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Here we will learn about GCSE number, including place value, arithmetic, factors, multiples, primes and many other number topics.

There are also GCSE number worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if youβre still stuck.

**Number **is one of the topic areas of GCSE mathematics that features across all of the other areas of mathematics. You will need to be confident with all of the number topics to be able to solve problems involving algebra, ratio, geometry, statistics and probability.

You will also be asked questions that are purely number problems involving topics such as factors and multiples, prime numbers, indices, rounding or arithmetic.

In this page we will link to lots of other pages that will help with a variety of number topics. Let’s look at some of the different ways your number skills could be assessed in GCSE Mathematics.

Over 75 GCSE number worksheets ready to download and give to your GCSE students. Each with functional and applied reasoning questions and exam style questions.

DOWNLOAD FREEOver 75 GCSE number worksheets ready to download and give to your GCSE students. Each with functional and applied reasoning questions and exam style questions.

DOWNLOAD FREE**Place value **is the value of each digit within a number.

A number is made up of digits. We need to be able to identify the location of each digit within the number as this can help to understand how large or small the number is.

To determine the value of a digit within a number we label each column with a title, like this,

For example,

Write down the value of the digit 5 in the number 4.563.

**Locate the digit within the number.**

β¬

2**Recall the place value of that column.**

The 5 is in the tenths column.

3**Write the value of the digit.**

The value of the digit 5 in the number 4.563 is five tenths, or 0.5.

**Step-by-step guide:** Place value

**Arithmetic** is the study of numbers and the operations between them.

To use arithmetic with positive and negative numbers, we need to be able to calculate using the four main operations.

For example,

A group of young people are donating money to some charities. Altogether they collect Β£5814 which will be split equally between three charities. How much does each charity receive?

**Determine which operator you need to use.**

We need to divide 5814 by 3.

2**Carry out the calculation.**

Use a method of division such as short division.

Each charity receives Β£1938.

**Step-by-step guide:** Arithmetic

**Rounding** is when we approximate numbers in order to make them easier to deal with. We need to round down or up depending on which value is nearer. There are several ways we can be asked to round a number.

For example,

1478.27 can be rounded in several ways, including the below methods.

**Step-by-step guide:** Rounding numbers

**Rounding to the nearest ones place,**\bf{10, \ 100}**and**\bf{1000}

1478.47 is 1500 rounded to the nearest 100.

**Step-by-step guide:** Rounding to the nearest 10, 100 and 1000

**Rounding to decimal places**

1478.2735 is 1478.3 rounded to 1 decimal place (the nearest tenth).

**Step-by-step guide:** Rounding decimals (coming soon)

**Rounding to significant figures**

1478.47 is rounded 1000 to 1 significant figure.

**Step-by-step guide:** Significant figures

**Truncation**is cutting a number off at a certain value with no need to round up.

Truncating a number is also linked with rounding.

34.278 is 34.2 truncated to 1 decimal place.

**Step-by-step guide:** Truncation

**Estimation**is using the approximated numbers for a calculation; we do not use the exact numbers.

Estimation is also linked with rounding.

34.278 is 34.3 rounded to 1 decimal place.

To estimate the answer to 367.1 \times 12.8 we would calculate 400 \times 10 to get an estimate of 4000.

**Step-by-step guide:** Estimation

**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.

We need to be able to find upper and lower bounds of rounded numbers and use them in calculations. To do this we think about what the smallest and biggest numbers that round to a value for a given degree of accuracy are and then use the correct values for the calculation required.

We can display the upper and lower bounds using **error intervals**.

For example,

A number x has been rounded to 1 decimal place and given as 6.4.

The smallest number that will round up to 6.4 is 6.35 and any number up to 6.45 will round down to 6.4.

Therefore we can write,

6.35 \leq x < 6.45.We have to use < for the upper bound as 6.45 would round up to 6.5, not down to 6.4.

However, any number up to 6.45 would round down to 6.4 (for example 6.44, \ 6.449, \ 6.44999999). This is why we write 6.45 as the upper bound.

**Step-by-step guide:** Upper and lower bounds

**Factors, multiples and primes** are different types of numbers.

In GCSE mathematics you may need to use factors, multiples and primes to solve problems. These may involve the following.

