AC9M5A02: Year 5 Mathematics Content Descriptor (AC v9) | null | Teacheese AC9M5A02: Year 5 Mathematics Content Descriptor (AC v9) | null | Teacheese
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AC9M5A02 Year 5 Mathematics

AC9M5A02 – Year 5 Mathematics: null

Strand
Algebra
Substrand
null

This Content Descriptor from Year 5 Mathematics provides the specific knowledge and skills students should learn. Use it to plan lessons, create learning sequences, and design assessments that align with the Australian Curriculum v9.

Content Description

find unknown values in numerical equations involving multiplication and division using the properties of numbers and operations

Elaborations

  • 1 using knowledge of equivalent number sentences to form and find unknown values in numerical equations; for example, given that \(3\times5=15\) and \(30\div2=15\) then \(3\times5=30\div2\) therefore the solution to \(3\times5=30\div\square\) is \(2\)
  • 2 using relational thinking, an understanding of equivalence and number properties to determine and reason about numerical equations; for example, explaining whether an equation involving equivalent multiplication number sentences is true, such as \(15 ÷ 3 = 30 ÷ 6\)
  • 3 using materials, diagrams and arrays to demonstrate that multiplication is associative and commutative but division is not; for example, using arrays to demonstrate that \(2 \times 3 = 3 \times 2\) but \(6 ÷ 3\) does not equal \(3 ÷ 6\); demonstrating that \(2 \times 2 \times 3 = 12\) and \(2 \times3 \times2 = 12\) and \(3 \times 2 \times 2 = 12\); understanding that \(8 ÷ 2 ÷ 2 = (8 ÷ 2) ÷ 2 = 2\) but \(8 ÷ (2 ÷ 2) = 8 ÷ 1 = 8\)
  • 4 using materials, diagrams or arrays to recognise and explain the distributive property; for example, where \(4 \times 13 = 4 \times 10 + 4 \times 3\)
  • 5 constructing equivalent number sentences involving multiplication to form a numerical equation, and applying knowledge of factors, multiples and the associative property to find unknown values in numerical equations; for example, considering \(3 \times 4 = 12\) and knowing \(2 \times 2 = 4\) then \(3 \times 4\) can be written as \(3\times\) (\(2 \times 2\)) and using the associative property (\(3 \times 2) \times 2\) so \(3 \times 4 = 6 \times 2\) and so \(6\) is the solution to \(3 \times 4 = \square\times 2\)

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