Revise
Standard Form
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:
What is Standard Form?
We always write a number in standard form exactly like this:
The Three Key Rules
1) must be any between and , in other words, .
2) Standard form is always to the power of something ()
3) must be a whole number, this is the number of places the decimal point moves.
e.g.
Skill 1: Standard Form into Large Numbers
Example: Express as a number not in standard form.
Firstly, recall that .
Then, we get
Multiplying by means moving the decimal place to the right, here we must do it times:
Step 1: Write out and move the decimal place jumps to the right.
Step 2: Add ‘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:
Skill 2: Standard Form into Small Numbers
Example: Write in decimal notation.
This is different because the power is negative, but it’s actually no harder. We know,
This means we are which means we move the decimal point spaces to the left.
Step 1: Write out and move the decimal place jumps to the left.
Step 2: Add ‘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
Skill 3: Writing Large Numbers in Standard Form
Example: Write in standard form.
Step 1: Move the decimal point to the left until the number becomes
Step 2: Count the number of times the decimal point has moved to the left, this will become our power (), in this case .
Step 3: We have moved to the left meaning it will be not
So,
Skill 4: Writing Small Numbers in Standard Form
Example: Write in standard form.
Step 1: Move the decimal point to the right until the number becomes
Step 2: Count the number of times the decimal point has moved to the right, this will become our power (), in this case .
Step 3: We have moved to the right meaning it will be not
So,
Skill 5: Multiplying Standard Form
Example: Find the standard form value of , without using a calculator.
Step 1: Change the order around of the things being multiplied.
Step 2: Multiply the numbers and the powers out separately.
Step 3: Convert the number at the front to standard form if necessary ().
This answer is not in standard form ( is not between and ), and we need it to be. Fortunately, if we recognise that , then we get that
Skill 6: Dividing Standard Form
Example: Find the standard form value of , without using a calculator.
Step 1: Break up the problem and change the order of how we divide things.
Step 2: Convert the number at the front to standard form if necessary ().
is between and , so this answer is in standard form, and so we are done.
(Remember: )
Standard Form Example Questions
Question 1: What is 1.15×10−61.15times10^{-6} in decimal notation?
[1 mark]
The power is negative, so this is going to be a very small number. As the power of ten is −6-6, we want to divide the number 1.151.15 by 1010 six times, and so we will move the decimal point six places to the left.
1.15×10−6=0.000001151.15times10^{-6}=0.00000115.
Question 2: What is 5,980,0005,980,000 in standard form?
[1 mark]
In this case, the power of 1010 is going to be positive.
So, if we move the decimal point in 5,980,0005,980,000 to the left six places it becomes 5.985.98. Therefore, we get that,
5,980,000=5.98×1065,980,000=5.98times10^{6}
Question 3: What is 0.00680.0068 in standard form?
[1 mark]
By considering the position where the first non-zero digit is compared to the units column we find,
0.0068=6.8×10−30.0068=6.8times10^{-3}
as the 66 is 33 places away from the units column.
Question 4: Calculate 5,600,000÷8005,600,000div800
Give your answer in standard form.
[3 marks]
First, write each of the numbers in standard form i.e. 5.6×1065.6times10^6 and 8×1028times10^2
(5.6×106)÷(8×102)=(5.6÷8)×(106÷102)(5.6times10^6)div(8times10^2)=(5.6div8)times(10^6div10^2)
Using the formula 10a÷10b=10a−b10^adiv10^b=10^{a-b} we can rewrite the eqaution as,
(5.6÷8)×106−2=0.7×104(5.6div8)times10^{6-2}=0.7times10^4
Standard form requires the number be between 11 and 1010, so adjusting by a factor of 1010, we have,
0.7×104=7×10−1×104=7×1030.7times10^4=7×10^{-1} times10^4=7times10^3
Question 5: Calculate (2.5×104)×(6×10−9)(2.5times10^{4})times(6times10^{-9}).
Give your answer in standard form.
[3 marks]
We will split up this multiplication, multiplying the initial numbers together and the powers of 1010 together separately. Firstly,
2.5×6=152.5times6=15.
Secondly, using the multiplication law of indices,
104×10−9=104+(−9)=10−510^4times10^{-9}=10^{4+(-9)}=10^{-5}
So, we get
(2.5×104)×(6×10−9)=2.5×6×104×10−9=15×10−5(2.5times10^{4})times(6times10^{-9}) =2.5times6times10^4times10^{-9}\ =15times10^{-5}
Standard form requires the number be between 11 and 1010, so adjusting by a factor of 1010, we have,
15×10−5=1.5×10×10−5=1.5×10−415times10^{-5}=1.5times10times10^{-5}=1.5times10^{-4}
Specification Points Covered
Number – 9. calculate with and interpret standard form , where and is an integer
Standard Form Worksheet and Example Questions
(NEW) Standard Form - Exam Style Questions - MME
Level 4-5GCSEKS3NewOfficial MMEStandard Form Drill Questions
Standard Form (1) - Drill Worksheet
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Standard Form (2) - Drill Worksheet
Level 4-5GCSEKS3