Proportion – Algebraic Problems

Topic Overview

In order to answer algebraic problems which involve proportions, you will first need to understand some related mathematical terms:

The term variable is used to describe the letter symbol within an algebraic equation, e.g. 'x' or 'y'. These symbols are used because we do not yet know the value of these variables.

For example, for the direct variation equation:

\[y = kx\]

You may be required to substitute values into this equation in order to solve the values of either 'y', 'x' or 'k' .

The mathematical term 'proportional' describes two quantities which always have the same relative size or 'ratio'. For example;

A shopkeeper earns £200 on the 1st day of the shop opening. If money earned is proportional to the days on which the shop is open, then the shopkeeper will earn £400 on the 2nd day, £600 on the 3rd day, £800 on the 4th day, £1000 on the 5th day and so on.

Key Concepts

In the new linear GCSE Maths paper, you will be required to solve various mathematical problems involving proportions. The specific questions you will be expected to answer will vary depending upon which examination board with which you are registered, but as a rule you will be required to:

  • Understand the constant of proportionality and use it to solve direct and inverse proportion related mathematical problems
  • Understand and use direct proportion
  • Understand and use inverse proportion

Listed below are a series of summaries and worked examples to help you solidify your knowledge about proportions.

Worked Examples

1-Constant of Proportionality/Variation
If a variable 'x' is directly proportional to another variable 'y', their relationship can be written as:

\[y = kx\]

This is equation often referred to as the 'direct variation equation' ; where 'k' is called the 'constant of proportionality' or the 'constant of variation'.

The 'constant of proportionality' is the value which relates the two variables.

For example, if you are paid £50 per day, the constant of proportionality 'k' would be 50. This is because:

The amount you are paid = £50 x the amount of days worked

This can be written in the form:

\[y = kx\]

Where 'y' is the amount you are paid, 'x' is the amount of days you have worked and 'k' is the constant of proportionality, which in this case is £50.

Example
(a) - The variable 'y' is directly proportional to the variable 'x' . When x = 6, y = 30. Using this information, find the constant of proportionality and the value of y when x =100

Solution
(a) In order to find the value of y when x = 100, you must find the constant of proportionality. By using the direct variation equation 'y = kx' , you can substitute the 'x' and 'y' values mentioned in the question above, and solve the equation to find 'k' :

y = kx

When x = 6, y = 30

Therefore:

30 = 6k

k = 30/6 = 5

Therefore: y = 5x

Using this equation, you can find the value of 'y' when 'x' = 100.

y = 5 x 100

y = 500

2 - Direct Proportion
Two variables are in 'direct proportion to one another' when they increase or decrease at the same rate. This can be written as:

\[y \propto x\]

or

\[y = kx\]

During your GCSE maths exam, you will be required to solve mathematical problems involving direct proportions:

Example
(a) - When x = 16, y = 4. If y is directly proportional to x, find the value of x when y = 5

Solution
(a) -
y is directly proportional to x . This can be written in the form :

y =kx

By substituting the known values for 'y' and 'x' , you can solve the equation to find the value of 'k' :

When x = 12, y = 3

Therefore:

3 = k x 12

k = 3/12 = 1/4

Now that you know the value of 'k', you can substitute your 'y' and 'k' values into the equation to find the value of 'x' when y = 5:

y = kx

8 = (1/4) x

x = 8 x 4 = 32

Therefore when y = 8, x = 32

Example
(b) - The variable 'y' is directly proportional to ' x^3 '. If y = 1 when x = 2, Find the value of y when

x = 4

Solution
(b) -
'y' is directly proportional to ' x^3 ' , which can be written as:

\[y = k{x^3}\]

As in the previous example, you must substitute your known values of y = 1 and x = 2 into y = kx^3 in order to find the value of k:

\[1 = k*{2^3}\]
\[1 = k*8\]
\[k = \frac{1}{8}\]

Now that you know the value of 'k', you can substitute your 'x' and 'k' values into the equation to find the value of 'y' when x = 4 :

\[y = \frac{1}{8}{(4)^3} = \frac{1}{8}*64 = 8\]

Therefore, when x = 4 , y = 8

3 - Inverse Proportion
Two variables are 'inversely proportional to one another' when one variable decreases at the same rate that the other variable increases.

If variable 'y' is inversely proportional to variable 'x', this can be written as:

\[y \propto \frac{1}{x}\]

During your GCSE maths exam, you will be required to solve mathematical problems involving inverse proportions.

Example
(a) - The variable 'y' is inversely proportional to the variable 'x' . When x = 12, y = 3. Using this information, find the value of x when y = 8 .

Solution
(a) -
y is inversely proportional to x. This can be written as:

\[y \propto \frac{1}{x}\]
\[y = \frac{k}{x}\]

or

\[xy = k\]

By substituting your known values of 'x' and 'y' into this equation, you can find the value of 'k' :

xy = k

When x = 12 , y = 3:

k = 12 x 3 = 36

Now that you know the value of 'k', you can substitute your 'y' and 'k' values into the equation to find the value of 'x' when y = 8 :

xy = k

8x = 36

x = 36/8 = 4.5

Therefore, when y = 8, x = 4.5

Example
(b) - The variable 'y' is inversely proportional to ' x^3 '. If y = 25 when x = 2, Find the value of x when y = 60

Solution
(b) -
'y' is inversely proportional to ' x^3 ' , which can be written as:

\[y = k/{x^3}\]
\[k = y({x^3})\]

As in the previous example, you must substitute your known values of y = 25 and x = 2 into this equation in order to find the value of k:

\[k = y({x^3})\]
\[k = 25*{2^3} = 200\]

Now that you know the value of 'k', you can substitute your 'y' and 'k' values into the equation to find the value of 'x' when y = 60 :

\[k = y({x^3})\]
\[60{x^3} = 200\]
\[{x^3} = 200/60\]
\[{x^3} = 3.333....\]
\[{x^3} = \sqrt[3]{{3.333}}... = 1.494\]

Therefore, when y = 60, x = 1.494

Exam Tips

  1. Remember that, if a variable 'x' is directly proportional to another variable 'y', their relationship can be written as: y = kx. This equation is often referred to as the 'direct variation equation' ; where 'k' is called the 'constant of proportionality' or the 'constant of variation'.
  2. Remember that, if variable 'y' is inversely proportional to variable 'x', this can be written as: y ∝ 1/x or xy = k.
  3. It is important to remember the difference between direct and indirect proportion. When two variables are in 'direct proportion to one another' they increase or decrease at the same rate.
  4. Alternatively, when two variables are 'inversely proportional to one another' one variable decreases at the same rate that the other variable increases.
  5. Write down ALL of your working out in order to earn maximum method marks.

Topic Summary

When solving proportional problems algebraically, you will find that the process is very similar to solving an algebraic equation. Therefore, it is advisable that you revise the revision guides on 'Solving Equations' and 'Change The Subject of a Formula'. If you can confidently manipulate and solve equations to calculate the values of 'y', 'x' and the constant of proportionality 'k', then you will be able to earn maximum marks for the proportions questions within your actual GCSE maths exam!

Related Topics

  • Repeated Proportional Change
  • Proportion: Numerical Problems
  • Problem Solving with Decimals and Percentages
  • Ratio Problems
  • Compound Measure Problems
  • Pre-calculus Skills