# GCSE Maths: Upper And Lower Bounds

## Introduction

We will introduce upper and lower bounds with a simple example.

Suppose that you have a piece of string that is 12cm long to the nearest cm. What are the possible lengths of the piece of string?

The string must be at least 11.5cm long as this rounds up to 12cm and it needs to be less than 12.5cm so that it can be rounded down to 12cm. But if the string is more than 12.5cm then it would be rounded up to 13cm.

The 11.5cm length is the least value and this is known as the lower bound

The 12.5cm length is the maximum value and this is known as the upper bound.

Upper and lower bound questions can appear on both the foundation and higher paper. Foundation paper questions generally tend to ask for the error interval whereas questions on the higher paper can be more complex.

## Upper and Lower Bounds Example

Take a look at the following question:

a) The smallest possible weight would be 24.5kg

b) The largest possible weight would be 25.5kg

## Example

Take a look at the following question:

For the length the upper bound is 340.5m and for the width the upper bound is 117.5m

So the upper bound for the perimeter of the field = 340.5 + 340. 5 + 117.5 + 117.5 = 916m

Example

Take a look at the following question:

The upper bound for the length is 35.5cm and the upper bound for the width is 26.5cm

So the upper bound for the area is 35.5 x 26.5 = 940.75 cm².

Example

Take a look at the following question:

This is not that straightforward because the measurements have been done to the nearest 5mm.

You can try and visualise the bounds with the aid of a number line as shown below:

So what is the lowest possible value that would be rounded to 645?

641 – rounds to 640

642 – rounds to 640

643 – rounds to 645, however you need to be half ways between 640mm and 645mm to round up to 645mm so the lower bound of the length would be 642.5mm and similarly the lower bound for the width would be 397.5mm

This means that the lower bound for the area would be 642.5 x 397.5 = 255,393.75 and this to 3 significant figures would be 255,000 mm².

Question Practice

Try the following questions on your own before looking at the solution:

Question Practice Solutions

The upper bound for V is 218.5 and the lower bound is 217.5

The upper bound for R is 12.65 and the lower bound is 12.55

In order to determine the lower bound of I, this variable needs to be made the subject:

I=\frac{V}{R}

Now because you want the lower bound of I you do not divide the lower bound numbers.

Doing so will give:

I=\frac{217.5}{12.55}=17.33 \mathrm{amps}

The lower bound is found by dividing by the upper bound for R which gives:

I=\frac{217.5}{12.65}=17.19 \mathrm{amps}

As you will see a smaller value has been obtained so it is important that when calculating lower bounds (or upper bounds) you are dividing the correct numbers.

For foundation GCSE maths you need to be able to calculate the error interval or in other words, the upper and lower bounds for a particular value and no more. The higher paper can see questions which are more complicated and to be more like the examples shown in this article.

Once you have got your head around upper and lower bounds, the questions do get easier with practice.

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