# Estimation

Here we will learn about estimation in maths including its definition and how to estimate a calculation. You’ll also learn about the differences between decimals and significant figures and how to round numbers to one significant figure.

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

## What is estimation in maths?

Estimation is when we use approximate values in a calculation to give an approximate, predicted answer rather than an exact answer.

Estimating calculations makes them more manageable by using round numbers.

E.g.
If we wanted to estimate the answer to 46.75 x 3.218 we could round both these numbers and use an estimated calculation of 50 x 3 = 150

## Estimating calculations

When we estimate the numbers in a calculation, we usually round the numbers to 1 significant figure.

E.g.
Let’s estimate 5.7 ÷ 1.8

First we estimate the value of 5.7 by rounding it to 1 significant figure (1.s.f).
We can see that the first significant figure is 5.
This means we need to use the digit to the right of the 5 which is 7, so we need to round 5.7 up to 6 which is the nearest whole number.

So 5.7 estimated is 6 as shown on the number line.

Then we follow the same process to estimate the value of 1.8 which rounded to 1 significant figure is 2.

Once we have estimated all the values in a calculation we can work out the estimated answer so

5.7 ÷ 1.8 ≈ 6 ÷ 2 = 3

Estimation is used in lots of different areas of maths in order to give an approximate answer to a question, such as estimating the perimeter of a shape or estimating the size of a number written in standard form. We also see examples of estimation all around us, such as estimating the number of bricks will be needed to build a garden wall, or working out if you have enough leftover from £5 in a supermarket to buy a bar of chocolate at the till.

### Key points about estimating calculations

• Estimation relies on using easier numbers that are near to the actual numbers involved
• To estimate we usually round to 1 significant figure
• We use estimation in many aspects of life and work – for a building job or planning a budget
• When we are estimating calculations we still have to consider using the order of operations or BIDMAS principles
•  means approximately equal to; = means exactly equal to

## How to estimate

In order to estimate a number:

1. Round each number to 1 significant figure.
2. Construct the new estimated calculation using the ‘approximately equal to’
symbol
.
3. Calculate using the rounded numbers to give an approximate answer and consider units in the answer.

## Estimation examples

### Example 1: (typical foundation GCSE question using the GCSE command word language and no units involved)

Estimate the value of:

$6.9\times 96$

1. Round numbers to one significant figure

6.9 rounded to 1 significant figure is 7

96 rounded to 1 significant figure is 100

2 Write out the new estimated calculation

$6.9\times 96\approx 7\times 100$

3 Complete the calculation using estimation

$7\times 100=700$

If we compare this to the exact answer which is 662.4 we can see that the estimated answer is an overestimate.

### Example 2: worded question

Brian buys 197 packets of crisps for a party. Each bag costs 45p. Use estimation to find the approximate cost in pounds £.

197 is 200, rounded to 1sf

45p is 50p, rounded to 1sf

$197 \times 45\approx 200 \times 50$

$200 \times 50 = 10000$

In this case we have to consider the units used

10000 is pence so we convert this to pounds (£)

10000 pence is £100

If we compare this to the exact answer which is £88.65 we can see that the estimated answer is an overestimate.

### Example 3: multi-step calculations

$\frac{42 \times 389}{81}$

42 rounded to 1sf is 40

389 rounded to 1sf is 400

81 rounded to 1sf is 80

$\frac{42 \times 389}{81} \approx \frac{40 \times 400}{80}$

$\frac{16000}{80}=200$

If we compare this to the exact answer which is 201.7 (1.d.p) we can see that the estimated answer is an underestimate.

### Example 4: estimate using BIDMAS principles

Estimate the value of 9.04 + 19.85 × 2.99 − 5.03

$9.04 + 19.85 \times 2.99 \hspace{1mm} – 5.03 \approx 9 + 20 \times 3 \hspace{1mm} – 5$

Multiply 20 and 3 together first

$9 + 60 \hspace{1mm} – 5$

Then follow the calculation from left to right, as it doesn’t matter which operation you do first with addition and subtraction.

$9 + 60 \hspace{1mm} – 5$

$9 + 60 \hspace{1mm} – 5 = 64$

If we compare this to the exact answer which is 63.3615 we can see that the estimated answer is an overestimate.

### Example 5: worded problem requiring reasoning

Estimate how many toys costing £8.48 each can be bought for £425? Is this an over or under-estimate?

£8.48 is £8, rounded to 1sf

£425 is £400, rounded to 1sf

$425 \div 8.48 \approx 400 \div 8$

$400 \div 8 = 50$

Consider units within a worded problem in context

The question considered both numbers in pounds but the answer should be a whole number of toys.

The estimate number of toys that could be bought is 50.

If we compare this to the exact answer which is 50.1179… we can see that the estimated answer is an underestimate. .

### Example 6: more complex calculation

Write down an estimate for:

$\sqrt{70}$

\begin{aligned} 7^{2}&=49 \\\\ 8^{2}&=64 \\\\ 9^{2}&=81 \\\\ 10^{2}&=100 \end{aligned}

70 lies between 64 and 81

$\sqrt{70}$

lies between

$\sqrt{64} \text { and } \sqrt{81}$

And

$\sqrt{64}=8, \sqrt{81}=9$

70 is closer to 64, being 6 units away, than 81, being 11 units away

So,

$\sqrt{70}$

is just under 8.5.

$\sqrt{70} \approx 8.4$

### Example 7: dividing by decimals

Estimate the value of:

$\frac{804 \times 2.86}{0.513}$

804 is 800, to 1sf

2.86 is 3, to 1sf

0.513 is 0.5, rounded to 1sf

$\frac{804 \times 2.86}{0.513} \approx \frac{800 \times 3}{0.5}$

\begin{aligned} 800 \times 3&=2400 \\\\ \frac{2400}{0.5}&=4800 \end{aligned}

If we compare this to the exact answer which is 4482.3 (1.d.p) we can see that the estimated answer is an overestimate.

