Repeated Proportional Change

Topic Overview

In order to answer exam questions related to repeated proportional change, you will first need to understand some related mathematical terms:

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

An object weighs 2kg on the 1st day. If weight is proportional to age, then the object will weigh 4kg on the 2nd day, 6kg on the 3rd day, 8kg on the 4th day, 10kg on the 5th day and so on.

The mathematical term 'ratio' defines the relationship between two numbers of the same kind. The relationship between these numbers is expressed in the form "a to b" or more commonly in the form:

a : b

A ratio is used to represent how much of one object or value there is in relation to another object or value. For example:

If there are 10 apples and 5 oranges in a bowl, then the ratio of apples to oranges would be 10 to 5 or 10:5.

This is equivalent to 2:1.

In contrast, the ratio of oranges to apples would be 1:2.

The mathematical term 'iteration' means to repeat an operation. Similarly to the worked examples outlined in the 'Functions' revision guide, iterations involve inputting a value into a function, receiving an output value, and then using this output value as the input value for the function. For example:

To iterate the function 2x + 3, first you place any value of x into the function:

So if x = 1, then 2x + 3 = 5

Now you must take this output value of 5 and put it into the original function.

Therefore, when x = 5 , then 2x + 3 = 13

By continuing these iterations, you will be presented with an infinite sequence of numbers:

5, 13, 29, 61, .......

Key Concepts

In the new linear GCSE Maths paper, you will be required to solve various mathematical problems involving repeated proportional change. 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:

  • Use percentages to compare proportions
  • Understand repeated proportional change and demonstrate how it can be used to solve compound interest problems

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

1 - Repeated Proportional Change: Growth and Decay
'Repeated proportional change' is the mathematical process of calculating a percentage change.

If we multiply a number over and over again with the same number, except 1, then we 'repeatedly change' it in the same 'proportion'.

There are two types of repeated percentage change: 'increase' and 'decrease', or 'growth' and 'decay' as they are often respectively known.

With repeated proportional 'increase' or 'growth', your original value gets larger after each proportional change:

Example
(a) - The number of rabbit in a field increases by a quarter every year. At the start of the first year there are 20 rabbits. How many rabbits will there be after 5 years?

Solution
(a) - The rate of increase is a quarter every year, which can be written as 25% or 0.25.

The initial number of rabbits is 100% or 1.

Therefore, you need to add this quarter to the initial number of rabbits to calculate the repeated rate of increase.

1 + 0.25 = 1.25

Therefore, after 1 year the number of rabbits will be:

20 x 1.25

For every year, you must raise 1.25 to another power, due to the fact that the number of rabbits will increase by another 25%. Therefore, after 5 years there will be:

\[20*{1.25^5} = 61.03515625....\]

= 61 rabbits

(Note: For repeated proportional change questions, you will be expected to approximate your answers to a suitable degree of accuracy. For example, in this question, it is most appropriate to round your answer down to a whole number)

Alternatively, with repeated proportional 'decrease' or 'decay', your original value reduces after each proportional change:

Example
(b) - The number of fish in a river are expected to decrease by 5% every year for the next 4 years. The river currently houses 200 fish. How many fish will be in the river after 4 years?

Solution
(b) - The rate of decrease is 5% which can also be written as 0.05. Due to the fact that the number of fish is expected to decrease, you must subtract 0.05 from the initial number of fish which is 100% or 1, in order to calculate the repeated rate of decrease:

1 -0.05 = 0.95

Therefore, after 1 year, the number of fish will be:

200 x (100% - 5%) = 200 x 0.95

For every year, you must rate 0.95 to another power, due to the fact that the number of fish will decrease by another 5%. Therefore, after 4 years there will be:

\[200*{0.95^4} = 162.90125\]

= 163 fish

2 - Compound interest
Repeated proportional change is an extremely useful mathematical process because it can be used to calculate real world financial problems such as compound interest.

'Compound interest' refers to the interest added to a deposit or loan. The added interest will also earn interest as time passes.

If you are sitting the GCSE Maths Higher Tier paper, you may be required to calculate the compound interest of a loan using repeated proportional change.

Example
(a) - £500 is borrowed for 6 years at 5 % compound interest. Calculate the amount of compound interest which will be paid

Solution
(a) - From the question, we know that 5% compound interest is added each year. This means that there will be 105% of the original amount borrowed at the end of the first year or 1.05.

With 1.05 as your multiplier, you can calculate the total amount of money borrowed after 6 years. The money was borrowed for 6 years, so you must raise 1.05 to the power of 6. Therefore:

Total amount of money borrowed is

\[500*{1.05^6} = 670.047...\]

= £670

The question has asked you to calculate the amount of compound interest. To do so, you must subtract the original amount borrowed (£500) from the value you have just generated:

670 - 500 = 170

As a result, the amount of compound interest which will be paid is £170

3 - Simple Interest
There is also 'simple interest' which means that the interest earned in the first year is the same for every year. During your GCSE maths exam, you may be asked to calculate the amount of simple interest on a loan:

Example
(b) - £600 is borrowed for 4 years with £120 interest per year. Calculate the amount of simple interest which will be paid

Solution
(b) - Calculating simple interest is much easier than compound interest. All you have to do is multiply the interest rate by the total number of years to calculate the total amount of money borrowed:

600 + (4 x 120) = 1080

Now you subtract the original amount borrowed from your generated value to calculate the amount of simple interest:

1080 - 600 = 480

Therefore the amount of simple interest which will be paid is £480

Exam Tips

  • Remember that repeated proportional change is the process of calculating a percentage increase or decrease.
  • Remember to approximate your answers to a suitable degree of accuracy.
  • Compound interest refers to the interest added to a deposit or loan. The added interest will also earn interest as time passes.
  • To calculate simple interest; multiply the interest rate by the total number of years.

Topic Summary

When tackling problems involving repeated proportional change, it is important that you have a comprehensive understanding of decimals and percentages. Therefore it is advisable that you revise the topics of decimals and percentages which are outlined in the Problem Solving with Decimals and Percentages revision guide. By doing so, you will be able to fully understand the principles of proportions and accurately calculate repeated proportional increase and decrease. Above all else, it is important that you write down ALL of your working out. By doing so, you can earn valuable method marks for your understanding of the correct repeated proportional change processes; even if you have calculated an incorrect value.

Related Topics

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