Bounds

What you'll learn

How to find lower and upper bounds from rounded measurements.
How to choose the correct bounds for area, perimeter, circles and formulae.
How to handle bounds with division, square roots and Pythagoras.
How to decide a “suitable degree of accuracy” using bounds.

1. Why bounds exist

When a number has been rounded, it is not usually the exact value. For example, a length written as 28 cm to the nearest cm could actually be a little smaller or a little larger.

Definition

Bounds

A lower bound is the smallest possible value before rounding.
An upper bound is the largest boundary value before rounding.
The real value is normally written as an interval, such as $\le l < 28.5$ .

The upper bound is written with < because the exact upper boundary would round up to the next value. In calculations, we still use that boundary value to find the upper bound of an answer.

Finding the half-gap

Look at the rounding amount, then halve it.

Nearest cm means a gap of 1 cm, so the half-gap is 0.5 cm.
Nearest 5 m means a gap of 5 m, so the half-gap is 2.5 m.
Correct to 2 decimal places means a gap of 0.01, so the half-gap is 0.005.
Nearest mm, when working in cm, means a gap of 0.1 cm, so the half-gap is 0.05 cm.

Example

Finding bounds from rounding

A plank is 28 cm to the nearest cm. A strip is 7.6 cm to the nearest mm. Find the bounds for each measurement.

Number-line intervals show how each rounded measurement sits halfway between its lower and upper boundaries.

For 28 cm to the nearest cm, subtract and add 0.5 cm.
Write the interval for the length:
$\le l < 28.5$
For 7.6 cm to the nearest mm, remember that 1 mm is 0.1 cm, so the half-gap is 0.05 cm.
Write the interval for the width:
$\le w < 7.65$

Common Mistake

Mixing up mm and cm

If a measurement is written in cm but rounded to the nearest mm, use 0.05 cm as the half-gap, not 0.5 cm.

2. Bounds in geometry

For positive lengths, areas and perimeters, the idea is usually straightforward:

Key Idea

Use small values for a lower answer and large values for an upper answer

For addition and multiplication with positive measurements, the lower bound comes from lower measurements, and the upper bound comes from upper measurements.

Example

Rectangle: upper perimeter and lower area

A rectangle has length 19 cm to the nearest cm and width 6.4 cm to the nearest mm. Find the upper bound for its perimeter and the lower bound for its area.

The rectangle diagram highlights that upper perimeter uses the largest possible length and width, while lower area uses the smallest possible length and width.

Write the bounds for both measurements, all in cm:
$18.5≤l<19.56.35≤w<6.45\begin{aligned} 18.5 &\le l < 19.5\\ 6.35 &\le w < 6.45 \end{aligned}$
For the upper bound of the perimeter, use the upper length and upper width:
$Pupper=2(19.5+6.45)=51.9P_{\text{upper}} = 2(19.5+6.45)=51.9$
For the lower bound of the area, use the lower length and lower width:
$Alower=18.5×6.35=117.475A_{\text{lower}} = 18.5 \times 6.35=117.475$
The answers are 51.9 cm for the upper perimeter and 117.475 cm² for the lower area.

Circles

For circles, the radius controls both circumference and area.

Use:

C=2πrC=2\pi r

and

A=πr2A=\pi r^2

Example

Circle bounds

A circle has radius 7.2 cm correct to 1 decimal place. Find the lower bound for its circumference and the upper bound for its area, in terms of $π\pi$ .

The circle’s radius interval controls both the lower circumference and the upper area.

Since 7.2 is correct to 1 decimal place, the half-gap is 0.05 cm.
Write the radius bounds:
$\le r < 7.25$
For the lower circumference, use the lower radius:
$Clower=2π(7.15)=14.3πC_{\text{lower}}=2\pi(7.15)=14.3\pi$
For the upper area, use the upper radius:
$Aupper=π(7.25)2=52.5625πA_{\text{upper}}=\pi(7.25)^2=52.5625\pi$

Tip

Keep $π$ exact if asked

If the question says “in terms of $π\pi$ ”, leave $π\pi$ in your answer. Do not multiply by 3.142 unless a decimal answer is requested.

3. Bounds in formulae

Many bounds questions use formulae such as $v=stv=\frac{s}{t}$ or $V = I R$ .

For positive values:

To make a product bigger, use upper bound times upper bound.
To make a product smaller, use lower bound times lower bound.
To make a fraction bigger, use the upper numerator and the lower denominator.
To make a fraction smaller, use the lower numerator and the upper denominator.

Common Mistake

Division direction

For an upper bound of a fraction, do not use the upper bound of the denominator. A smaller denominator makes the fraction larger.

Example

Upper bound for speed

A distance is 6.48 m correct to 2 decimal places. A time is 2.37 s correct to 2 decimal places. Given $v=stv=\frac{s}{t}$ , find the upper bound for $v$ to 3 decimal places.

