Upper and Lower Bounds | Questions and Revision | MME

Upper and Lower Bounds Worksheets, Questions and Revision

Level 4 Level 5

What you need to know

Upper and Lower Bounds

Upper and lower bounds are the upper and lower values that a number can have as a result of rounding or approximating that number. So if a table is measured to be 100 cm long to the nearest 10 cm, it means it is correct within a range of 10 cm, so the lower bound would be 95 cm and the upper bound 105 cm.

Having a good knowledge of rounding is essential for upper and lower bounds so if needed please visit our rounding revision page.

Note: Upper and lower bounds can also be called an error interval which is often given as an inequality. Visit our inequalities revision page for further notes on this topic.

Truncating and Rounding

Upper and lower bounds is about the consequences of two processes – rounding and truncating. When you truncate a number it is effectively always rounded down. Truncating is just chopping part of the number off.

For example, if we truncate 35.7289 to 1dp, we just count along 1 decimal place and then cut off the rest of the number. So, 35.7289 truncated to 1dp is 35.7. Same goes for 3 decimal places, so 35.7289 truncated to 3dp would 35.728.

Rounding and truncating are useful tools, but we must be aware that they come with consequences. If you only gave the number 57.7 to someone and told them it had been rounded to 1 decimal place, they couldn’t possibly know whether the original number was 57.653 or 57.72 or 57.6834 or 57.7468 or… and this is where bounds help.

Calculating Upper and Lower Bounds

So if 57.7 has been rounded to 1 decimal place we can work out the upper and lower bound (or error interval) of this value.

Firstly, the lower bound. We know by the rules of rounding that 5 is the cut-off point – specifically, having a 5 in the digit after the one we’re rounding to. So, we know that any number between 57.65 and 57.7 will round up to 57.7 just fine, but any number below 57.65 will round down. This makes 57.65 the lower bound.

Secondly, the upper bound. By the rules of rounding again, any number which is between 57.7 and 57.75 will round down to 57.7, but anything which is greater than (or equal to!) 57.75 will round up. This makes 57.75 the upper bound.

So, we express our error interval for $x$ like

$57.65 \leq x < 57.75$

Note: whilst 57.65 is included in the interval because it does round to 57.7, 57.75 is not included because it does not itself round down to 57.7. This will always be the case, i.e. the inequality symbols will always be this way round when expressing an error interval.

Example 1: Basic Bounds

The weight of a cat has been measured at 209 ounces to the nearest ounce. Work out the interval within which $w$, the weight of the cat, lies.

Firstly, the lower bound. In order to round up to 209 to the nearest whole number, the original weight of the cat must have been bigger or equal to 208.5. Any lower, and we’ll be rounding down to 208.

Next, the upper bound. In order to round down to 209, the weight of the cat must have been strictly less than 209.5. Any value which was equal to 209.5 or higher and it would’ve rounded up to 210.

So, the error bounds are 208.5 and 209.5, and we express the interval like

$208.5 \leq w < 209.5$

Note:when we truncate, the rules for finding error bounds are different. see the next example.

Example 2: Bounds and Truncation

The weight of a dog has been truncated to 402.3 ounces to 1dp. Work out the interval within which $w$, the weight of the dog, lies.

In order for some number to be truncated to 402.3, it must have begun 402.3 followed by some other digits. Maybe 402.31, 402.395, 402.35467, and so on. As long as the number begins “402.3” then we’re fine.

In other words, anything that is greater than or equal to 402.3 but less than 402.4 (at which point it would just truncate to 402.4). We express this error interval like

$402.3 \leq w < 402.4$

This will always be the case for truncations – the first part will be the number itself, and the second part will be that number with the last digit increased by 1.

Example 3: Upper and Lower Bounds

The base and height of a triangle have been measured to 2 significant figures to be 12cm and 5.0cm respectively. Find the lower limit for area of the triangle.

The formula for the area of a triangle is $\dfrac{1}{2}bh$, so for the minimum area of the triangle, we will need the smallest possible values of both $b$ and $h$.

The lower bound for $b$ when rounding to 12cm to 2 significant figures is 11.5cm.

The lower bound for $h$ when rounding to 5.0cm to 2 significant figures is 4.95cm.

Therefore, the lower bound for the area of the triangle is

$\dfrac{1}{2} \times 11.5 \times 4.95 = 28.4625 \text{cm}^2$.

Example Questions

First, consider the lower bound.

In order for the capacity of the jug to be measured as 5.43 litres to 2 decimal places, the lower bound must be greater than or equal to 5.425 litres. Any lower and it would’ve rounded down to 5.42 litres.

Next, the upper bound.

In order to round down to 5.43, the capacity of the jug must have been strictly less than 5.435. Any value which was equal to 5.435 or higher and it would’ve rounded up to 5.44.

Our lower bound is 5.425 and our upper bound is 5.435, so we express our answer as,

$5.425 \leq C < 5.435$

The interval is determined by the bounds $\pm 0.5$ cm of her measured height.

We use $-0.5$ cm for the lower bound and $+0.5$ cm for the upper bound.

$175-0.5=174.5$
$175+0.5=175.5$

We can now create our interval, remembering that the lower bound has a non-strict inequality and the upper bound has a strict inequality,

$174.5\leq h<175.5$

The interval is determined by the bounds $\pm £0.005$ billion of the projected cost

We use $-0.005$ for the lower bound and $+0.005$ for the upper bound.

$5.45 -0.005=5.445$
$5.45+0.005=5.455$

We can now create our interval, remembering that the lower bound is a non-strict inequality and the upper bounds is a strict inequality.

$£5.445 \leq C<£5.455$

In order to make the fraction as small as possible, we want the numerator to be as a small as possible but the denominator to be as large as possible.

The lower bound for $P$ is 144.5

The upper bound for $v$ is 23.45

Therefore, the lower bound for the driving force of the car is,

$F = \dfrac{144.5}{23.45} = 6.16\text{ Newtons (3sf)}$

This is a suitable degree of accuracy as it is the same degree to which the values are given in the question.

To find the upper bound for the average speed, we will need the upper bound for the distance and the lower bound for the time. Whereas to find the lower bound for the average speed we will need the lower bound for the distance and the upper bound for the time.

The upper bound for distance is 80.35, whilst the lower bound is 80.25

The upper bound for time is 1.875, whilst the lower bound is 1.865.

Therefore, we get:

\begin{aligned}\text{Max average speed } &= \dfrac{\text{distance upper bound}}{\text{time lower bound}} = \dfrac{80.35}{1.865} = 43.1 (3sf)\\ \\ \text{Min average speed } &= \dfrac{\text{distance lower bound}}{\text{time upper bound}} = \dfrac{80.25}{1.875} = 42.8\end{aligned}

Level 5-7

Level 5-7

Level 5-7

GCSE MATHS

GCSE MATHS

GCSE MATHS

Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.