GCSE Maths Number Rounding Numbers

Upper and Lower Bounds

# Upper and Lower Bounds

Here we will learn about upper and lower bounds including how to find upper and lower bounds and how to use them to solve problems.

There are also upper and lower bounds worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What are upper and lower bounds?

Upper and lower bounds are the maximum and minimum values that a number could have been before it was rounded. They can also be called limits of accuracy.
The upper and lower bounds can be written using error intervals

E.g.
A rectangle has a width of 4.3 cm rounded to 1 decimal place and a length of 6.4 cm rounded to 1 decimal place.

Let’s look at the length:
The smallest number that will round up to give 6.4 is 6.35, this is the lower bound.
The largest number that will round down to give 6.4 is 6.44999… so we say that 6.45 is the upper bound.

Therefore we can write:

$6.35 \mathrm{~cm} \leq l<6.45 \mathrm{~cm}$

Let’s look at the width:
The smallest number that will round up to 4.3 is 4.25, this is the lower bound.
The largest number that will round down to give 4.3 is 4.34999… so we say that 4.35 is the upper bound.

Therefore we can write

$4.25 \mathrm{~cm} \leq w<4.35 \mathrm{~cm}$

We usefor the lower bound as 4.25 would round up to 4.3 but we have to use < for the upper bound as 4.35 would round up to 4.4, not down to 4.3.

These upper and lower bounds of the length and width can then be used to find the upper and lower bounds of the perimeter and area of the rectangle.

Step by step guide: Error intervals

## How to find upper and lower bounds

In order to find the upper and lower bounds of a rounded number:

1. Identify the place value of the degree of accuracy stated.
2. Divide this place value by 2.
3. Add this amount to the given value to find the upper bound, subtract this amount from the given value to find the lower bound.

You can then write these as an error interval,
$\text{Lower bound} \leq x < \text{Upper bound}$

## Upper and lower bound examples

### Example 1: finding upper and lower bounds

A number was given as 38.6 to 3 significant figures. Find the upper and lower bounds of the number.

1. Identify the place value of the degree of accuracy stated.

The place value of the degree of accuracy is 0.1.

2 Divide this place value by 2.

$0.1 ÷ 2 = 0.05$

3 Add this amount to the given value to find the upper bound, subtract this amount from the given value to find the lower bound.

$\text{Lower bound} \leq x < \text{Upper bound}$

Upper bound =  38.6 + 0.05 = 38.65.

Lower bound = 38.6 – 0.05 = 38.55.

$\text{Error interval: } 38.55\leq y< 38.65$

### Example 2: finding upper and lower bounds

The population of a town, y, is given as 280000 people rounded to the nearest 10000. Find the upper and lower bounds of y and state the error interval.

The place value of the degree of accuracy is 10000.

$10000 ÷ 2 = 5000$

$\text{Lower bound} \leq x < \text{Upper bound}$

Upper bound =  280000 + 5000 = 285000

Lower bound = 280000 – 5000 = 275000

$\text{Error interval: } 275000\leq y< 285000$

## How to use upper and lower bounds in calculations

In order to use the upper and lower bounds in calculations:

1. Find the error intervals of the rounded numbers.
2. If adding or multiplying – group together corresponding bounds, if subtracting or dividing – group together opposite bounds.

\begin{aligned} \text{Addition: } X_{LB}+Y_{LB}\leq X+Y < X_{UB}+Y_{UB} \\\\ \text{Multiplication: } X_{LB} \times Y_{LB}\leq X \times Y < X_{UB} \times Y_{UB} \\\\ \text{Subtraction: } X_{LB}-Y_{UB}\leq X-Y < X_{UB}-Y_{LB} \\\\ \text{Division: } X_{LB} \div Y_{UB}\leq X \div Y < X_{UB} \div Y_{LB} \end{aligned}

3 Perform the required calculation and write an error interval if required.

## Calculations with upper and lower bound examples

### Example 3: using upper and lower bounds in an addition

A rectangle has a width w of 3 cm to the nearest centimetre and a length l of 8.6 cm to the nearest millimetre. Find the upper and lower bounds of the perimeter, P, of the rectangle.

