IGCSEMathsEdexcelMaths A (Modular)Higher Unit 2Revision NotesNumberStandard FormOperations with Standard Form

Operations with Standard Form (Edexcel IGCSE Maths A (Modular))

Revision Note

Flashcards
Naomi C

Written by: Naomi C

Reviewed by: Dan Finlay

Edexcel Maths A (Modular): Higher Unit 2Cambridge (CIE) International Maths: Extended

Did this video help you?

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

How do I perform calculations with numbers in standard form using index laws?

Multiplication and division

  • Consider the "number parts" separately to the powers of 10

    • E.g. open parentheses 3 cross times 10 squared close parentheses space cross times space open parentheses 4 cross times 10 to the power of 5 close parentheses

      • Can be written as open parentheses 3 cross times 4 close parentheses cross times open parentheses 10 squared cross times 10 to the power of 5 close parentheses

    • Then calculate each part separately

      • Use laws of indices when combining the powers of 10

      • 12 cross times 10 to the power of 7

    • This can then be rewritten in standard form

      • 1.2 cross times 10 cross times 10 to the power of 7equals 1.2 cross times 10 to the power of 8

  • This process is the same for a division

    • E.g. open parentheses 8 cross times 10 to the power of negative 5 end exponent close parentheses space divided by space open parentheses 2 cross times 10 to the power of negative 3 end exponent close parentheses

      • Can be written as fraction numerator 8 cross times 10 to the power of negative 5 end exponent over denominator 2 cross times 10 to the power of negative 3 end exponent end fraction equals 8 over 2 cross times 10 to the power of negative 5 end exponent over 10 to the power of negative 3 end exponent

    • Then calculate each part separately

      • Use laws of indices when combining the powers of 10

      • Be careful with negative powers -5 -(-3) is -5 + 3

      • 4 cross times 10 to the power of negative 2 end exponent

Addition and subtraction

  • One strategy is to write both numbers in full, rather than standard form, and then add or subtract them

    • E.g. open parentheses 3.2 cross times 10 cubed close parentheses plus open parentheses 2.1 cross times 10 squared close parentheses

    • Can be written as 3200 space plus thin space 210 space equals space 3410

    • Then this can be rewritten in standard form if needed, 3.41 cross times 10 cubed

  • However this method is not efficient for very large or very small powers

  • For very large or very small powers:

    • Write the values with the same, highest, power of 10

    • And then calculate the addition or subtraction, keeping the power of 10 the same

    • Consider open parentheses 4 cross times 10 to the power of 50 close parentheses plus open parentheses 2 cross times 10 to the power of 48 close parentheses

      • Rewrite both with the highest power of 10, i.e. 50

      • Changing 1048 to 1050 has made it 102 times larger, so make the 2 smaller by a factor of 102 to compensate

      • open parentheses 4 cross times 10 to the power of 50 close parentheses plus open parentheses 0.02 cross times 10 to the power of 50 close parentheses

      • These can now be added

      • 4.02 cross times 10 to the power of 50

    • Consider open parentheses 8 cross times 10 to the power of negative 20 end exponent close parentheses minus open parentheses 5 cross times 10 to the power of negative 21 end exponent close parentheses

      • Rewrite both with the higher power of 10, i.e. -20

      • Changing 10-21 to 10-20 has made it 101 times larger, so make the five 101 times smaller to compensate

      • open parentheses 8 cross times 10 to the power of negative 20 end exponent close parentheses minus open parentheses 0.5 cross times 10 to the power of negative 20 end exponent close parentheses

      • These can now be subtracted

      • 7.5 cross times 10 to the power of negative 20 end exponent

Worked Example

Show how open parentheses 45 cross times 10 to the power of negative 3 end exponent close parentheses space divided by space open parentheses 0.9 cross times 10 to the power of 5 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.

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

fraction numerator 45 cross times 10 to the power of negative 3 end exponent over denominator 0.9 cross times 10 to the power of 5 end fraction equals fraction numerator 45 over denominator 0.9 end fraction cross times 10 to the power of negative 3 end exponent over 10 to the power of 5

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 3 end exponent over 10 to the power of 5 using laws of indices

10 to the power of negative 3 end exponent over 10 to the power of 5 equals 10 to the power of negative 3 minus 5 end exponent equals 10 to the power of negative 8 end exponent

Combine back together

open parentheses 45 cross times 10 to the power of negative 3 end exponent close parentheses space divided by space open parentheses 0.9 cross times 10 to the power of 5 close parentheses equals 50 cross times 10 to the power of negative 8 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 cross times 10 to the power of negative 8 end exponent equals 5 cross times 10 to the power of negative 7 end exponent

5 cross times 10 to the power of negative 7 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.

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

Last updated:

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Flashcards

    Did this page help you?

    Previous:Converting To & From Standard FormNext:Percentage Increases & Decreases
    Author
    Reviewer
    Naomi C

    Author: Naomi C

    Expertise: Maths

    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.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    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.