Operations with Standard Form (Edexcel IGCSE Maths A): Revision Note

Exam code: 4MA1

Operations with standard form

How do I perform calculations in standard form using a calculator?

  • Make use of brackets around each number, and use the box enclose cross times 10 to the power of x end enclose button to enter numbers in standard form

    • e.g. open parentheses 3 cross times 10 to the power of 8 close parentheses cross times open parentheses 2 cross times 10 to the power of negative 3 end exponent close parentheses 

    • You can instead use the standard multiplication and index buttons

  • If your calculator answer is not in standard form, but the question requires it:

    • Either rewrite it using the standard process

      • e.g. 3 820 000 = 3.82 × 106

    • Or rewrite numbers in standard form, then apply the laws of indices

      • e.g.  243 × 1020 = (2.43 × 102) × 1020 = 2.43 × 1022

Examiner Tips and Tricks

Exam questions tend to use large numbers as indices. This is so that the question can test whether you know how to calculate using standard form.

How do I multiply and divide numbers in standard form using index laws?

Multiplication

  • Consider the example open parentheses 3 cross times 10 to the power of 112 close parentheses space cross times space open parentheses 4 cross times 10 to the power of 235 close parentheses

  • STEP 1
    Multiply the ordinary numbers

    • e.g. 3 cross times 4 equals 12

  • STEP 2
    Multiply the powers of 10 by adding the indices

    • e.g. 10 to the power of 112 cross times 10 to the power of 235 equals 10 to the power of 112 plus 235 end exponent equals 10 to the power of 347

  • STEP 3
    Multiply the new ordinary number and the new power of 10

    • e.g. 12 cross times 10 to the power of 347

  • STEP 4 (if needed)
    Write the ordinary number in standard form and simplify the powers of 10 by adding the indices

    • e.g. 12 equals 1.2 cross times 10 to the power of 1 and 1.2 cross times 10 to the power of 1 cross times 10 to the power of 347 equals 1.2 cross times 10 to the power of 348

Division

  • Consider the example open parentheses 2 cross times 10 to the power of negative 150 end exponent close parentheses space divided by space open parentheses 8 cross times 10 to the power of negative 131 end exponent close parentheses

  • STEP 1
    Divide the first ordinary number by the second ordinary number

    • e.g. 2 divided by 8 equals 0.25

  • STEP 2
    Divide the first power of 10 by the second power of 10 by subtracting the indices

    • e.g. 10 to the power of negative 150 end exponent divided by 10 to the power of negative 131 end exponent equals 10 to the power of negative 150 minus open parentheses negative 131 close parentheses end exponent equals 10 to the power of negative 19 end exponent

      • Be careful with negatives!

  • STEP 3
    Multiply the new ordinary number and the new power of 10

    • e.g. 0.25 cross times 10 to the power of negative 19 end exponent

  • STEP 4 (if needed)
    Write the ordinary number in standard form and simplify the powers of 10 by adding the indices

    • e.g. 0.25 equals 2.5 cross times 10 to the power of negative 1 end exponent and 2.5 cross times 10 to the power of negative 1 end exponent cross times 10 to the power of negative 19 end exponent equals 2.5 cross times 10 to the power of negative 20 end exponent

How do I add and subtract numbers in standard form using index laws?

  • Consider the example open parentheses 4 cross times 10 to the power of 150 close parentheses minus open parentheses 2 cross times 10 to the power of 148 close parentheses

  • STEP 1
    Rewrite the number with the biggest power of 10 so that it has the same power of 10 as the number with the lowest power of 10

    • e.g. 4 cross times 10 to the power of 150 equals 4 cross times 10 squared cross times 10 to the power of 148 equals 400 cross times 10 to the power of 148

      • The ordinary gets bigger as the power of 10 gets smaller

  • STEP 2
    Collect like terms by adding or subtracting the ordinary numbers

    • e.g. open parentheses 400 cross times 10 to the power of 148 close parentheses minus open parentheses 2 cross times 10 to the power of 148 close parentheses equals 398 cross times 10 to the power of 148

      • Do not change the power of 10

  • STEP 3
    Write the ordinary number in standard form and simplify the powers of 10 by adding the indices