**Step-by-step guide:** Factors, multiples and primes

**Factors**

A **factor** is a number which divides into another number exactly with no remainders.

For example, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

**Step-by-step guide:** Factors

**Multiples**

A **multiple** of a number is a number in its times table.

For example, the first 5 multiples of 13 are 13, 26, 39, 52 and 65.

**Step-by-step guide:** Multiples

**Primes**

A **prime number** is a number that only has two factors, 1 and itself.

For example, the first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

**Step-by-step guide:** Prime numbers

**Prime factors**

**Prime factors** are factors of a number that are also prime numbers.

For example, the prime factors of 60 are 2, 3 and 5.Β

60 can be written as a product of its prime factors,

60=2\times 2\times 3\times 5.**Step-by-step guide:** Prime factors

**Lowest common multiple**

The **lowest common multiple (LCM)** of two numbers is the **lowest** value that is a **multiple** of each number.

For example, the lowest common multiple of 12 and 18 is 36.

**Step-by-step guide:** Lowest common multiple

**Highest common factor**

The **highest common factor (HCF) **of two numbers** **is the **highest** value that is a **factor** of each number.

For example, the highest common factor of 12 and 18 is 6.

**Step-by-step guide:** Highest common factor

**Negative numbers** are any numbers less than zero. They have a negative or minus sign (-) in front of them. Numbers greater than zero are referred to as **positive numbers**. If there is no sign in front of a number the number is positive.

On the **number line** below we can see some positive and negative integers (whole numbers).

The numbers in orange are negative values and the blue numbers are positive values.

Zero is neither positive nor negative.

Just like you can add, subtract, multiply and divide positive numbers, you can do the same with negative numbers whether they are integers, decimals or fractions.

When adding and subtracting negative numbers we need to remember some rules.

**If the signs are the same, replace them with a positive sign.**

**If the signs are different, replace them with a negative sign.**

The chart below summarises this.

As a rule of thumb – same signs add, different signs subtract.

For example, 5-(-2)=5+2=7.

When multiplying and dividing negative numbers we need to remember some rules.

**If the signs are the same, the answer is positive.**

**If the signs are different, the answer is negative.**

When multiplying negative numbers,

The same rules apply for dividing negative numbers.

For example, 5\times \left( -2 \right)=-10, \ \left( -45 \right)\div \left( -15 \right)=3.

**Step-by-step guide:** Negative numbers

**Fractions, decimals and percentages** are different ways of representing a proportion of the same amount. There is equivalence between fractions, decimals and percentages.

For example,

\frac{43}{100}=0.43=43\%There are various methods of using fractions, decimals and percentages. They are important to understand as they are used in everyday life.

To succeed in GCSE mathematics you will need to be familiar with the following fraction, decimal and percentage topics,

- Adding and subtracting fractions
- Multiplying and dividing fractions
- Equivalent fractions, improper fractions and mixed numbers
- Ordering fractions
- Fractions of an amount
- Adding and subtracting decimals
- Multiplying and dividing decimals
- Percentage of an amount
- Percentage multipliers
- Percentage increase and decrease
- Percentage change
- Reverse percentages
- Converting and comparing fractions, decimals and percentages

**Step-by-step guide:** Fractions, decimals and percentages

**Fractions**

**Fractions **are a way of writing equal parts of one whole.

There are two parts to a fraction, a numerator (top number) and a denominator (bottom number). The denominator shows how many equal parts the whole has been divided into. The numerator shows how many of the equal parts we have.

For example,

This shape has 9 equal parts and 4 of them are shaded.

This represents four ninths.

\frac{4}{9}**Step-by-step guide:** Fractions

**Decimals**

**Decimals **are a way of writing numbers that are not whole.

For example,

This shows the fraction \frac{7}{10}.

\frac{7}{10} can also be written as \bf{0.7} .

Decimal numbers can be recognised as they have a **decimal point.**

A decimal place is a position after the decimal point.

For example,

0.37 has two decimal places.

There is a 3 in the tenths place and 7 in the hundredths place.

**Step-by-step guide:** Decimals

**Percentages**

**Percentages** are numbers which are expressed as parts of 100.

Percent means βnumber of parts per hundredβ and the symbol we use for this is the percent sign \%.

For example,

43\%.There are 100 equal parts and 43 of them are shaded.

