Tutorial ID | Title | Tutorial | Video Tutorial | Revision Notes | Revision Questions | |
---|---|---|---|---|---|---|

4.1 | Ratios |

In this Math tutorial, you will learn:

- What are ratios? Why ratios are called so?
- How many quantities can be present in a ratio?
- Why it is better to use a ratio rather than subtraction to compare two quantities?
- What is the procedure used to calculate a ratio?
- How to express ratios R in a number line?
- How to calculate the inverse ratio 1:R?
- How to calculate the fraction of a quantity out of a total?
- How to scale up ratios?
- How to calculate the part of a whole?
- How to calculate part of another part.
- How to find the new ratio when one of quantities changes?
- How to divide in a given ratio?

Imagine the following scenario: A friend tells you that he owns $50,000 less than his brother. Is this information sufficient to draw a conclusion about their wealth? In other words, is it certain that your friend is poor and his brother is rich?

Obviously, the answer is NO. Your friend may be broke; in this case, his brother owns $50,000. This means none of them is rich but your friend's brother is economically stable. However, another option is that the two brothers may be both rich, as your friend may own 10 million USD of wealth and his brother just $50,000 more.

Therefore, it is not sufficient to know how much a quantity is greater than another to draw conclusions about their respective sizes. Therefore, we need other types of comparison to have a better information and understanding about quantities.

There are many situations in practice when we need to compare two quantities. Basically, we use two methods for comparison of two quantities:

**By subtraction;**for example, when we calculate how much taller a child is compared to one year ago, so that we can calculate the increase in height of that child; and**By division;**for example, when we calculate how much more fruit a large box has compared to a small box, so that we can calculate the number of boxes.

Often, it is not enough to simply find which quantity is greater but a quantitative comparison is needed (as seen in the example above), in order to know how much a quantity is bigger than another. In these scenarios we use the concept of a ratio - a method used to compare two quantities of the same type through division.

By definition, a ratio is a comparison of two or more numbers by means of division.

Thus, if we consider the second assumption about the wealth of the two brothers discussed in the introduction, it is obvious that they have almost the same wealth as 10,050,000 / 10,000,000 = 1.005. This means that your friend's brother is only 0.5 percent richer than your friend, so they can be considered as equally rich. This conclusion would be impossible to draw if you simply considered the difference between their wealth.

We know that rational numbers are all those numbers that can be expressed as fractions, it is evident that the term "rational number" derives from "ratio".

In other words, ratios are numbers of the "same type of quantity" written as a fraction or division. This means ratios compare by divisions only quantities of the same type, such as in the example discussed earlier where we compared through division two amounts of money. Hence, ratios have no unit as all units are simplified.

Another form of representing ratios is through the colon symbol (:). This is because the colon was the old symbol used for division. The colon is pariculalry suitable when representing ratios of more than two quantities.

Two objects have the lengths: 40 cm and 60 cm respectively. What is the ratio of their lengths expressed in the simplest form?

We have to write the ratio as a fraction (or division) in the simplest terms, that is in the smallest numbers possible. Thus, since GCF of 40 and 60 is 20, we have

Ratio = *40 cm**/**60 cm*

=*40 cm ÷ 20**/**60 cm ÷ 20*

=*2**/**3* (or 2:3)

=

=

We can use the same approach as when calculating the part of a whole if we know the total and the ratio, and when the amounts represented by the ratio are required. In such situations, it is better to operate in terms of the greatest common factor, GCF. Thus, in the previous example, we could have expressed the greatest common factor (20) by k and the ratio therefore is expressed as

Ratio = *40 cm**/**60 cm*

=*2 × 20 cm**/**3 × 20 cm*

=*2k**/**3k*

=*2**/**3*

=

=

=

Let's see the utility of the greatest common factor k through another example.

The female-male ratio in a company is 5:4 and the company has 63 employees in total. What is the number of employees from each gender?