For the upper bound of speed, choose the largest possible distance and the smallest possible time.

Find the bounds:
$6.475≤s<6.4852.365≤t<2.375\begin{aligned} 6.475 &\le s < 6.485\\ 2.365 &\le t < 2.375 \end{aligned}$
To maximise $v$ , use the largest distance and the smallest time.
Calculate the upper bound:
$vupper=6.4852.365=2.742071…v_{\text{upper}}=\frac{6.485}{2.365}=2.742071\ldots$
To 3 decimal places, the upper bound is 2.742.

4. Bounds with Pythagoras

Definition

Pythagoras' theorem

In a right-angled triangle, if $c$ is the hypotenuse, then $c^2=a^2+b^2$ . The hypotenuse is the longest side, opposite the right angle.

In a right-angled triangle, the hypotenuse is the side opposite the right angle.

If you are finding the lower bound for the hypotenuse, use the lower bounds of the two shorter sides.

But if you are finding a shorter side, rearrange first:

b=c2−a2b=\sqrt{c^2-a^2}

Now subtraction matters. To make $b$ as small as possible, use the lower bound of $c$ and the upper bound of $a$ .

Example

Lower bound for a missing side

In a right-angled triangle, $a = 4.6$ cm correct to the nearest mm and $c = 10$ cm correct to the nearest cm. Find the lower bound for $b$ to 1 decimal place, where $c$ is the hypotenuse.

To minimise the missing shorter side, use the smallest possible hypotenuse and the largest possible known shorter side.

Write the relevant bounds:
$4.55≤a<4.659.5≤c<10.5\begin{aligned} 4.55 &\le a < 4.65\\ 9.5 &\le c < 10.5 \end{aligned}$
Rearrange Pythagoras:
$b=c2−a2b=\sqrt{c^2-a^2}$
For the lower bound of $b$ , use $c = 9.5$ and $a = 4.65$ .
Calculate:
$blower=9.52−4.652=68.6275=8.284…b_{\text{lower}}=\sqrt{9.5^2-4.65^2}=\sqrt{68.6275}=8.284\ldots$
To 1 decimal place, the lower bound is 8.3 cm.

5. Suitable degree of accuracy

Sometimes you are asked to find a value “to a suitable degree of accuracy”. This means you must calculate both the lower and upper bounds of the final answer, then see what rounding is guaranteed.

Definition

Suitable degree of accuracy

A suitable degree of accuracy is a rounding level where every possible value between the lower and upper bounds rounds to the same answer.

Example

Choosing a suitable accuracy

Given $p=FAp=\frac{F}{A}$ , where $F = 30.42$ correct to 2 decimal places and $A = 5.67$ correct to 3 significant figures, find $p$ to a suitable degree of accuracy.

The final answer is only suitable if every value between the lower and upper bounds rounds to the same value.

Write the bounds:
$30.415≤F<30.4255.665≤A<5.675\begin{aligned} 30.415 &\le F < 30.425\\ 5.665 &\le A < 5.675 \end{aligned}$
For the lower value of $p$ , use lower $F$ and upper $A$ . For the upper value of $p$ , use upper $F$ and lower $A$ .
Calculate both bounds:
$plower=30.4155.675≈5.359pupper=30.4255.665≈5.371\begin{aligned} p_{\text{lower}}&=\frac{30.415}{5.675}\approx 5.359\\ p_{\text{upper}}&=\frac{30.425}{5.665}\approx 5.371 \end{aligned}$
These do not both round to the same value to 2 decimal places, because they give 5.36 and 5.37.
They do both round to 5.4 to 2 significant figures, so a suitable answer is $p = 5.4$ to 2 significant figures.

Exam technique

In the exam

Underline the rounding phrase first: nearest cm, nearest 5 m, 2 d.p., or 3 significant figures.
Write the bounds for each given value before substituting into a formula.
For fractions and subtractions, think carefully about which choice makes the final answer bigger or smaller.

Self review

Check yourself

What are the bounds for 58.0 cm correct to the nearest millimetre?
For $Q=xyQ=\frac{x}{y}$ , with positive $x$ and $y$ , which bounds give the upper value of $Q$ ?
In $b=c2−a2b=\sqrt{c^2-a^2}$ , which bound of $a$ would you use to find the lower bound of $b$ ?

Bounds

What you'll learn

How to find lower and upper bounds from rounded measurements.
How to choose the correct bounds for area, perimeter, circles and formulae.
How to handle bounds with division, square roots and Pythagoras.
How to decide a “suitable degree of accuracy” using bounds.

1. Why bounds exist

When a number has been rounded, it is not usually the exact value. For example, a length written as 28 cm to the nearest cm could actually be a little smaller or a little larger.