**Step-by-step guide:** Percentages

**Power and roots** are terms that express a **repeated multiplication** of a number.

**Powers**

For example,

We say this as 2 **to the power of** 5 because the number 2 is being **multiplied by itself** 5 times.

**Roots**

For example,

As 2 \times 2 \times 2 \times 2 \times 2=32, we can say that the 5th **root of** 32 is 2.

**Step-by-step guide:** Powers and roots

In GCSE mathematics you will need to be familiar with the following.

**Negative powers**

When a number is raised to a negative power, we calculate the positive reciprocal of that number.

For example,

10^{-2}=\frac{1}{10^{2}}=\frac{1}{100}**Step-by-step guide:** Negative powers

**Fractional powers**

Fractional powers are a type of index that represents the nth root of a number.

For example,

8^{\frac{1}{3}}=\sqrt[3]{8}=2**Step-by-step guide:** Fractional powers

**Expressing a power with a different base**

We can sometimes express a power with a different base in order to help to simplify expressions.

For example,

8 \times 2^{2}=2^{3} \times 2^{2}=2^{5}**Step-by-step guide:** Expressing power with a different base (coming soon)

**The laws of indices**

Powers and roots are used in the five laws of indices,

Multiplying | To multiply two powers with the same base, add the indices. | x^{a}\times{x^{b}}=x^{a+b} |

Dividing | To divide two powers with the same base, subtract the indices. | x^{a}\div{x^{b}}=x^{a-b} |

Brackets | To raise one power to another power, multiply the indices. | (x^{a})^{b}=x^{ab} |

Fractions | To raise a base to a fraction \frac{m}{n}, calculate the nth root of the base, raised to the power m. | x^{\frac{m}{n}}=\sqrt[n]{x^m} |

Negatives | To raise a base to a negative number, take the positive power of the reciprocal. | x^{-a}=\frac{1}{x^a} |

**Step-by-step guide:** Laws of indices

**Standard form** is a way of writing very large or very small numbers by using powers of ten. It is also known as scientific notation.

Numbers in standard form are written in this format,

a\times10^{n} .

Where a is a number 1 \leq a < 10 and n is an integer.

To write a number in standard form we need to understand the place value of the number.

For example,

Letβs look at the number 8290000 and write the digits in a place value table.

10^6 | 10^5 | 10^4 | 10^3 | 10^2 | 10^1 | 10^0 |
---|---|---|---|---|---|---|

8 | 2 | 9 | 0 | 0 | 0 | 0 |

So, 8290000 written in **standard form** is

8.29\times10^{6} .

In GCSE mathematics you will be expected to convert between ordinary numbers and standard form as well as perform calculations with numbers in standard form with and without a calculator.

For example,

Calculate (5\times10^{4}) \times (7\times10^{8}) .

Write your answer in standard form.

**Multiply the non-zero digits.**

2**Multiply the powers of ten by adding the powers.**

3**Put these two parts together and check whether the answer is in standard form.**

This number is not in standard form as 35 is not between 1 and 10.

To convert 35 to a decimal between 1 and 10 you need to divide it by 10. To maintain the value of the number, you need to multiply the power of ten by 10, which adds one to the exponent.

\hspace{1.3cm} 35 \times 10^{12} \div \ 10 \ β \hspace{2cm} \times \ 10 \ β \hspace{1.3cm} 3.5\times10^{13}**Step-by-step guide**: Standard form

**Simple and compound interest** are two ways of calculating interest.

**Simple interest** is calculated on the original (principal) amount, whereas **compound interest** is calculated on the original amount and on the interest already accumulated on it.

The difference between simple and compound interest is that **simple interest** is calculated using **only **the** original amount **whereas **compound interest** works out the **interest** on a **previous amount **as well.

Check out the comparative example below to see the similarities and differences between the two forms of interest.

\hspace{1.5cm} \bf{Β£100} is invested for \bf{3} **years** at \bf{2\%} per year. Find the final value.

