Start. A paper diagram has four sections of length 3/4 m. Write 4 × 3/4 m = 3 m. Reverse the question: how many 3/4 m sections fit in 3 m? Write 3 m ÷ (3/4 m) = 4 sections. Contrast this with 3 m ÷ 4 = 3/4 m per section. The divisor and its units determine what the answer counts. Introduce cancellation in (3/5) × (10/9) = 2/3 by cancelling common factors before multiplication.
Route 28A · Explain the size of a product
On a unit square, shade 2/3 of its width and 3/4 of its height. Find the overlap as a fraction of the whole square. Calculate 2/3 × 3/4, simplify and explain why this product is smaller than either positive factor. Then contrast multiplication by 3/2, which enlarges a positive starting amount.
Teacher next move: overlap 6/12 = 1/2; a factor between 0 and 1 scales down. “Multiplication always makes larger” needs conditions. Zero and negative factors are outside this particular area explanation.
Route 28B · Count full pieces and keep the remainder
A fictional length stock has 2½ m. Each complete section needs 3/8 m. Find the exact quotient, the number of complete sections, the used length and the remaining length. Explain why a quotient of 6⅔ does not mean seven complete sections.
Teacher next move: (5/2) ÷ (3/8) = 20/3 = 6⅔; six complete sections use 18/8 = 9/4 m, remaining 1/4 m. The count of complete sections is discrete; the length is continuous. Rounding up would require 21/8 m, more than stock.
Route 28C · Cancel before multiplying
Compare two exact methods for (7/12) × (18/21), then calculate (5/6) ÷ (10/9). Explain which common factors can be cancelled and why cancelling across addition would be invalid. Give a small counterexample for cancelling in (2 + 3)/3.
Teacher next move: product 1/2; quotient 3/4. Cancellation divides common factors of a product numerator and denominator. (2 + 3)/3 = 5/3, whereas 2 + 1 = 3 is not the same expression.
Close: say whether the answer measures an amount, an amount per group or a count of groups.