Definition

Bounds

A lower bound is the smallest possible value before rounding.
An upper bound is the largest boundary value before rounding.
The real value is normally written as an interval, such as $\le l < 28.5$ .

The upper bound is written with < because the exact upper boundary would round up to the next value. In calculations, we still use that boundary value to find the upper bound of an answer.

Finding the half-gap

Look at the rounding amount, then halve it.

Nearest cm means a gap of 1 cm, so the half-gap is 0.5 cm.
Nearest 5 m means a gap of 5 m, so the half-gap is 2.5 m.
Correct to 2 decimal places means a gap of 0.01, so the half-gap is 0.005.
Nearest mm, when working in cm, means a gap of 0.1 cm, so the half-gap is 0.05 cm.

Example

Finding bounds from rounding

A plank is 28 cm to the nearest cm. A strip is 7.6 cm to the nearest mm. Find the bounds for each measurement.

Number-line intervals show how each rounded measurement sits halfway between its lower and upper boundaries.

For 28 cm to the nearest cm, subtract and add 0.5 cm.
Write the interval for the length:
$\le l < 28.5$
For 7.6 cm to the nearest mm, remember that 1 mm is 0.1 cm, so the half-gap is 0.05 cm.
Write the interval for the width:
$\le w < 7.65$

Common Mistake

Mixing up mm and cm

If a measurement is written in cm but rounded to the nearest mm, use 0.05 cm as the half-gap, not 0.5 cm.

2. Bounds in geometry

For positive lengths, areas and perimeters, the idea is usually straightforward:

Key Idea

Use small values for a lower answer and large values for an upper answer

For addition and multiplication with positive measurements, the lower bound comes from lower measurements, and the upper bound comes from upper measurements.

Example

Rectangle: upper perimeter and lower area

A rectangle has length 19 cm to the nearest cm and width 6.4 cm to the nearest mm. Find the upper bound for its perimeter and the lower bound for its area.

The rectangle diagram highlights that upper perimeter uses the largest possible length and width, while lower area uses the smallest possible length and width.

Write the bounds for both measurements, all in cm:
$18.5≤l<19.56.35≤w<6.45\begin{aligned} 18.5 &\le l < 19.5\\ 6.35 &\le w < 6.45 \end{aligned}$
For the upper bound of the perimeter, use the upper length and upper width:
$Pupper=2(19.5+6.45)=51.9P_{\text{upper}} = 2(19.5+6.45)=51.9$
For the lower bound of the area, use the lower length and lower width:
$Alower=18.5×6.35=117.475A_{\text{lower}} = 18.5 \times 6.35=117.475$
The answers are 51.9 cm for the upper perimeter and 117.475 cm² for the lower area.

Circles

For circles, the radius controls both circumference and area.

Use:

C=2πrC=2\pi r

and

A=πr2A=\pi r^2

Example

Circle bounds

A circle has radius 7.2 cm correct to 1 decimal place. Find the lower bound for its circumference and the upper bound for its area, in terms of $π\pi$ .

The circle’s radius interval controls both the lower circumference and the upper area.

Since 7.2 is correct to 1 decimal place, the half-gap is 0.05 cm.
Write the radius bounds:
$\le r < 7.25$
For the lower circumference, use the lower radius:
$Clower=2π(7.15)=14.3πC_{\text{lower}}=2\pi(7.15)=14.3\pi$
For the upper area, use the upper radius:
$Aupper=π(7.25)2=52.5625πA_{\text{upper}}=\pi(7.25)^2=52.5625\pi$

Tip

Keep $π$ exact if asked

If the question says “in terms of $π\pi$ ”, leave $π\pi$ in your answer. Do not multiply by 3.142 unless a decimal answer is requested.

3. Bounds in formulae

Many bounds questions use formulae such as $v=stv=\frac{s}{t}$ or $V = I R$ .

For positive values:

To make a product bigger, use upper bound times upper bound.
To make a product smaller, use lower bound times lower bound.
To make a fraction bigger, use the upper numerator and the lower denominator.
To make a fraction smaller, use the lower numerator and the upper denominator.

Common Mistake

Division direction

For an upper bound of a fraction, do not use the upper bound of the denominator. A smaller denominator makes the fraction larger.

Example

Upper bound for speed

A distance is 6.48 m correct to 2 decimal places. A time is 2.37 s correct to 2 decimal places. Given $v=stv=\frac{s}{t}$ , find the upper bound for $v$ to 3 decimal places.

For the upper bound of speed, choose the largest possible distance and the smallest possible time.

Find the bounds:
$6.475≤s<6.4852.365≤t<2.375\begin{aligned} 6.475 &\le s < 6.485\\ 2.365 &\le t < 2.375 \end{aligned}$
To maximise $v$ , use the largest distance and the smallest time.
Calculate the upper bound:
$vupper=6.4852.365=2.742071…v_{\text{upper}}=\frac{6.485}{2.365}=2.742071\ldots$
To 3 decimal places, the upper bound is 2.742.