**Simple interest**

A=P(1+rt) \

Here:

- P=100
- r=0.02 (as 2% = 0.02)
- t=3

Substituting these values into the **simple interest** formula

A=P(1+rt)

We get:

A = 100(1+0.02\times{3})\

A = 100(1+0.06)\

A = 100(1.06)\

A = 100\times{1.06}\

A = Β£106

**Compound interest**

A=P(1+\frac{r}{100})^{n} \

Here:

- P=100
- r=2
- n=3

Substituting these values into the **compound interest** formula

A=P(1+\frac{r}{100})^{n} \

We get:

A = 100(1+(\frac{2}{100})^{3}

A = 100(1+0.02)^3\

A = 100(1.02)^3\

A = 100 x 1.02^3\

A = Β£106.12

As well as compound interest, you will also need to be able to work out compound depreciation. When the value of an asset reduces in price, it is known as **depreciation**.

Depreciation can be calculated the same way as compound interest but we must remember that the multiplier being used will be less than 1.

For example,

A car worth Β£20 \ 000 depreciates by 10\% per year for 2 years.

To find the value after 2 years we calculate using the multiplier of 0.9.

20 \ 000 \times 0.9^{2}=16 \ 200The car will be worth Β£16 \ 200.

**Step-by-step guide:** Simple interest and compound interest

**Types of numbers **are different sets of numerical values.

You should already be familiar with **odd** and **even** numbers, as well as **positive** and **negative** numbers, whole numbers (**integers)** and **decimals**.

You should also be familiar with other sets of numbers including **square** and **cube** numbers, **prime** numbers, and **triangular** numbers.

Below is a description of each of these types of numbers, along with a few examples.

**Step-by-step guide:** Types of numbers

**Surds **are a root that gives an irrational number. An irrational number canβt be written as a fraction, and in decimal form is infinitely long with no recurring pattern.

For example,

\sqrt{5} \approx 2.23606 , which is an irrational number.

The square root of 5 is a surd.

In GCSE mathematics you need to be able to use surds in calculations without a calculator. You will need to be able to do the following.

You will also be expected to expand brackets involving surds and use surds in other algebra and geometry problems.

**Step-by-step guide:** Surds

**Simplify surds**

For example,

\begin{aligned} \sqrt{60} &=\sqrt{4 \times 15} \\\\ &=\sqrt{4} \times \sqrt{15} \\\\ &=2 \times \sqrt{15} \\\\ &=2 \sqrt{15} \end{aligned}**Step-by-step guide:** Simplifying surds

**Add and subtract surds**

For example,

\begin{aligned} 2 \sqrt{5}+\sqrt{45} &=2 \sqrt{5}+3 \sqrt{5} \\\\ &=5 \sqrt{5} \end{aligned}**Step-by-step guide:** Adding and subtracting surds

**Multiply and divide surds**

**Step-by-step guide:** Multiplying and dividing surds

**Rationalise denominators**

For example,

\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}**Step-by-step guide:** Rationalise the denominator

**Money problems **are types of questions that involve finance. Examples could include,

- Best buys
- Using exchange rates
- Budgeting for a holiday
- Calculating the amount spent on a shopping trip

To be able to solve money problems in GCSE mathematics you will need to be confident with decimal arithmetic and rounding.

When calculating with money it will be important to leave answers to two decimal places if the answer is in pounds, Β£.

For example,

A pack of 6 rolls of toilet paper costs Β£2.45. A pack of 9 toilet rolls costs Β£3.59.

How much is the cheapest price for 36 toilet rolls?

We need to multiply the cost of 6 rolls by 6 and multiply the cost of 9 rolls by 4.

Β£2.45\times 6 = Β£14.70 Β£3.59\times 4 = Β£14.36The cheapest price for 36 rolls of toilet paper is Β£14.36.

**Step-by-step guide:** Money problems maths

**Calculator skills **are important for the GCSE mathematics calculator papers and will help you later in life in many careers.

You may be asked to perform calculations involving powers and roots, fractions and trigonometric functions.

For example,

Work out \frac{\sqrt[3]{340}-{{2.6}^{5}}}{\sin 25{}^\circ }, write down all the numbers on your calculator display.

= -264.6223272**Step-by-step guide:** Calculator skills

**Sometimes we need a zero in as a place holder when rounding with decimal numbers**

Round 3.4973 to 2 decimal places will be 3.50. We need the zero in the hundredths column. So, 3.4 \colorbox{yellow}{9}73 \approx 3.50 we round up.

