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Day 18 · Prime factors can simplify a comparisonYear 7 Maths · T1 W3–4 · Day 18 lesson

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Year 7 / Mathematics / Term 1 / Weeks 03 04

Part of the full two-week lesson sequence. Check the pack guide and taught point before teaching.

Open for this lesson: Pack guide · Sources and materials · Daily routes · Worked swaps · Fresh learner checks · Aids and text routes.

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Teacher copy: This page may include teaching prompts or answer keys. Answer keys in this public library can be viewed by anyone. Give learners a clean prompt, use checks as formative evidence, and change a case locally when prior access matters.

Learner prompts

Codes: AC9M7N02, AC9M7N04. Goal: factorise 90 and 150, then explain a shared factor and equivalent fraction. Prepare: tree mat, blank ratio/fraction line.

  1. Launch · 2 min. Two invented collections have 90 and 150 paper cards. Ask what common equal group size might simplify the comparison.
  2. Model · 5 min. 90=2×3²×5; 150=2×3×5². Both include 2×3×5=30, so 90/150=(90÷30)/(150÷30)=3/5. Verify 3/5=0.6; do not infer that the collections have the same total.
  3. Guide · 6 min. Draw two factor trees, then circle one matching 2, 3 and 5 in each. Check 90=2×9×5; 150=2×3×25. Discuss why a leftover 3 versus 5 matters.
  4. Choice · 6 min. D18-A/B/C: compare original cards, factor-leaf match or critique a false reduction. Each keeps both totals and explains the division by 30.
  5. Exit · 4 min. Complete 90/150=__/__ in simplest form: 3/5. Name 30 as a shared factor and check 150×3/5=90.
  6. Note · 2 min. Record whether cancellation paired equal prime factors, not arbitrary digits.

Optional home/extension: Factor 90 and 150 on paper. Extension: find another common factor and show it reaches the same 3/5 after further reduction.

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