Standard Form Questions | Worksheets and Revision | MME

Standard Form Questions, Worksheets and Revision

Level 4-5

Standard Form

Standard form is a shorthand way of expressing VERY LARGE or VERY SMALL numbers. There are 6 key skills that you need to learn.

Standard from to ordinary numbers and ordinary numbers to standard form are two skills you will be taught. 

Make sure you are happy with the following topics before continuing:

KS3 Level 4-5

What is standard form?

We always write a number in standard form exactly like this: 

\textcolor{red}{A}\times\textcolor{blue}{10}^\textcolor{limegreen}{n}

The Three Key Rules

1)   \textcolor{red}{A} must be a whole number between \mathbf{1} and \mathbf{10}, in other words, 1\leq A<10.

2)   Standard form is always \textcolor{blue}{10} to the power of something (\textcolor{limegreen}{n})

3)   \textcolor{limegreen}{n} must be a whole number, this is the number of places the decimal point moves

e.g. \textcolor{red}{4.2}\times\textcolor{blue}{10}^\textcolor{limegreen}{5}

KS3 Level 4-5

Skill 1: Standard Form into Large Numbers

Example: Express 4.2\times\textcolor{limegreen}{10^5} as a number not in standard form.

Standard Form to Large Numbers

Firstly, recall that \textcolor{limegreen}{10^5}=\textcolor{limegreen}{10}\times\textcolor{limegreen}{10}\times\textcolor{limegreen}{10}\times\textcolor{limegreen}{10}\times\textcolor{limegreen}{10}.

Then, we get  4.2\times10^5=4.2\times10\times10\times10\times10\times10

Multiplying by 10 means moving the decimal place to the right, here we must do it 5 times:

Standard Form to Large Numbers

Step 1: Write 4\textcolor{red}{.}2 out and move the decimal place \textcolor{red}{5} jumps to the right

Step 2: Add \textcolor{limegreen}{\mathbf{0}}‘s to fill in the spaces created as the decimal point has moved. 

Step 3: Remove the original decimal point.

This gives the answer to be:

4.2\times10^{5}=420000

KS3 Level 4-5

Skill 2: Standard Form into Small Numbers

Example: Write 2.8\times\textcolor{limegreen}{10^{-4}} in decimal notation.

Standard Form to Small Numbers

This is different because the power is negative, but it’s actually no harder. We know, 

\textcolor{limegreen}{10^{-4}}=\textcolor{limegreen}{\dfrac{1}{10^4}}

This means we are \textcolor{limegreen}{\text{dividing by }10} \textcolor{limegreen}{\text{four times}} which means we move the decimal point 4 spaces to the left.

Step 1: Write 2\textcolor{red}{.}8 out and move the decimal place \textcolor{red}{4} jumps to the left

Standard Form to Small Numbers

Step 2: Add \textcolor{limegreen}{\mathbf{0}}‘s to fill in the space created as the decimal point has moved. 

Step 3: Remove the original decimal point.

Therefore, we have concluded that

2.8\times10^{-4}= 0.00028

KS3 Level 4-5
KS3 Level 4-5

Skill 3: Writing Large Numbers in Standard Form

Example: Write 56,700,000 in standard form.

Step 1: Move the decimal point to the left until the number becomes 5.67 (1\leq A<10)

Large Numbers to Standard Form

Step 2: Count the number of times the decimal point has moved to the left, this will become our power (\textcolor{limegreen}{n}), in this case \textcolor{limegreen}{7}.

Step 3: We have moved to the left meaning it will be \textcolor{limegreen}{+7} not -7

So,

56,700,000 = 5.67\times10^{\textcolor{limegreen}{7}}

KS3 Level 4-5

Skill 4: Writing Small Numbers in Standard Form

Example: Write 0.0000099 in standard form.

Step 1: Move the decimal point to the right until the number becomes 9.9 (1\leq A<10)

Small Numbers to Standard Form

Step 2: Count the number of times the decimal point has moved to the right, this will become our power (\textcolor{limegreen}{n}), in this case \textcolor{limegreen}{6}

Step 3: We have moved to the right meaning it will be \textcolor{limegreen}{-6} not +6

So,

0.0000099 = 9.9\times10^{\textcolor{limegreen}{-6}}

KS3 Level 4-5
KS3 Level 4-5

Skill 5: Multiplying Standard Form

Example: Find the standard form value of (3\times10^8)\times(7\times10^4), without using a calculator. 