The error interval of the width, w, is

$2.5\mathrm{~cm} \leq w < 3.5\mathrm{~cm}$

The error interval of the length, l, is

$8.55\mathrm{~cm} \leq l < 8.65\mathrm{~cm}$

For an addition, group together corresponding bounds.

$X_{LB}+Y_{LB}\leq X+Y < X_{UB}+Y_{UB}$

By adding the lower bounds for the sides, we find the lower bound for the perimeter.

By adding the upper bounds for the sides, we find the upper bound for the perimeter.

For the perimeter of the rectangle, this becomes:

\begin{aligned} 2 \left ( w_{LB} + l_{LB} \right ) &\leq P < 2 \left ( w_{UB} + l_{UB} \right )\\\\ 2 \left ( 2.5 + 8.55 \right ) &\leq P < 2 \left ( 3.5 +8.65 \right ) \end{aligned}

Lower bound of P = 2(2.5+8.55) = 22.1 cm

Upper bound of P = 2(3.5+8.65) = 24.3 cm

### Example 4: using upper and lower bounds in a multiplication

A parallelogram has a base b of 5.64 m to 2 decimal places and a perpendicular height h of  2.3 m to 2 significant figures. Find the upper and lower bounds of the area A of the parallelogram.

The error interval of the base, b, is

$5.635 \mathrm{~m} \leq b < 5.645 \mathrm{~m}$

The error interval of the perpendicular height, h, is

$2.25 \mathrm{~m} \leq h < 2.35 \mathrm{~m}$

For a multiplication, group together corresponding bounds.

$X_{LB} \times Y_{LB}\leq X \times Y < X_{UB} \times Y_{UB}$

By multiplying the lower bounds, we find the lower bound for the area.
By multiplying the upper bounds, we find the upper bound for the area.

For the area of the parallelogram, this becomes:

\begin{aligned} b_{LB} \times h_{LB} & \leq A < b_{UB} \times h_{UB} \\\\ 5.635 \times 2.25 & \leq A < 5.645 \times 2.35 \end{aligned}

Lower bound of A = 5.635 × 2.25 = 12.67875 m2

Upper bound of A = 5.645 × 2.35 = 13.26575 m2

### Example 5: using upper and lower bounds in a subtraction

A plank of wood is 3 metres in length to the nearest centimetre. A carpenter cuts a section from the plank measuring 1.2 metres to the nearest centimetre. Find the error interval for the length x of the remaining wood.

The error interval of the plank, is

$2.995 \mathrm{~m} \leq \text{plank} < 3.005 \mathrm{~m}$

The error interval of the cut section is

$1.195 \mathrm{~m} \leq \text{cut section} < 1.205 \mathrm{~m}$

For a subtraction, group together opposite bounds.

$X_{LB}-Y_{UB}\leq X-Y < X_{UB}-Y_{LB}$

To find the lower bound of the remaining piece of wood we need to start with the lower bound of the plank and subtract the upper bound of the cut section. This will make the answer as small as possible.

To find the upper bound for the remaining piece of wood we start with the upper bound for the plank and subtract the lower bound for the cut section. This will make the answer as big as possible.

For the length of the remaining wood, this becomes:

\begin{aligned} \text{plank}_{LB} – \text{cut section}_{UB} &\leq x < \text{plank}_{UB} – \text{cut section}_{LB} \\\\ 2.995-1.205 &\leq x < 3.005-1.195 \end{aligned}

Lower bound of x = 2.995 – 1.205 = 1.79m

Upper bound of A = 3.005 – 1.195 = 1.81m

$\text{Error interval: } 1.79 \mathrm{~m} \leq x < 1.81\mathrm{~m}$

### Example 6: using upper and lower bounds in a division

A car travels 620 kilometres in 8.4 hours. Both values have been rounded to 2 significant figures. Find the upper and lower bounds of the average speed of the car, give your values to 2 decimal places.