First, we identify the GCF of the number of employees from each gender. We denote this GCF by k. Thus, from the ratio given in the clues, it is obvious that the number of female employees is 5k and that of males is 4k. Hence, the total number of employee is 5k + 4k = 9k, which corresponds to 63 employees. Hence,

9k = 63

k =*63**/**9* = 7

k =

Now, we can find the exact number of each category by multiplying the corresponding part of ratio (share) by k. Hence,

Number of female employees = 5 × k = 5 × 7 = 35

and

Number of male employees = 4 × k = 4 × 7 = 28

The same procedure is also used when dealing with ratios containing more than two quantities. Let's see another example to clarify this point.

Two knots are formed in a 24 m long rope producing three segments or rope a, b and c as shown in the figure, which have the ratio 3:4:5 to each other. Calculate the length of each segment.

Given the ratio in the clues, we denote the pieces by 3k, 4k and 5k as usual. Thus, we have

a = 3k

b = 4k

c = 5k

b = 4k

c = 5k

In addition, since

a + b + c = 24 m

we can write the above clue in terms of k. Thus,

3k + 4k + 5k = 24

This allows us to find the GCF of the three pieces' lengths, here represented by k. We have

12k = 240 m

k =*240**/**12* = 20 m

k =

Therefore, we obtain for the length of each piece:

a = 3k = 3 × 20 m = 60 m

b = 4k = 4 × 20 m = 80 m

c = 5k = 5 × 20 m = 100 m

b = 4k = 4 × 20 m = 80 m

c = 5k = 5 × 20 m = 100 m

The proof gives 60 m + 80 m + 100 m = 240 m, as the clues suggest.

Number lines can help us express ratios. We need at least two number lines to represent each quantity involved in a ratio. The units are not the same but they correspond to the quantities they represent when viewed vertically. This allows us to very easily find other equivalent ratios, not only the one expressed in the simplest terms. In addition, expressing ratios in a number line allows us identify smaller groups of elements formed, based on their given relationship. Let's see an example to clarify this point.

The ratio between the number of workers in a factory and the T-shirts they can produce in one day is shown in the figure below.

Calculate the daily production of T-shirts in the factory if 150 workers are employed in it.

The numbers written in the same position of units are helpful in determining the ratio R, which in this case represents the daily production of a single worker. (We call this type of ratio the "rate", we will discuss this type of ratio in the next tutorial). Thus, we have

R = Daily production of a single worker = *No.of T-shirts produced**/**No.of workers hired in the factory*

=*80**/**5* = *160**/**10* = *240**/**15* = ⋯ = 16 (T-shirts)/day

=

Therefore, following this rule, we can work out the daily production if 150 workers are hired in the factory. Thus,

R = *No.of T-shirts produced**/**No.of workers hired in the factory*

Hence,

No.of T-shirts produced = R × No.of workers hired in the factory

=*16 (T-shirts)**/**day* × 150 workers

= 2,400 T-shirts

=

= 2,400 T-shirts

Practical situations as the one described in the question above may be also reconsidered, in order to express them as ratios of type 1:R. In other words, we may want to calculate how much from the quantity b is needed for every a. We silently did this in the above example, where we didn't start calculating the ratio from the first quantity (quantity a) but from the quantity b instead. This is because it is not very suitable to calculate how many workers are needed to produce one T-shirt in a day, as the result of ratio will be a decimal (5/80 = 1/16 = 0.0625 T-shirts/worker). Hence, we inverted the fraction derived from the ratio and instead of calculating the ratio R, we calculated its inverse, ** 1/R** instead, this is purely for convenience.

Most modern bronze is made with 88% copper and 12% tin. This means that in 100 g bronze, for every 12 g tin there are 88 g copper. How many grams of copper are needed for every kilogram of tin?

The tin : copper ratio for 100 g bronze is

R = *12 g tin**/**88 g copper* = 0.1__36__

Since this value is not suitable because it is very small, we deal with the inverse ratio. Hence, the inverse ratio ** 1/R** which shows how much copper is needed for every kilogram of tin is

This result means we need 7.333 kg of copper for every kg of tin to produce modern bronze.

Sometimes, we need to calculate what part of the total is one component involved in the ratio. In these situations we first calculate in how many parts the total is made and then, we find the fraction that shows what part of the total is the quantity required.

Three sisters: Stacy, Rebecca and Wendy share a certain amount of money in the ratio 2:5:7.

- What part of the total each sister share?
- How much money does each sister has if they own in total $630?

