Variation. Types of Variation

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4.5Variation. Types of Variation

In this Math tutorial, you will learn:

  • What is variation?
  • What are the three types of variation?
  • Where does variation differ from proportion?
  • Where to use proportional and variation approach in graph problems?
  • What is the constant of proportionality used in variation?
  • What is the shape of the graph that illustrates direct/inverse variation?
  • Which are the properties of variation?

Introduction

In previous tutorials we explained ratio, rate and proportion, in this tutorial we will discuss variation - a common concept encountered when dealing with proportion-related situations. Despite the fact that in many textbooks the terms that represent these two concepts (proportion and variation) are considered as synonyms of each other, there are a few important differences between them, which we will explain here. In addition we provide many examples that depict various situations where variation is used to illustrate the theory.

Definition of Variation

Variation is a simple concept to understand. Basically, if two quantities are dependent on each other in one of the three ways described in this tutorial, we say that one will vary according to the other.

In other words, there are numerous ways in which two quantities are related to each other but only three of them are considered as variations. They are direct, inverse and joint variation, which we will explain in this tutorial.

"Variation" defines a concept that deals with variability in mathematics - a more advanced concept we will explain in future tutorials. By definition, variation represents any change in quantity due to change in another if the two quantities have a form of relationship with each other.

The Difference between Proportion and Variation

As we said above, the concepts of proportion and variation are very similar. However, despite the similarity, they have their differences that make these two concepts distinguishable from each other.

The first difference lies in the fact that a proportion does not necessarily involve related terms, i.e. the existence of one term of proportion does not necessarily imply the existence of the other term. For example, when we say that three sisters A, B and C share everything the family owns in the ratio 3:4:5, we can form the proportion A:B:C = 3:4:5, but this proportion does not necessarily involve the existence of money to share. They can share everything else but money. Hence, when we say that an amount of $1200 has to be shared among the three sisters in the way described above, we are not referring to a variation but to a proportion instead, as the existence of the three sisters does not necessarily imply the existence of money to share.

The second difference lies in the fact that proportion is limited to a finite number of terms while variation includes a much wider range of terms related with each other through a formula. This leads to the plotting of a graph that includes all possible ordered pairs, where the first of values in each pair is always from the first set of values, while the second term from the second set. Variation therefore is a more inclusive concept than proportion, as practically it includes an infinite number of combinations which meet the condition indicated in the formula.

Many situations we have used in the previous tutorial of this chapter to illustrate proportions, are in fact examples of variation. For example, the number of workers needed to complete a job versus the time needed to do the job can be considered either as an inverse proportion when we use a real-life approach (as the number of workers involved is practically finite), but also as a variation if we approach the situation mathematically, where the number of people involved in the process can be assumed as infinite (or when we consider not only the whole values). In the second case, we can plot a graph where the line is continuous, while if we consider the situation from the proportion viewpoint, the graph represents a set of dots representing the number of people versus the time taken to complete the job. Look at the figures below, where the relationship Number of people = 12/Time taken is shown based on the two above-mentioned approaches.

  1. Proportion approach
    Here, the number of people is discrete, i.e. it can only be a whole number. As a result, we cannot extend the graph more on the right of 12, as there cannot be less than one person working. As for the left part of the graph, it contains more points as the number of people can increase as much as we want - an action that brings a reduction in the working time. Math Tutorials: Variation. Types of Variation Example
  2. Variation approach
    This time, we neglect the restriction of the discrete number of people and we assume the two values as real numbers. Therefore, the graph will be an uninterrupted line (in tutorial 4.3 we called it "hyperbola"). Math Tutorials: Variation. Types of Variation Example

Types of Variation - Direct Variation

A direct variation is a variation that can be expressed through the general formula

y = k × x

where x and y represent possible values of the two quantities involved in the variation while k is the constant of proportionality that provides the value of each individual ratio or rate involved in variation. More specifically, the constant k represents the quotient of division between any vertical coordinate to the corresponding horizontal coordinate in the variation graph.

Obviously, direct variation is very similar to direct proportion we discussed in tutorial 4.3. The graph that geometrically represents the direct variation is a straight line that passes through the origin.

Example 1

An employee is paid $25/hour. Plot the graph indicating the variation of his daily salary from the working hours.

Solution 1

This is a situation involving a direct variation as the daily salary is directly proportional to the number of working hours (the more working hours are committed, the higher the daily salary).