**Upper bound is incorrect**

It is very common for students to put an incorrect upper bound.

For example, if a length has been rounded to 56 \ cm to the nearest cm, they may 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 56.499999999β¦. The upper bound in this case is 56.5 \ cm.

- \bf{1}
**and prime numbers**

1 is not a prime number. This is because it only has one factor, rather than the 2 factors needed to be a prime number.

**Factors and multiples**

Factors and multiples are easily mixed up. Remember multiples are the multiplication table, whereas factors are the numbers that go into another number without a remainder.

**Raising a negative number to a power greater than one**

Remember when raising a negative number to a power greater than 1 the resulting answer could be positive or negative. When you raise a negative number to an odd exponent the resulting answer is negative, when you raise a negative number to an even exponent the resulting answer is positive.

**Raising a number to a negative power**

A common error is thinking that raising a number to a negative power makes the answer negative.

For example, the value of 3^{-2} is sometimes incorrectly given as -9.

The correct answer is \frac{1}{9} as the negative power gives the reciprocal to the base number.

1. Round 0.0365 to 2 significant figures.

0.03

0.3

0.037

0.036

We are rounding to 2 significant figures. The first significant figure is 3. The 5 in the ten thousandths column indicates we round the 6 up.

2. 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 = 42.9 \ cm, \ lower bound = 42.7 \ cm

Upper bound = 42.94 \ cm, lower bound = 42.75 \ cm

Upper bound = 42.8 \ cm, \ lower bound = 42.9 \ cm

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.

3. Find the lowest common multiple of 8 and 10.

20

2

80

40

Write down the multiples of 8 and the multiples of 10 and see which is the lowest number that occurs in both lists.

Multiples of 8: Β 8, 16, 24, 32, 40, 48, β¦

Multiples of 10: 10, 20, 30, 40, 50, …

So, 40 is the LCM of 8 and 10.

Alternatively you can use prime factorisation and Venn diagrams.

\text{LCM}=2\times 2\times 2\times 5=40

4. 65\% of a number is 520.

What is the original number?

800

858

338

700

5. Evaluate \sqrt[4]{81} \div 9^{\frac{1}{2}} .

3

1

4.5

9

\sqrt[4]{81}\div 9^{\frac{1}{2}}=3\div{3}=1

6. Expand and simplify, 4(3+ \sqrt{28}) .

12+2\sqrt{7}

12+4\sqrt{28}

12+4\sqrt{7}

12+8\sqrt{7}

Simplify \sqrt{28}=\sqrt{4 \times 7}=2 \sqrt{7}.

\begin{aligned} &4 \times 3=12 \\\\ &4 \times 2 \sqrt{7}=8 \sqrt{7} \end{aligned}

So the final answer is 12+8\sqrt{7}.

1. Work out an estimate for \frac{67.3 \times 21.6}{0.46}.

**(3 marks)**

Show answer

Rounding one original number to 1 significant figure 70 or 20 or 0.5.

**(1)**

Calculation involving all three rounded numbers.

**(1)**

**(1)**

2. (a) Find the highest common factor of 36 and 84.

(b) Two buses leave the depot at 12.00.

Bus A takes 40 minutes to complete its journey and return to the depot.

Bus B takes 25 minutes to complete its journey and return to the depot.

What time is it when both buses are back at the depot together?

**(5 marks)**

Show answer

(a)

List of factors or prime factor decomposition of one or both.

**(1)**

HCF of 12 .

**(1)**

(b)

Prime factorisation or list of multiples of one or both.

**(1)**

LCM of 200 minutes.

**(1)**

15.20 or 3.20 \ pm

**(1)**

3. (a) Expand and simplify (3+\sqrt{7})(3-\sqrt{7}).

(b) Hence, or otherwise, show that \frac{16}{3+\sqrt{7}} can be written as 24-8\sqrt{7}.

**(5 marks)**

Show answer

(a)

Any two correct terms, 9+3\sqrt{7}-3\sqrt{7}-7 .

**(1)**

All four correct terms, 9+3\sqrt{7}-3\sqrt{7}-7 .

**(1)**

Simplified to 2 .

**(1)**

(b)

\frac{16(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}

**(1)**

\frac{48-16\sqrt{7}}{2} leading to 24-8\sqrt{7} as required

**(1)**

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