### Common misconceptions

• It is common to find the actual calculation rather than use estimation

Estimation relies on rounding to significant figures to simplify the calculation.

An estimate of 5.8 × 2.3 is 12. The actual answer is 13.34.

• The first significant figure is the first non-zero digit

E.g.
The first significant figure of 0.00123 is the fourth digit which is 1.

E.g.
The first significant figure of 0.023 is the third digit which is 2.

E.g.
The first significant figure of 0.0000547 is the sixth digit which is 5

• Dividing by a decimal gives a larger number

E.g.

$120 \div 0.2 = 600$

• Significant figures and decimal places

Rounding to one decimal place relies on what happens after the first digit after the decimal point, whereas the first significant figure is the 1st non-zero number. It is important to have a strong understanding of place value to do this.

E.g.
Although 0.02 has 2 decimal places, the 2 is the 1st significant figure.

• Estimate with fractions

We can also estimate with fractions by first converting the fraction to a decimal and then rounding the numbers to 1 significant figure.

• Estimation with graphs

We can use graphs to estimate values by using a line of best fit and extrapolating.

### Did you know?

The word estimation comes from the Middle English word aestimatio which means a valuation.

Estimation is part of our series of lessons to support revision on rounding numbers. You may find it helpful to start with the main rounding numbers lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

### Practice estimation questions

1.  Estimate:

7.2 \times 98

720

700

705.6

706

Remember to round each number to one significant figure, then calculate

\begin{aligned} 7.2 \times 98 &\approx 7 \times 100\\\\ 7 \times 100 &= 700 \end{aligned}

2. Estimate the value of:

\frac{907}{3.02}

300.33

305

300

302

Remember to round each number to one significant figure, then calculate

\begin{aligned} \frac{907}{3.02} & \approx \frac{900}{3} \\\\ \frac{900}{3} &=300 \end{aligned}

3. Estimate the cost of 7.4 litres of petrol costing 123p per litre

\pounds 8.00

\pounds 7.00

\pounds 9.10

\pounds 7.40

We round the number of litres to one significant figure, and the price per litre to one significant figure.

\begin{aligned} 7.4 \times 123 &\approx 7 \times 100 \\ 7 \times 100&=700 p \text { or } \pounds 7.00 \end{aligned}

4. Estimate the value of:

\frac{23 \times 494}{54}

210.407

200

225.4

210

Remember to round each number to one significant figure, then calculate

\begin{aligned} \frac{23 \times 494}{54} & \approx \frac{20 \times 500}{50} \\\\ \frac{20 \times 500}{50} &=\frac{10000}{50} \\\\ \frac{10000}{50} &=200 \end{aligned}

5. Estimate how many packets of seeds costing \pounds 1.84 each can be bought for \pounds 20 ?

11

10

9

10.87

Remember to round each number to one significant figure, then calculate

\begin{array}{l} 1.84 \approx 2\\20 \div 2 = 10 \text{ packets} \end{array}

6. Estimate:

\frac{609 \times 4.63}{0.51}

6000

3000

5529

5528

Remember to round each number to one significant figure, then calculate

\begin{aligned} \frac{600 \times 5}{0.5}&=\frac{3000}{0.5} \\\\ \frac{3000}{0.5}&=6000 \end{aligned}

7. Estimate:

\sqrt{30}

5.1

5.25

5.87

5.4

The nearest square numbers are 25 and 36

\sqrt{25}=5 \text { and } \sqrt{36}=6

so the estimate lies between both.

30 is slightly closer to 25 than 36 so an estimate will be 5.4 .

### EstimationGCSE questions

1. Anna says that:

472 \div 5 = 9.44

Use estimation to show that Anna cannot be correct.

(3 Marks)

472 \approx 500 \\

(1)

500 \div 5 = 100

(1)

The answer should be near 100

(1)

2. Oliver wants to buy the following items for his garden:

5 packs of turf at \pounds 91 per pack

30 paving slabs at \pounds 5.81 each

4 fruit trees at \pounds 16.50 each

Estimate the total cost.

(3 Marks)

\pounds 91 \approx \pounds 90, \pounds 5.81 \approx \pounds 6, \pounds 16.50 \approx \pounds 20

(1)

\begin{aligned} 5 \times \pounds 90 &= \pounds 450 \\30 \times \pounds 6 &= \pounds 180 \\4 \times \pounds 20 &= \pounds 80 \end{aligned}

(1)

\pounds 450 + \pounds 180 + \pounds 80 = £710

(1)

3. (a) Estimate:

\frac{31 \times 3.96}{0.47}

(b) Estimate:

\sqrt{5.3 \times 8.9}

(6 Marks)

(a)

\frac{31 \times 3.96}{0.47} \approx \frac{30 \times 4}{0.5}

(1)

\frac{30 \times 4}{0.5} = \frac{120}{0.5}

(1)

\frac{120}{0.5} = 240

(1)

(b)

\sqrt{5.3 \times 8.9} \approx \sqrt{5 \times 9}

(1)

\sqrt{5 \times 9} = \sqrt{45}

(1)

\begin{array}{l} \sqrt{36} = 6, \sqrt{49} = 7\\\\ \sqrt{45} \approx 6.7 \end{array}

(accept 6.6-6.8)

(1)

## Learning checklist

You have now learned how to:

• Estimate by rounding using significant figures
• Estimate within the context of a worded problem, considering units
• Estimate using multiple steps
• Estimate considering BIDMAS principles
• Estimate using square roots

## Still stuck?

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