Step 1: Change the order around of the things being multiplied.

(\textcolor{red}{3}\times\textcolor{limegreen}{10^8})\times(\textcolor{red}{7}\times\textcolor{limegreen}{10^4})=\textcolor{red}{3}\times\textcolor{red}{7}\times\textcolor{limegreen}{10^8}\times\textcolor{limegreen}{10^4}

Step 2: Multiply the numbers and the powers out separately.

(\textcolor{red}{3}\times\textcolor{red}{7})\times(\textcolor{limegreen}{10^8}\times\textcolor{limegreen}{10^4})=\textcolor{red}{21}\times\textcolor{limegreen}{10^{12}}

Step 3: Convert the number at the front to standard form if necessary (1 \leq A < 10).

This answer is not in standard form (21 is not between 1 and 10), and we need it to be. Fortunately, if we recognise that \textcolor{red}{21}=\textcolor{red}{2.1}\textcolor{limegreen}{\times10}, then we get that

\textcolor{red}{21}\textcolor{limegreen}{\times10^{12}}=\textcolor{red}{2.1}\textcolor{limegreen}{\times10}\textcolor{limegreen}{\times10^{12}}=\textcolor{red}{2.1}\textcolor{limegreen}{\times10^{13}}

KS3 Level 4-5

Skill 6: Dividing Standard Form

Example: Find the standard form value of (\textcolor{red}{8}\times\textcolor{limegreen}{10^{-5}})\div(\textcolor{red}{2}\times\textcolor{limegreen}{10^6}), without using a calculator. 

Step 1: Break up the problem and change the order of how we divide things. 

\dfrac{\textcolor{red}{8}\times\textcolor{limegreen}{10^{-5}}}{\textcolor{red}{2}\times\textcolor{limegreen}{10^6}}=\dfrac{\textcolor{red}{8}}{\textcolor{red}{2}}\times\dfrac{\textcolor{limegreen}{10^{-5}}}{\textcolor{limegreen}{10^6}}=\textcolor{red}{4}\times\textcolor{limegreen}{10^{-11}}

Step 2: Convert the number at the front to standard form if necessary (1 \leq A < 10).

\textcolor{red}{4} is between 1 and 10, so this answer is in standard form, and so we are done.

(Remember: 10^{-5} - 10^6 = 10^{-5-6} = 10^{-11})

KS3 Level 4-5

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Example Questions

The power is negative, so this is going to be a very small number. As the power of ten is -6, we want to divide the number 1.15 by 10 six times, and so we will move the decimal point six places to the left.

 

1.15\times10^{-6}=0.00000115.

In this case, the power of 10 is going to be positive.

So, if we move the decimal point in 5,980,000 to the left six places it becomes 5.98. Therefore, we get that,

 

5,980,000=5.98\times10^{6}

By considering the position where the first non-zero digit is compared to the units column we find,

 

0.0068=6.8\times10^{-3}

 

as the 6 is 3 places away from the units column.

First, write each of the numbers in standard form i.e.  5.6\times10^6 and 8\times10^2

 

(5.6\times10^6)\div(8\times10^2)=(5.6\div8)\times(10^6\div10^2)

 

Using the formula 10^a\div10^b=10^{a-b} we can rewrite the eqaution as,

 

 (5.6\div8)\times10^{6-2}=0.7\times10^4

 

Standard form requires the number be between 1 and 10, so adjusting by a factor of 10, we have, 

 

0.7\times10^4=7×10^{-1} \times10^4=7\times10^3

We will split up this multiplication, multiplying the initial numbers together and the powers of 10 together separately. Firstly,

 

2.5\times6=15.

 

Secondly, using the multiplication law of indices,

 

10^4\times10^{-9}=10^{4+(-9)}=10^{-5}

So, we get

 

(2.5\times10^{4})\times(6\times10^{-9}) =2.5\times6\times10^4\times10^{-9}\\ =15\times10^{-5}

 

Standard form requires the number be between 1 and 10, so adjusting by a factor of 10, we have, 

 

15\times10^{-5}=1.5\times10\times10^{-5}=1.5\times10^{-4}

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