Upper and Lower Bounds | Questions and Revision | MME

# Upper and Lower Bounds Worksheets, Questions and Revision

Level 4-5

## 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. There are 3 key skills that you need for upper and lower bound questions.

Make sure you are happy with the following topics before continuing.

KS3 Level 4-5

## Skill 1: Upper and Lower Bounds

Upper and lower bounds can also be represented as an error interval which is often given as an inequality.

Example: $57.7$ has been rounded to $1$ decimal place. Work out the upper and lower bounds (or error interval) of this value.

To calculate the upper and lower bound we need to use the size of the interval which is $1$ decimal place ($0.1$).

Next we need to divide the size of the interval by $2$ to get half the interval:

$\dfrac{0.1}{2} = \textcolor{red}{0.05}$

Step 1: Calculate the lower bound.

For the lower bound, we subtract half the interval.

Lower bound $= 57.7 - 0.05 = 57.65$

Step 2: Calculate the upper bound.

For the upper bound, we add half the interval.

Upper bound $= 57.7 + 0.05 = 57.75$

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 rounds up to $57.7$, $57.75$ is not included because it does not round down to $57.7$. This will always be the case in an error interval.

Revise Inequalities here

KS3 Level 4-5
Level 6-7

## Skill 2: Maximum and Minimum Values

When we do a calculation using rounded numbers, there will be a difference between the calculated value and the actual value. We use the upper and lower bounds to find the maximum and minimum values the calculation can be.

Example: A field is measured to be $34$m long and $28$m wide, to the nearest metre. Calculate the minimum and maximum values for the area of the field.

Step 1: Find the upper and lower bounds (error interval) for the length and width:

$33.5 \leq \text{length} < 34.5$

$27.5 \leq \text{width} < 28.5$

Step 2: Find the minimum area, by multiplying the lower bounds for length and width:

$\text{Minimum Area} = 33.5 \times 27.5 = 921.25$ m$^2$

Find the maximum area, by multiplying the upper bounds for length and width:

$\text{Minimum Area} = 34.5 \times 28.5 = 983.25$ m$^2$

So, our error interval for the area of the field is

$921.25$ m$^2 \leq \text{Area } < 983.25$ m$^2$

Level 6-7

## Skill 3: Truncation

When you truncate a number it is effectively always rounded down, so if a number is truncated to a certain unit, the actual value of the number can be up to a whole unit bigger than the truncated number, but no smaller. Truncating is just chopping part of the number off.

Example: What is $9.876$ truncated to $1$ decimal place?

Step 1: We determine our cut-off point, which is after the $1$st decimal place, between the $8$ and the $7$.

Step 2: Any numbers after the cut-off point we ignore (chop off), and what ‘s left is our final answer. So,

$9.876$ truncated to $1$ decimal place is $9.8$

Level 4-5
Level 6-7

## Example 1: Maximum and Minimum

A motorbike travels a distance of $110$m to the nearest $10$ metres, in a time of $5$ seconds to the nearest second. Calculate the maximum and minimum values for the speed, $s$, in metres per second (m/s), using the formula

$s = \dfrac{d}{t}$

where $d$ is the distance and $t$ is the time.

[4 marks]

The error intervals for the distance and time are:

$105 \leq \text{distance} < 115$

$4.5 \leq \text{time} < 5.5$

The minimum speed is found by dividing the lower bound for the distance by the upper bound for the time. So

$\text{Minimum Speed} = \dfrac{105}{5.5} = 19.0909...$ m/s

The maximum speed is found by dividing the upper bound for the distance by the lower bound for time. So

$\text{Maximum Speed} = \dfrac{115}{4.5} = 25.555...$ m/s

So the error interval for the speed of the motorbike is

$19.0909...$ m/s $\leq \text{Speed} < 25.555...$ m/s

Level 6-7

## Example 2: Truncation

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

[1 mark]

In order for some number to be truncated to $402.3$, it must have began $402.3$ 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 $402.4$. We express this error interval like

$402.3 \leq w < 402.4$

Level 4-5

## GCSE Maths Revision Cards

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### 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,

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

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:

$\text{Max average speed } = \dfrac{\text{distance upper bound}}{\text{time lower bound}} = \dfrac{80.35}{1.865} = 43.1$

$\text{Min average speed } = \dfrac{\text{distance lower bound}}{\text{time upper bound}} = \dfrac{80.25}{1.875} = 42.8$

both to ($3$ sf)

### Worksheets and Exam Questions

#### (NEW) Upper And Lower Bounds Exam Style Questions - MME

Level 5-7 New Official MME

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