- First, we find the total number of shares. If we take only the first letter of each name, we have: S:R:W = 2:5:7We can write this ratio asThus,
=*S**/**2*=*R**/**5*= k*W**/**7*S = 2kTherefore, the total number of shares is

R = 5k

W = 7k2k + 5k + 7k = 14kHence, the part of money each sister owns isStacy =of the total.=*2k**/**14k*=*2**/**14**1**/**7*

Rebecca ==*5k**/**14k**5**/**14*

Wendy ==*7k**/**14k*=*7**/**14**1**/**2* - From the above ratios, we can find how much money each sister owns as follows: Stacy =The solution is correct as $90 + $225 + $315 = $630.
of $630 =*1**/**7*× $630 =*1**/**7*= $90*1 × $630**/**7*

Rebecca =of $630 =*5**/**14*× $630 =*5**/**14*= $225*5 × $630**/**14*

Wendy =of $630 =*1**/**2*× $630 =*1**/**2*= $315*1 × $630**/**2*

We will use another example to illustrate the difference between these two types of ratios.

A company has 120 employees where 45 are male and ** 3/5** of female employees have attended university.

- What part of the staff are female?
- How many women from the staff have attended university?
- What part of the staff are female employees who have attended university?
- What is the maximum number of groups that can be formed by preserving the same gender and education structure of the whole staff?

- The number of female workers in the company is 120 - 45 = 75. This number makes Female∶Total = 75∶120

=*75**/**120*

=*75 ÷ 15**/**120 ÷ 15*

=of the staff*5**/**8* - We have 3 out of 5 women from the staff who have attended university. Thus, the total number of female staff N
_{fu}who have attended university isN_{fu}=of 75*3**/**5*

=× 75*3**/**5*

=*3 × 75**/**5*

= 45 women - The above number divided by the total gives the part of the staff that are both female employee and have attended university. Thus, R(N
_{fu}) =*N*_{fu}*/**N*_{tot}

=*45**/**120*

=*45 ÷ 15**/**120 ÷ 15*

=of the staff*3**/**8* - The number of male employees is also 45. This number makes
of the staff, as found above. Therefore, a group of 8 people must contain 5 women (where 3 of them have attended university) and 3 men. This is the smallest group we can form by preserving the original structure of the company staff. Therefore, the maximum number of groups possible is*3**/**8*N_{groups}=*120 employees**/**8**employees**/**group*

= 15 groups

We can use the ratio between two or more quantities given in the simplest form to find other equivalent ratios, this can help us find the number of elements in a group. We saw this property when explaining ratios expressed in a number line, where we could find other ratios besides the required one by looking which numbers were aligned vertically. This time however, we can use not two horizontal separate axes placed one under the other but two perpendicular axes instead, which allows us to draw a straight line which gives all possible relationships between the quantities involved through a straight line drawn from the origin.

The advantage of this method known as scaling up ratios consists in the fact that we can obtain a larger number of possible combinations between the quantities involved, which follow the rule given in the ratio. However, it also has a disadvantage: we cannot include more than three quantities in the calculations, as the maximum number of axes we can use is three (the space is 3D).

From algebra, it is known that if two quantities (we often call them "variables", as the values of one quantity affect the corresponding values of the other) are in a constant ratio, the line that shows all possible relations of the two quantities (we call it the "graph" of relation), is linear. The best feature of this method consists on the need for only two corresponding values; one for each quantity. These values act like coordinates of the graph, expressed by number pairs. Then, we can draw a straight line that connects the origin and the given point expressed by the above values (coordinates). This line can be extended further in that direction to obtain other pairs of numbers.

It is better to have the ratio expressed in this form plotted in a millimetre paper, as in the example below.

A factory produces hats and shirts in the ratio 3:4.

