Operations with Decimals (Cambridge (CIE) IGCSE Maths)

Revision Note

Operations with Decimals

How do I add or subtract decimals without a calculator?

  • If the numbers involve decimals, the column methods for addition and subtraction can be used

    • Line the place value columns up carefully

    • Make sure the decimal points are all in the same column

      •  Writing zeros (often called place value holders) can help keep everything in line

    • e.g.  2.145 + 13.02 would be written as

Worked Example

Johnny has £32.50 and spends £1.74. 

Calculate how much money Johnny has left.

This is a subtraction question as Johnny has spent money

Make a quick estimate

33 - 2 = 31

Align the digits by place value, ensuring £32.50 is the top number and using the decimal points as a starting point
Using place value holding zeros is optional

The column furthest right is the hundredths column
0 is smaller than 4 so borrow from the next (tenths) column

Next is the tenths column, but again 4 is smaller than 7 so borrow 10 from the ones column

 

14 - 7 = 7 in the tenths column and continue working 'right to left'

Check the final answer is similar to the estimate
£31 and £30.76 are reasonably close

Johnny has £30.76 left

How do I multiply decimals without a calculator?

  • The standard methods for multiplication can be easily adapted for use with decimal numbers

    • E.g. column method, lattice method and grid method

  • You can make a problem easier by converting to integer values then undoing the action to the answer

    • Multiply each value by a power of 10 to make it an integer

    • Perform the easier multiplication

    • Undo the initial action by dividing by the same powers of 10

    • E.g. 2.5 x 4.01

      • Multiply 2.5 by 10 to give 25 and 4.01 by 100 to give 401

      • 25 x 401 = 10 025

      • Divide the answer by 10 and 100 to undo the initial changes

      • So the answer is 10.025

  • An alternative method is to ignore the decimal point whilst multiplying

    • then put it back in the correct place for the final answer using estimation

    • E.g. 1.3 × 2.3

      • Ignoring the decimals this is 13 × 23, which works out to 299

      • 1.3 × 2.3 is approximately 1.5 × 2, which is 3

      • So we know the answer should be close to 3, rather than 0.3 or 30

      • So the answer is 2.99

Examiner Tips and Tricks

A good way to check your answer without a calculator is to estimate it by rounding everything to 1 or 2 significant figures.

Worked Example

Without using a calculator, find 1.57 × 0.78 .

Method 1

Multiply both values by powers of 10 to create a calculation using integer values

1.57 x 100 = 157
0.78 x 100 = 78

157 x 78

Using the grid method

100

50

7

70

7000

3500

490

8

800

400

56

stack attributes charalign center stackalign right end attributes 7000 row plus 3500 end row 490 800 400 56 horizontal line 12246 end stack

157 × 78 = 12 246

Undo the initial action by dividing the answer by 10 000
(This is dividing by 100 and then 100 again)

12 246 ÷ 10 000

1.2246

Method 2

Ignore decimal point to form a different, non-decimal calculation

157 × 78

Using the grid method

100

50

7

70

7000

3500

490

8

800

400

56

stack attributes charalign center stackalign right end attributes 7000 row plus 3500 end row 490 800 400 56 horizontal line 12246 end stack

157 × 78 = 12 246

Estimate the original calculation to write this as the correct answer

1.57 × 0.78 is approximately 2 × 1 = 2

So the correct answer is between 1 and 10

1.2246

How do I divide a decimal without a calculator?

  • Use a similar approach to when multiplying decimals

  • Make a problem easier by converting to integer values then undoing the action to the answer

    • Write the division as a fraction

    • Multiply each value by a power of 10 to make it an integer

    • Perform the easier division

    • Undo the initial action

      • divide by the same power of 10 the numerator was multiplied by

      • multiply by the same power of 10 the denominator was multiplied by

    • E.g. 37.5 ÷ 0.25

      • 37.5 divided by 0.25 space rightwards arrow space fraction numerator 37.5 over denominator 0.25 end fraction

      • fraction numerator 37.5 cross times 10 over denominator 0.25 cross times 100 end fraction equals 375 over 25

      • 375 over 25 equals 15

      • Undo the initial actions fraction numerator 15 cross times 100 over denominator 10 end fraction

      • So the answer is 150

  • Alternatively, ignore the decimal point whilst carrying out the division

    • then put it back in the correct place for the final answer using estimation

    • E.g. 4.68 ÷ 6

      • 4.68 ÷ 6 is approximately 5 ÷ 6, which is between 1 and 0.5

      • So we know the answer should be 0.78, rather than 7.8 or 0.078

Worked Example

Without using a calculator, find 69.02 ÷ 0.7 .

Method 1

Write as a fraction

fraction numerator 69.02 over denominator 0.7 end fraction

Multiply both numbers by powers of 10 to create a calculation with integer values

fraction numerator 69.02 cross times 100 over denominator 0.7 cross times 10 end fraction equals 6902 over 7

Use short division (bus stop method)

long division by 7 yields 0986 pile 6 to the power of 6 9 to the power of 6 0 to the power of 4 2 end pile end long division

6902 over 7 equals 986

Convert the answer
Divide by 100 to cancel out 69.02 being multiplied by 100
Multiply by 10 to cancel out 0.7 being multiplied by 10

fraction numerator 986 cross times 10 over denominator 100 end fraction

98.6

Method 2

Ignore the decimal point to form a different, non-decimal calculation

6902 ÷ 7

Use short division (bus stop method)

long division by 7 yields 0986 pile 6 to the power of 6 9 to the power of 6 0 to the power of 4 2 end pile end long division

6902 ÷ 7 = 986

Estimate the original calculation to write this as the correct answer

69.02 ÷ 0.7 is approximately 70 ÷ 1 = 70

So the correct answer is between 10 and 100

98.6

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Jamie Wood

Author: Jamie Wood

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Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

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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.

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