The error interval of the distance, is

$615\mathrm{~km} \leq \text{distance} < 625\mathrm{~km}$

The error interval of the time is

$8.35 \text{ hours} \leq \text{time} <8.45 \text{ hours}$

For a division, group together opposite bounds.

$X_{LB} \div Y_{UB}\leq X \div Y < X_{UB} \div Y_{LB}$

To find the lower bound of the speed we need to start with the lower bound of the distance and divide by the upper bound of the time. This is because if you divide a bigger number, you are left with a smaller answer.

To find the upper bound for the speed, we start with the upper bound for the distance and divide by the lower bound for the time. This will leave us with a bigger answer.

For the average speed of the car, this becomes:

\begin{aligned} \text{distance}_{LB} \div \text{time}_{UB} &\leq \text{speed} < \text{distance}_{UB} – \text{time}_{LB} \\\\ 615 \div 8.45 &\leq \text{speed} < 625 \div 8.35 \end{aligned}

Lower bound of the average speed = 615 ÷ 8.45= 72.78 km/h (2 d.p.)

Upper bound of the average speed  = 625 ÷ 8.35 = 74.85 km/h (2 d.p.)

### Example 7: using upper and lower bounds to find a suitable degree of accuracy

The mass m of a rock is given as 53 kg to the nearest kilogram. The volume v of the rock is measured to be 0.026 cubic metres correct to 2 significant figures. By considering bounds, find the density, d, of the rock to a suitable degree of accuracy.

The error interval of the mass, is

$52.5\mathrm{~kg} \leq m < 53.5\mathrm{~kg}$

The error interval of the volume is

$0.0255\mathrm{~m}^{3} \leq v < 0.0265\mathrm{~m}^{3}$

For a division, group together opposite bounds.

$X_{LB} \div Y_{UB}\leq X \div Y < X_{UB} \div Y_{LB}$

For the density of the rock, this becomes:

\begin{aligned} \text{mass}_{LB} \div \text{volume}_{UB} &\leq d < \text{mass}_{UB} \div \text{volume}_{LB} \\\\ 52.5 \div 0.0265 &\leq d < 53.5 \div 0.0255 \end{aligned}

Lower bound of the density = 52.5 ÷ 0.0265 = 1981.13 kg/m3 (2 d.p.)

Upper bound of the density = 53.5 ÷ 0.0255 = 2098.04 kg/m3 (2 d.p.)

Both bounds round to 2000, therefore the density of the rock is 2000 kg/m3.

### Common misconceptions

• Upper bound is incorrect
It is very common for students to put an incorrect upper bound
E.g.
If a length has been rounded to 56 cm to the nearest cm, an error maybe to write the upper bound as 56.4 cm.
The upper bound must be bigger than 56.4 cm as 56.49 also rounds down to 56, so does 56.499999999….
The upper bound in this case is 56.5 cm.

### Practice upper and lower bound questions

1. A door was measured as 92 cm wide to the nearest centimetre.

Find the upper and lower bounds of the width of the door.

Upper bound = 92.49 cm , lower bound = 91.49 cm

Upper bound = 93 cm , lower bound = 91 cm

Upper bound = 92.5 cm , lower bound = 91.5 cm

Upper bound = 92.55 cm , lower bound = 91.55 cm

The degree of accuracy is the units. We can divide this place value by two; add to the given measure for the upper bound and subtract from the given measure for the lower bound.

2. The number of pupils, n , in a school was approximated to 1400 to the nearest 100 .

Find the upper and lower bounds of n and write the error interval.

1300 \leq n < 1500

1350 \leq n < 1450

1399 \leq n < 1449

1400 \leq n < 1499

The degree of accuracy is the hundreds. We can divide this place value by two; add to the given measure for the upper bound and subtract from the given measure for the lower bound.

3. Two boxes of height 21.4 cm , measured to the nearest millimetre, are stacked on top of each other.

Find the upper and lower bounds of the combined height.