4. Bounds with Pythagoras

Definition

Pythagoras' theorem

In a right-angled triangle, if $c$ is the hypotenuse, then $c^2=a^2+b^2$ . The hypotenuse is the longest side, opposite the right angle.

In a right-angled triangle, the hypotenuse is the side opposite the right angle.

If you are finding the lower bound for the hypotenuse, use the lower bounds of the two shorter sides.

But if you are finding a shorter side, rearrange first:

b=c2−a2b=\sqrt{c^2-a^2}

Now subtraction matters. To make $b$ as small as possible, use the lower bound of $c$ and the upper bound of $a$ .

Example

Lower bound for a missing side

In a right-angled triangle, $a = 4.6$ cm correct to the nearest mm and $c = 10$ cm correct to the nearest cm. Find the lower bound for $b$ to 1 decimal place, where $c$ is the hypotenuse.

To minimise the missing shorter side, use the smallest possible hypotenuse and the largest possible known shorter side.

Write the relevant bounds:
$4.55≤a<4.659.5≤c<10.5\begin{aligned} 4.55 &\le a < 4.65\\ 9.5 &\le c < 10.5 \end{aligned}$
Rearrange Pythagoras:
$b=c2−a2b=\sqrt{c^2-a^2}$
For the lower bound of $b$ , use $c = 9.5$ and $a = 4.65$ .
Calculate:
$blower=9.52−4.652=68.6275=8.284…b_{\text{lower}}=\sqrt{9.5^2-4.65^2}=\sqrt{68.6275}=8.284\ldots$
To 1 decimal place, the lower bound is 8.3 cm.

5. Suitable degree of accuracy

Definition

Suitable degree of accuracy

A suitable degree of accuracy is a rounding level where every possible value between the lower and upper bounds rounds to the same answer.

Example

Choosing a suitable accuracy

Given $p=FAp=\frac{F}{A}$ , where $F = 30.42$ correct to 2 decimal places and $A = 5.67$ correct to 3 significant figures, find $p$ to a suitable degree of accuracy.

The final answer is only suitable if every value between the lower and upper bounds rounds to the same value.

Write the bounds:
$30.415≤F<30.4255.665≤A<5.675\begin{aligned} 30.415 &\le F < 30.425\\ 5.665 &\le A < 5.675 \end{aligned}$
For the lower value of $p$ , use lower $F$ and upper $A$ . For the upper value of $p$ , use upper $F$ and lower $A$ .
Calculate both bounds:
$plower=30.4155.675≈5.359pupper=30.4255.665≈5.371\begin{aligned} p_{\text{lower}}&=\frac{30.415}{5.675}\approx 5.359\\ p_{\text{upper}}&=\frac{30.425}{5.665}\approx 5.371 \end{aligned}$
These do not both round to the same value to 2 decimal places, because they give 5.36 and 5.37.
They do both round to 5.4 to 2 significant figures, so a suitable answer is $p = 5.4$ to 2 significant figures.

Exam technique

In the exam

Underline the rounding phrase first: nearest cm, nearest 5 m, 2 d.p., or 3 significant figures.
Write the bounds for each given value before substituting into a formula.
For fractions and subtractions, think carefully about which choice makes the final answer bigger or smaller.

Self review

Check yourself

What are the bounds for 58.0 cm correct to the nearest millimetre?
For $Q=xyQ=\frac{x}{y}$ , with positive $x$ and $y$ , which bounds give the upper value of $Q$ ?
In $b=c2−a2b=\sqrt{c^2-a^2}$ , which bound of $a$ would you use to find the lower bound of $b$ ?

Bounds

What you'll learn

1. Why bounds exist

Finding the half-gap

2. Bounds in geometry

Circles

3. Bounds in formulae

4. Bounds with Pythagoras

5. Suitable degree of accuracy

Recap questions

You've reached the end

Bounds

What you'll learn

1. Why bounds exist

Finding the half-gap

2. Bounds in geometry

Circles

3. Bounds in formulae

4. Bounds with Pythagoras

5. Suitable degree of accuracy

Recap questions

You've reached the end

Bounds

What you'll learn

1. Why bounds exist

Finding the half-gap

2. Bounds in geometry

Circles

3. Bounds in formulae

4. Bounds with Pythagoras

5. Suitable degree of accuracy

Recap questions

You've reached the end

Bounds Revision Guide

Bounds

What you'll learn

1. Why bounds exist

Finding the half-gap

2. Bounds in geometry

Circles

3. Bounds in formulae

4. Bounds with Pythagoras

5. Suitable degree of accuracy

Recap questions

You've reached the end

Bounds Revision Guide