# KS3 Bitesize

Maths

Approximation

This Revision Bite looks at ways of simplifying calculations to find approximate solutions.

## Introduction

This Revision Bite covers:

## Rounding to the nearest whole number

It is not always necessary to give the exact number. For example, if there were 54 785 inhabitants in a town, you could say that its population was approximately 55 000.

By doing this we have rounded 54 785 to the nearest thousand to give 55 000.

### Rounding to the nearest 1000

What is 1484.5 to the nearest 1000?
It is between 1000 and 2000, but it is closer to 1000, so round down.
1484.5 rounded to the nearest thousand is 1000.

### Rounding to the nearest 100

What is 1484.5 to the nearest 100?
1484.5 is between 1400 and 1500, but it is closer to 1500, so round up.
1484.5 rounded to the nearest hundred is 1500.

### Rounding to the nearest 10

1484.5 lies between 1480 and 1490, but it is closer to 1480, so round down.
1484.5 rounded to the nearest ten is 1480.

### Rounding to the nearest whole number

1484.5 lies between 1484 and 1485 and it is exactly halfway between them. In this situation round up.
1484.5 rounded to the nearest whole number is 1485.

## Making estimates

### Rounding prices

Imagine that you are buying a t-shirt for £9.99, a pair of socks for £1.49 and a belt for £8.99 The cashier charges you £23.47. You feel that this is too much - but how do you know?

One way of finding out whether you have been over-charged is to estimate what the total amount should be. Round the different prices into easier numbers - £9.99 is approximately £10, £1.49 is approximately £1.50 and £8.99 is approximately £9 - and you can do the calculation quickly in your head.

£9.99 + £1.49 + £8.99 £10 + £1.50 + £9 = £20.50

This is almost £3 less than the cashier asked for, so obviously you have been over-charged.

Note: the symbol means 'approximately equal to'.

#### Examples

By rounding the actual values to more manageable numbers, you can estimate the answers to many problems:

£2.99 + £3.10 + 99p £3 + £3 + £1 = £7

29 × 9 30 × 10 = 300

61 ÷ 6 60 ÷ 6 = 10

## Rounding to a given number of places

### Counting decimal places

Decimal places are counted from the decimal point:

So, the number 5.1492 has four decimal places and 4.34 has two decimal places.

To round a number to a given number of decimal places, look at the number in the next decimal place.

• If it's less than 5 round down
• If it's 5 or more, round up
Question

Q1. Round 7.2648 to 2 decimal places (d.p.)

Q2. Round 8.352 to 1 decimal place

A1.

To round to two decimal places, look at the number in the third decimal place. It's a 4, so round down.

Therefore, 7.2648 = 7.26 (2 d.p.)

A2.

To round to one decimal place, look at the number in the second decimal place. It's a 5, so round up.

Therefore, 8.352 = 8.4 (1 d.p.)

To round a number to a given number of decimal places then the answer must have that number of decimal places, even if you have to add some zeros.

For example, rounding 3.40021 to two decimal places gives 3.40 (2 d.p.).

You need to write both decimal places, even though the second number is a zero, to show you rounded to two decimal places.

Remember to look at the number after the one you're interested in. If it's less than 5, round down. If it's 5 or more, round up.

## Trial and improvement

When using trial and improvement guess what the answer might be, then improve on it until you get close to the correct answer.

### Example

A square has an area of 20 cm2. Use a method of trial and improvement to find the length of its side, correct to 1 decimal place (1 d.p.).

1. Start by trying 4:

4 × 4 = 16 (too small)

2. 4 is too small, so try 5:

5 × 5 = 25 (too big)

3. 4 is too small, and 5 is too big, so the answer lies between 4 and 5.

Try 4.5:

4.5 × 4.5 = 20.25 (too big)

Try 4.4:

4.4 × 4.4 = 19.36 (too small)

Therefore, the answer lies between 4.4 and 4.5, but it is closer to 4.5.

So the answer is 4.5cm to 1 d.p.

The question asked us to find the length of the side correct to 1 d.p., so you only need to try values with 1 d.p.

If, however, the question had asked for values correct to 2 d.p. you would have got a more accurate answer. For example:

4.45 × 4.45 = 19.80 (too small)

## Significant figures - decimals

Rounding 12.756 or 4.543 to 1 decimal place (d.p.) seems sensible, as the rounded figures are very close to the actual value.

12.756 = 12.8 (1 d.p.)
4.543 = 4.5 (1 d.p.)

But what happens if you round a very small number to 1 d.p?

0.00546 = 0.0 (1 d.p.)
0.00213 = 0.0 (1 d.p.)

This is not a useful answer. Another way to find an approximate answer with very small numbers is to use significant figures.

### Counting significant figures

Significant figures start at the first non-zero number, so ignore the zeros at the front, but not the ones in between. Look at the following examples:

From the first significant figure onwards, all zeros are included. It's only the zeros at the beginning that don't count.

Question

How many significant figures do the following numbers have?

a) 0.3007
b) 2.01
c) 0.001023

a) 0.3007 has four significant figures.
b) 2.01 has three significant figures.
c) 0.001023 has four significant figures.

## Rounding significant figures

The method for rounding to a given number of significant figures is almost identical to the method used for rounding to a given number of decimal places.

You look at the number after the one you're interested in, to see whether it is greater or less than 5. If it's less than 5, round down. If it's 5 or more, round up.

Question

Q1. Round 0.0724591 to 3 significant figures (s.f.)

Q2. Round 0.2300105 to 4 significant figures (s.f.)

A1.

To round to three significant figures, look at the fourth significant figure. It's a 5, so round up.

Therefore, 0.0724591 = 0.0725 (3s.f.)

A2.

To round to four significant figures, look at the fifth significant figure. It's a 1, so round down.

Therefore, 0.2300105 = 0.2300 (4s.f.)

Even though 0.2300 is the same as 0.23, include the zeros to show that you have rounded to 4 significant figures.

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