- Plot a graph to show this relationship
- From the graph, find the number of shirts produced for every 21 hats
- Again, use the graph to find the number of hats produced for every 24 shirts

- We express the hats on the horizontal axis and shirts on the vertical one - all this in a millimetre paper, as the one shown below. Since for every 3 hats we have 4 shirts produced, then for every 30 hats there are 40 shirts produced. In this way, we use the pair (30, 40) as a reference for our graph (which starts from the origin). Then, we continue further in that direction.
- If we look at the graph carefully, we notice that for 21 hats there are 28 shirts produced. This is because the point (21, 28) is a point of the graph.
- Again, by observing the graph carefully, we notice that for 24 shirts, there are 18 hats produced. This is because the point (18, 24) is also a point of the graph. As you see, if we have the graph of a ratio plotted, no further calculations are necessary; we can find all values needed by looking at the graph.

Knowing how to find the new ratio when one of the quantities involved in the original ratio changes is particularly useful in chemistry, as this phenomenon is very common when dealing with chemical solutions, when the amount of one of the quantities of an element used in a mixture changes to provide a new mixture. Let's consider an example to explain this situation.

The ratio X : Y for two substances contained in a mixture is 3 : 5 and the amount of mixture is 240 g.

- What will the new ratio between X and Y be if we add in the mixture other 20 g from the substance X?
- What are the original and the new ratio of the substance X to the total?

- First, we have to find how many grams from each substance were originally in the mixture. For this, we use the known procedure of expressing both quantities in terms of their GCF. Thus, since X = 3k and Y = 5k, we haveX + Y = 240Thus,

3k + 5k = 240

8k = 240

k =*240**/**8*

k = 30 gX = 3 × 30 g = 90gandY = 5 × 30 g = 150 gWith the new quantity of 50 g added, the substance X becomes 90 g + 20 g = 110 g, while the quantity Y doesn't change (it remains 150 g). Hence, the ratio between X and Y in the new mixture isR_{new}==*X*_{new}*/**Y*=*110 g**/**150 g**11**/**15* - The original ratio R
_{1}of the substance X to the total can be calculated in two ways: (1) through the greatest common factors k, and (2) through the masses in grams. Thus, we haveRor_{1}==*3k**/**3k + 5k*=*3k**/**8k**3**/**8*RAs for the new mixture, we no longer have the same GCF, so we use only the second method to find the new ratio R_{1}==*90 g**/**240 g*=*90 g ÷ 30**/**240 g ÷ 30**3**/**8*_{2}of the substance X to the total. Thus,RWe can use the same procedure when removing a quantity from one of the components involved in a ratio as well. Let's consider another example._{2}==*110 g**/**110 g + 150 g*=*110 g**/**260 g**11**/**26*

The ratio of salt to the total in a new alimentary product was 3:200. However, consumers left negative feedback because it was too salty. Therefore, the producers decided to remove 5 g salt from each kilogram of the actual product without adding other things in it. What is the new ratio of salt to the new alimentary product?

The ratio 3 : 200 means that in one kilogram of the original product there were 5 × 3 g = 15 g of salt, as 1 kg = 1000 g = 5 × 200 g.

When removing 5 g salt from each kilogram of the product, the total of food in each package becomes 1000 g - 5 g = 995 g and the salt in each package becomes 15 g - 5 g = 10 g. Therefore, the new ratio of salt to the food is

R = *10 g**/**995 g* = *10 g ÷ 5**/**995 g ÷ 5* = *2**/**249*

In this way, the food became less salty than it was before.

Ratios can be used to divide a given quantity into unequal amounts according the numbers of the ratio. We discussed an example of this at the beginning of this ratio tutorial, when an amount of money has to be shared among three sisters according to a given ratio. Now, let's see another example, this time with geometry.

The ratio between the interior angles of a triangle is 2:3:4. Calculate the mass of the greater angle.

From geometry, it is known that the mass of the interior angles in any triangle is 180°. We can denote the angles by a, b and c. Therefore, using the known procedure discussed earlier, we have:

2k + 3k + 4k = 180^{0}

9k = 180^{0}

k =*180*^{0}*/**9* = 20^{0}

9k = 180

k =

Hence, the mass of each angle is

a = 2k = 2 × 20^{0} = 40^{0}

b = 3k = 3 × 20^{0} = 60^{0}

c = 4k = 4 × 20^{0} = 80^{0}

b = 3k = 3 × 20

c = 4k = 4 × 20

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- Continuing learning ratio and proportion - read our next math tutorial: Rates. Applications of Ratios and Rates in Practice

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