2.1.6

Method F - Constant Difference Method

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Constant Difference Theory: 753 − 491

Step 1

Step 1

  • We are trying to avoid any carrying or exchanging – the problem is the middle digit in the subtrahend being a 9.
  • So we could add 10 to both the minuend and the subtrahend to make the subtraction easier.
Step 2

Step 2

  • We write out the new subtraction in the usual column format.
  • This isn’t totally necessary though because we just need to subtract the hundreds from the hundreds, the tens from the tens and the ones from the ones.
  • We can work left to right or right to left (because no exchanges are required), with or without the column format.
Answer

Answer

  • So if 763 − 501 = 262, then 753 − 491 = 262.
Jump to other topics
1

Course Overview

1.1

Course Structure & Commonly Asked Questions

1.1.1Introduction to the Course & Common Questions
2

Subtraction: KS2/3

2.1

Subtraction Methods

2.1.1Method A - Decomposition Method2.1.2Method B - Equal Addition Method2.1.3Method C - Expanded Form Method2.1.4Method D - Partitioning Method2.1.5Method E - Counting-Up Method2.1.6Method F - Constant Difference Method2.1.7Method G - Partial Differences Method2.1.8Method H - Complementary Method2.1.9Method I - Nines Complement Method
3

Expanding Double Brackets: KS3/4

3.1

Expansion Methods

3.1.1Method A - Grid Method3.1.2Method B - Each by Each3.1.3Method C - Distributive Law3.1.4Method D - Inspection3.1.5Method E - Column Method3.1.6Method F - Vertical & Crosswise

Practice questions on Method F - Constant Difference Method

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  1. 1
    What is the Constant Difference Method also known as?Multiple choice
  2. 2
    What should students add or subtract from the minuend and subtrahend in this calculation to make it simpler?<br><br>5500 &minus; 4174Multiple choice
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Method E - Counting-Up Method

Method G - Partial Differences Method