If we denote the working hours by x and the daily salary by y, we obtain the variation

y = 25 · x

where 25 is the value of the constant of proportionality k given in the general formula of direct variation

y = k · x

We can stop to the number 8 in the values of working hours as an employee usually works 8 hours a day when employed full time. Thus, the graph showing this (direct) variation is

Math Tutorials: Variation. Types of Variation Example

From the graph, it is evident that at the end of the working day, the employee earns $200.

We can find any missing value when we know the variation formula, as in the example below.

Example 2

The quantity x and y vary directly and y = 6 when x = 18. What is the value of x when y = 102?

Solution 2

First, we find the constant k of variation (proportionality). Since the general formula of direct variation is y = k · x, we obtain

6 = k × 18
k = 6/18
= 1/3

Thus, the formula of this specific variation is

y = 1/3 × x

Now, substituting y = 102 in the above formula yields

102 = 1/3 × x
x = 102 ÷ 1/3
= 102 × 3/1
= 102 × 3
= 306

Types of Variation - Inverse Variation

Inverse variation is similar to inverse proportion discussed in tutorial 4.3. By definition, two quantities vary inversely when their product is always constant.

Mathematically, we express an inverse variation as

x × y = k

or

y = k/x

where x and y are the quantities involved in the variation and k is the constant of proportionality.

For example, the number of workers hired to do a job when multiplied by the time taken always provides a constant number, which indicates the size of the job to be done.

As a special case of inverse variation, we can mention inverse proportion, which is an inverse variation between two quantities in two different instants (1) and (2). In this case, we can write

x1 × y1 = x2 × y2 = k

or

x1/x2 = y2/y1

Obviously, the graph of an inverse variation is an uninterrupted hyperbola, while that of inverse proportion shows a finite number of points in the graph - the points whose coordinates represent the quantities involved in the proportion. Earlier we discussed both of these graphs in the example regarding the number of workers hired to do a job versus time taken, where the proportion and variation approach were used to express the data.

Example 3

Four identical pipes can fill a tank in 2 hours.

  1. Plot the graph of the hours needed to fill the tank depending on the number of pipes available.
  2. From the graph, find out how long it takes if we use 16 of the same pipes to fill the tank completely.

Solution 3

  1. This is an example of inverse variation, where the relationship between the number of pipes used to fill the tank versus the time taken is
    Number of pipes used × Time taken to fill the tank = Constant
    We express the time taken to fill the tank by x and the number of pipes by y. hence, from the clues, we can calculate the constant k first, as
    x × y = k
    2h × 4pipes = k
    k = 8
    Therefore, the formula that expresses this inverse variation is
    y = 8/x
    Now, let's plot the graph by taking a number of values as shown in the table below. Math Tutorials: Variation. Types of Variation Example
  2. From the graph (and table) it is clear that if there are 16 pipes available, they can fill the tank in 0.5 h. Therefore, the more pipes that are open, the shorter the time needed to fill the tank.

Another situation relevant to inverse variation is when we have to divide an amount into more than two shares where the criterion is determined by the inverse relationship between the quantities involved in the variation. In these situations we express the relationship as

a:b:c = k/d:k/e:k/f

This is because when we consider each quantity on the left to the corresponding quantity on the right separately, we obtain the following relations:

a × d = k
b × e = k
c × f = k

Let's consider an example to clarify this point.

Example 4

$420 are divided among three siblings aged 6, 8 and 12 respectively, inversely proportional to their ages. Find the amount each sibling receives.

Solution 4

First, let's calculate the constant k of proportionality. If we express the siblings as A, B and C, we obtain

A:B:C = k/12:k/8:k/6

In addition, we know that

A + B + C = $420

Therefore, we can write

k/12 + k/8 + k/6 = $420
2k/24 + 3k/24 + 4k/24 = $420
9k/24 = $420
k = $420 × 24/9
= $1,120

Hence, sibling A, who is 6 years old receives

A = $1,120/6 = $186.666

Sibling B, who is 10 years old receives

B = $1,120/8 = $140

And sibling C, who is 12 years old receives

C = $1,120/12 = $93.3

Types of Variation - Joint Variation

If there are more than two quantities involved in a variation where both types of variation are present, then we are dealing with a joint variation.

For example, the quantity x may vary inversely with a quantity y and directly with another quantity z to form a joint variation. In such cases, we can write for two different situations (1) and (2):

z1 = k × x1 × y1

and

z2 = k × x2 × y2

where k is a constant of proportionality as usual.

Dividing the two above equations side by side, yields

z1/z2 = k × x1 × y1/k × x2 × y2

Simplifying k from both terms, yields

z1/z2 = x1 × y1/x2 × y2

The general rule of joint variation is that the two quantities that are at the same side of the formula vary inversely, while each of them varies directly with the quantity that is present at the other side of the formula.