Upper bound = 42.94 cm , lower bound = 42.75 cm

Upper bound = 42.8 cm , lower bound = 42.9 cm

Upper bound = 42.9 cm , lower bound = 42.7cm

Upper bound = 42 cm , lower bound = 43 cm

The error interval of the height, h , is

21.35 cm \leq h < 21.45 cm

By adding the lower bounds for the height, we find the lower bound for the combined height.

By adding the upper bounds for the height, we find the upper bound for the combined height.

4. A triangle has a base of 6 cm and a height of 5 cm , both are rounded to the nearest centimetre.

Find the upper and lower bounds of the area of the triangle.

Upper bound = 17.85 cm^2 , lower bound = 12.35 cm^2

Upper bound = 15 cm^2 , lower bound = 10 cm^2

Upper bound = 17.875 cm^2 , lower bound = 12.375 cm^2

Upper bound = 17.8 cm^2 , lower bound = 12.3 cm^2

The error interval of the base, b , is

5.5 cm \leq b < 6.5 cm

The error interval of the height, h , is

4.5 cm \leq h < 5.5 cm

By using the lower bounds for the base and height, we find the lower bound for the area.

By using the upper bounds for the base and height, we find the upper bound for the area.

5. Sarah thinks that she had about \pounds 50 in her purse at the start of the day, to the nearest \pounds 10 . After a shopping trip she estimates that she has spent \pounds 34 to the nearest \pounds 1 .

Find the upper and lower bounds of the money remaining in her purse.

Upper bound = \pounds 15 , lower bound = \pounds 6

Upper bound = \pounds 21.50 , lower bound = \pounds 10.50

Upper bound = \pounds 16 , lower bound = \pounds 6

Upper bound = \pounds 20.50 , lower bound = \pounds 11.50

The error interval of the starting amount, P , is

£45 \leq P < £55

The error interval of the money spent, S , is

£33.50 \leq S < £34.50

For the upper bound of money remaining, we subtract the lower bound of S from the upper bound of P .

For the lower bound of money remaining, we subtract the upper bound of S from the lower bound of P .

6. Dean cycles approximately 68 km , to the nearest kilometre. Deans cycling app recorded his average speed as 28.00 km/h to the nearest hundredth.

By considering bounds, find the time taken for Dean to complete his journey to a suitable degree of accuracy.

2.4 hours

41 minutes

25 minutes

2.5 hours

Upper bound = 2.447 hours (3 d.p.), lower bound = 2.410 hours (3 d.p.).

Both round to 2.4 to 2 s.f., therefore, the journey took 2.4 hours.

### Upper and lower bounds GCSE questions

1. The weight, w of a parcel is recorded as 440g to the nearest 10g . Write down the error interval for the weight of the parcel.

(2 Marks)

Lower bound 435g , Upper bound 445g

(1)

435g \leq w < 445g 1 mark

(1)

2. (a) A set of books each have a width x of 1.4cm to the nearest mm . Write down the error interval for the width of one book.

(b)  Kevin’s bookshelf has a length of 36cm to the nearest cm . Calculate the maximum number of these books that Kevin can fit on his shelf.

(5 Marks)

(a)

Lower bound 1.35cm , upper bound 1.45cm

(1)

1.35cm \leq x < 1.45cm 1 mark

(1)

(b)

Upper bound of shelf 36.5cm

(1)

Maximum number of books:

36.5 \div 1.35 = 27.03 1 mark

(1)

Maximum number of books is 27 .

(1)

3.  A room has a width of 4.55m correct to 2 dp and a length of 3.984m correct to 3 d.p.

By considering bounds, find the area of the room to a suitable degree of accuracy.

(5 Marks)

Lower bound of width 4.545m , upper bound of width 4.555m

(1)

Lower bound of length 3.9835m , upper bound of length 3.9845m

(1)

Lower bound of area:

4.545 \times 3.9835 = 18.1050 1 mark

(1)

Upper bound of area:

4.555 \times 3.9845 = 18.1494 1 mark

(1)

Area is:

18.1m^{2} 1 mark

(1)

## Learning checklist

You have now learned how to:

• apply and interpret limits of accuracy when rounding
• find upper and lower bounds of rounded values

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