, we were dealing with problems involving three values, where the ratios were written in the form π‘₯ to 𝑦 to 𝑧. The questions that we answered included simplifying ratios, dividing a quantity using a ratio, and algebraic problems. Important topics that help us solve problems involving ratio include factors, multiples, and basic arithmetic.

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