Example 5

10 workers can build 5 houses in 12 months. How long it takes to 15 workers to build 8 houses?

Solution 5

Increasing the number of workers brings an increase in the number of houses built during a certain time. Therefore, the number of workers varies directly with the number of houses built.

Yet, increasing the number of workers brings a decrease in the time taken to build a certain number of houses. Hence, the number of workers is inversely proportional to the time taken.

Finally, increasing the number of houses increases the time taken to build them. Therefore, the number of houses is directly proportional to the time.

Since the number of houses is directly proportional to each from the other two quantities, it must appear alone in the formula, while the other two quantities must be together on the other side of the formula, multiplied with each other. Hence, we have

No.of Houses 1/No.of Houses 2 = No.of workers 1 × Time 1/No.of workers 2 × Time 2

Substituting the known values yields

5 houses/8 houses = 10 workers × 12 months/15 workers × Time 2

Applying the cross product, yields

5 × 15 × T2 = 8 × 10 × 12
75 × T2 = 960
T2 = 960/75
= 12.8 months

If there are more than three quantities involved in a variation, we must carefully analyse which of them vary directly, and which inversely. Then, the appropriate expression is written by observing the above-mentioned rules.

Example 6

18 men can build 5 houses in 3 months by working 6 hours a day. How long will it take to 27 men building 12 houses by working 8 hours a day?

Solution 6

Let's express the information in a table for a better understanding.

Math Tutorials: Variation. Types of Variation Example

Let's focus on the missing quantity z and check what kind of variation it has with the other quantities.

  1. More people working = less months needed to build a fixed number of houses. Thus, x and z vary inversely.
  2. More houses built = more months needed IF the other variables remain unchanged. Thus, y and z vary directly.
  3. More hours a day committed = less time needed to build a fixed number of houses IF the number of people is constant. Therefore, z and t vary inversely.

Combining all the above findings, we write

z = k × y/x × t

where k is the usual constant of proportionality.

Writing the above expression for the two situations (1) and (2) yields:

z1/z2 = k × y1/x1 × t1/k × y2/x2 × t2

or

z1/z2 = x2/x1 × y1/y2 × t2/t1

Substituting the known values, we obtain

3/z2 = 27/18 × 5/12 × 8/6 = 1080/1296

Thus,

z2 = 3 × 1296/1080
= 3.6 months

Summarizing the Properties of Variation

In the previous paragraphs, we explained the three types of variation by illustrating them with examples. In this part, we will summarize the main properties of variation, so you will find it easier to take them as a reference when trying to solve examples.

1. x and y vary directly as a and b

If x and y vary directly as a and b, then the following formula is true.

x/a = y/b = k

Example 7

20 oranges are divided between two children, where each share vary directly as 2 and 3. Calculate the number of oranges each child receives.

Solution 7

If we denote the children as x and y respectively, we obtain

x/a = y/b = k

Or

x/2 = y/3 = k

This means that

x = 2k and y = 3k

Giving that

x + y = 20

we obtain

2k + 3k = 20
5k = 20
k = 4

Hence,

x = 2 × 4 = 8 oranges
y = 3 × 4 = 12 oranges

(We have solved similar problems to this one in previous guides but using the proportion approach.)

2. x and y vary inversely as a and b

If x and y vary inversely as a and b, then the following formula is true.

a × x = b × y = k

Example 8

Four people can paint a building in 10 days. How many people are needed to paint the same building in 8 days?

Solution 8

This is an example of an inverse variation where the number of workers is inversely proportional to the time needed to paint the building. Thus, if expressing the workers by x and y and the number of days by a and b respectively, yields

a × x = b × y = k
10 days × 4 workers = 8 days × y workers
y = 10 × 4/8 = 5 workers

3. Joint Variation

In a joint variation where x and y vary inversely as a and b while z varies directly with either x or y as c with a or b, the following formula is true.

z/c = x × y/a × b

Example 9

y varies directly as x2 + 1 and inversely as x + 2. Find the relation between x and y if y = 5 when x = 3.

Solution 9

The quantities that vary inversely are written in the same side of formula while those varying inversely on different sides. Thus, we have

x2 + 1 = k × y × (x + 2)

where k is the usual constant of proportionality.

Substituting the known values yields

32 + 1 = k × 5 × (3 + 2)
9 + 1 = k × 5 × 5
10 = k × 25
k = 10/25 = 2/5

Thus, we obtain for the relationship between x and y:

x2 + 1 = 2/5 × y × (x + 2)
y = 5 × (x2 + 1)/2 × (x + 2)

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