[discia] GCSE Maths › Number › BIDMAS › Practise
  1. Either paper 1 mark

    Work out \(12 + 3 \times 9\)

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    \(39\)

    Multiplication is done before addition.

    \(3 \times 9 = 27\)

    \(12 + 27 = 39\)

  2. Either paper 2 marks
    (a)

    Work out \(7 + 5 \times 7\)

    (b)

    Work out \((7 + 5) \times 7\)

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    (a) \(42\)
    (b) \(84\)

    Multiplication first: \(5 \times 7 = 35\)

    Then add: \(7 + 35 = 42\)

    Brackets first: \(7 + 5 = 12\)

    Then multiply: \(12 \times 7 = 84\)

  3. Either paper 2 marks

    Find the value of \(4 \times 9 - 90 \div 9\)

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    \(26\)

    Multiplication and division are done first, before the subtraction.

    \(4 \times 9 = 36\) and \(90 \div 9 = 10\)

    Then subtract: \(36 - 10 = 26\)

    • M1multiplication and division done first: \(36\) and \(10\)
    • A1subtracting to get \(26\)
  4. Either paper 2 marks
    (a)

    Work out \((3 \times 6)^2\)

    (b)

    Work out \(3 \times 6^2\)

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    (a) \(324\)
    (b) \(108\)

    The bracket is worked out first, and the index then applies to the whole of it.

    \(3 \times 6 = 18\)

    \(18^2 = 324\)

    There is no bracket, so the index applies only to the 6, and the index comes before the multiplication.

    \(6^2 = 36\)

    \(3 \times 36 = 108\)

  5. Either paper 2 marks

    Work out \(\sqrt{ 15 + 10 } \times 9\)

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    \(45\)

    The square root sign groups everything underneath it, so work that out first.

    \(15 + 10 = 25\)

    \(\sqrt{ 25 } = 5\)

    Then multiply: \(5 \times 9 = 45\)

    • M1working under the root first: \(15 + 10 = 25\), root \(5\)
    • A1multiplying by \(9\) to get \(45\)
  6. Either paper 2 marks
    (a)

    Work out \(19 - 12 + 16\)

    (b)

    Work out \(75 \div 5 \times 3\)

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    (a) \(23\)
    (b) \(45\)

    Addition and subtraction have equal priority, so they are worked from left to right rather than the addition first.

    \(19 - 12 = 7\), then \(7 + 16 = 23\)

    Doing the addition first would give \(-9\), which is not the same answer.

    Multiplication and division also have equal priority, so again work from left to right rather than doing the multiplication first.

    \(75 \div 5 = 15\), then \(15 \times 3 = 45\)

    Doing the multiplication first would give \(75 \div 15 = 5\), which is not the same answer.

  7. Either paper 2 marks

    Work out the number that the box is covering.

    \(4 + \square \times 7 = 25\)

    Show answer
    \(3\)

    The multiplication happens before the addition, so the box multiplied by 7 is whatever has to be added to 4 to reach 25.

    \(25 - 4 = 21\)

    So the box times 7 is 21, and the box is

    \(21 \div 7 = 3\)

    • M1subtracts 4 from 25 to get \(21\)
    • A1divides by 7 to get \(3\)
  8. Either paper 3 marks
    Here are three calculations.
    A\(15 + 6 \times 9\)
    B\(15 \times 6 - 9\)
    C\(9 \times (15 - 6) + 15\)
    Work out the answer to each calculation. Write the letters A, B and C in order of their answers, starting with the smallest.
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    A, B, C

    Work each one out using BIDMAS.

    A: multiplication first, \(6 \times 9 = 54\), then \(15 + 54 = 69\).

    B: multiplication first, \(15 \times 6 = 90\), then \(90 - 9 = 81\).

    C: the bracket first, \(15 - 6 = 9\), then \(9 \times 9 = 81\), then \(81 + 15 = 96\).

    Putting \(69\), \(81\) and \(96\) in order, smallest first, gives

    A, B, C

    • M1A, \(15 + 6 \times 9 = 69\), and B, \(15 \times 6 - 9 = 81\)
    • M1C, \(9 \times (15 - 6) + 15 = 96\)
    • A1order smallest to largest, A, B, C
  9. Either paper 2 marks

    Work out the value of \((3 + 6)^2 - 3 \times 8\)

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    \(57\)

    Brackets first: \(3 + 6 = 9\)

    Then the index: \(9^2 = 81\)

    Then the multiplication: \(3 \times 8 = 24\)

    Finally the subtraction: \(81 - 24 = 57\)

    • M1brackets then index: \(9\) squared is \(81\)
    • A1subtracting \(24\) to get \(57\)
  10. Either paper 2 marks

    Insert one pair of brackets into the calculation below to make it correct.

    \(2 + 6 \times 6 - 8 = 40\)

    Show answer
    \((2 + 6) \times 6 - 8 = 40\)

    With no brackets, BIDMAS gives \(2 + 6 \times 6 - 8 = 30\), which is too small.

    Bracketing the addition makes it multiply by 6 as well:

    \((2 + 6) \times 6 - 8 = 8 \times 6 - 8 = 48 - 8 = 40\)

    • M1evaluating with the addition bracketed: \(8 \times 6 - 8\) or \(48 - 8\)
    • A1correct statement \((2 + 6) \times 6 - 8 = 40\)