    • e.g. 398 equals 3.98 cross times 10 squared and 3.98 cross times 10 squared cross times 10 to the power of 148 equals 3.98 cross times 10 to the power of 150

  • This method works for negative powers too

    • e.g. consider open parentheses 8 cross times 10 to the power of negative 120 end exponent close parentheses minus open parentheses 5 cross times 10 to the power of negative 121 end exponent close parentheses

      • 8 cross times 10 to the power of negative 120 end exponent equals 8 cross times 10 to the power of 1 cross times 10 to the power of negative 121 end exponent equals 80 cross times 10 to the power of negative 121 end exponent

      • open parentheses 80 cross times 10 to the power of negative 121 end exponent close parentheses minus open parentheses 5 cross times 10 to the power of negative 121 end exponent close parentheses equals 75 cross times 10 to the power of negative 121 end exponent

      • 75 equals 7.5 cross times 10 to the power of 1 and 7.5 cross times 10 to the power of 1 cross times 10 to the power of negative 121 end exponent equals 10 to the power of negative 120 end exponent

Worked Example

Show how open parentheses 45 cross times 10 to the power of negative 123 end exponent close parentheses space divided by space open parentheses 0.9 cross times 10 to the power of 105 close parentheses can be written in the form A cross times 10 to the power of n, where 1 less or equal than A less than 10 and n is an integer.

Answer:

Rewrite the division as a fraction, then separate out the powers of 10

fraction numerator 45 cross times 10 to the power of negative 123 end exponent over denominator 0.9 cross times 10 to the power of 105 end fraction equals fraction numerator 45 over denominator 0.9 end fraction cross times 10 to the power of negative 123 end exponent over 10 to the power of 105

Work out fraction numerator 45 over denominator 0.9 end fraction

fraction numerator 45 over denominator 0.9 end fraction equals 450 over 9 equals 50

Work out 10 to the power of negative 123 end exponent over 10 to the power of 105 using laws of indices

10 to the power of negative 123 end exponent over 10 to the power of 105 equals 10 to the power of negative 123 minus 105 end exponent equals 10 to the power of negative 228 end exponent

Combine back together

open parentheses 45 cross times 10 to the power of negative 123 end exponent close parentheses space divided by space open parentheses 0.9 cross times 10 to the power of 105 close parentheses equals 50 cross times 10 to the power of negative 228 end exponent

Rewrite in standard form, where a is between 1 and 10

50 cross times 10 to the power of negative 8 end exponent equals 5 cross times 10 to the power of 1 cross times 10 to the power of negative 228 end exponent equals 5 cross times 10 to the power of negative 227 end exponent

5 cross times 10 to the power of negative 227 end exponent

Worked Example

Given that a cross times b equals c, where

a equals 5 cross times 10 to the power of 20
b equals 6 cross times 10 to the power of n
c equals 3 cross times 10 to the power of 33

Find the value of n.

Answer:

Substitute the values into the given equation, a cross times b equals c

5 cross times 10 to the power of 20 space cross times space 6 cross times 10 to the power of n space equals space 3 cross times 10 to the power of 33

Rearrange so the powers of 10 are grouped together

5 cross times 6 cross times 10 to the power of 20 cross times 10 to the power of n space equals space 3 cross times 10 to the power of 33

Calculate the 'number' part and use laws of indices to combine the powers of 10

30 cross times 10 to the power of 20 plus n end exponent equals 3 cross times 10 to the power of 33

The two sides of the equation are almost in the same format, but we need to change the 30 to a 3

Dividing the number by 10, means the power of 10 must be increased by 1 to compensate

3 cross times 10 to the power of 21 plus n end exponent equals 3 cross times 10 to the power of 33

The two sides of the equation now match, so the powers of 10 on each side must be equal

21 plus n equals 33

Subtract 21 from both sides to find n

n equals 12

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Naomi C

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Naomi graduated from Durham University in 2007 with a Masters degree in Civil Engineering. She has taught Mathematics in the UK, Malaysia and Switzerland covering GCSE, IGCSE, A-Level and IB. She particularly enjoys applying Mathematics to real life and endeavours to bring creativity to the content she creates.

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Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.