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Task 13

Mixed Numbers to Improper Fractions

Replace each whole with unit fractions to see why whole × denominator + numerator counts the total number of parts.

Mathematical focus

A mixed number contains whole units and fractional parts.

To rewrite it as an improper fraction, replace every whole with the number of unit fractions that make that whole, then count all the parts.

This makes sense of the familiar calculation whole number × denominator + numerator: the multiplication counts the fractional parts hidden inside the wholes.

We do — Two and a half

Let the red rod represent one whole. White is one-half of red:

2w = r

Build :

r + r + w

Now replace each red whole with two white halves.

r + r + w = 2w + 2w + w = 5w

So:

2½ = 5⁄2

Why do we calculate 2 × 2? Because there are two wholes and each whole contains two halves. That gives four halves, and the extra half makes five.

2 × 2 + 1 = 5

We do — Two and two-thirds

Now let the light green rod represent one whole. White is one-third of light green:

3w = g

Build 2⅔:

g + g + w + w

Replace each whole with three thirds:

g + g + 2w = 3w + 3w + 2w = 8w

Therefore:

2⅔ = 8⁄3

The formal calculation records exactly what happened:

2 × 3 + 2 = 8

The denominator tells us how many unit fractions are hidden inside each whole.

Your turn — Count all the parts

Write each mixed number as an improper fraction.

Predict first. Then use the rods to model and explain your answer.

Build the mixed number. Replace every whole with the correct number of unit fractions. Then count all the parts.

  1. 1⅔
  2. 2⅔
  3. 2⅕

For at least two examples, write a sentence explaining why the whole number is multiplied by the denominator.

Present task ↗
Show answers
  1. 1½ = 3⁄2
  2. 3½ = 7⁄2
  3. 1⅔ = 5⁄3
  4. 2⅔ = 8⁄3
  5. 1¾ = 7⁄4
  6. 2⅕ = 11⁄5

Useful whole rods: red for halves, light green for thirds, pink for quarters and yellow for fifths.

Feedback / teacher prompts

  • What is the multiplication actually counting?
  • Why do we multiply the whole number by the denominator rather than the numerator?
  • Where is the original numerator in the rod model?
  • Why is it added after we count the parts inside the wholes?
  • Can pupils explain the calculation before carrying it out?
  • At what point can the rods be withdrawn?

Connect to the formal method

The familiar rule is now a record of something pupils can see:

whole number × denominator + numerator

For , three wholes contain six halves, then the extra half gives seven:

3 × 2 + 1 = 7    therefore    3½ = 7⁄2

Exit strategy: the aim is not permanent dependence on rods. The model gives meaning to the algorithm so pupils can later use it with understanding.

Rehearse the teaching

For a task worth practising, work in pairs. One person teaches while the other responds as a pupil using the rods. Model enough to get the pupil started, then stop explaining and question their thinking. Swap roles.

Reflect: What was harder to explain than you expected? Which wording or question worked well?

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