Ratio Problems

Activities to try

1. 1

   Scale a mixture and compare adding versus multiplying

   Why it works: The two quantities in a ratio are linked by a constant multiplier, not a constant difference: scaling both parts by the same factor preserves the relationship, while adding the same amount to both breaks it whenever the two parts are unequal.

   1. Meet the problem 'red:blue is 3:5 and there are 15 blue — how many red?' alongside a visible outcome such as rows of beads or a paint shade produced by the ratio.
   2. Scale the ratio up a ratio table (3:5, 6:10, 9:15) and watch two bars (one per part) grow together, stating the multiplier that turns 5 into 15 (×3) and applying the same ×3 to red to get 9.
   3. Try two contrasting moves on the mixture: 'add 10 to each part' (giving 13:15) and 'multiply each part by 3' (giving 9:15), and observe which move keeps the mixture/shade the same and which changes it.
   4. Say in words why multiplying by 3 keeps the same relationship but adding 10 to both does not.
   5. As an optional extension beyond the integer-fact objective, keep red:blue 3:5 but switch to a measurable paint mixture: with 12 ml blue the multiplier is 12÷5 = 2.4, so red = 3×2.4 = 7.2 ml. This shows that building up in whole groups of 3:5 cannot reach the answer. (A bead count could not be 7.2, so use continuous amounts here.)
2. 2

   Build a bar model to find one share

   Why it works: Finding the value of a single equal 'share' or block reveals the constant multiplier and makes the equal-grouping structure of a ratio explicit before any symbolic arithmetic.

   1. Draw a bar model with equal blocks for each part of the ratio (3 blocks red, 5 blocks blue).
   2. Match the known quantity to its blocks and work out the value of one block (15 blue ÷ 5 blocks = 3 per block).
   3. Multiply the value of one block by the number of blocks in the unknown part (3 × 3 = 9 red).
   4. Explain, pointing at the model, why the red count is that number because every block holds the same amount.
3. 3

   Solve varied missing-value problems and catch the additive trap

   Why it works: The same constant multiplier applies to both parts regardless of which quantity is known, and multiplicative reasoning works even when the multiplier is not a whole number — unlike repeated adding.

   1. Solve a set of missing-value problems where the known quantity varies between the two parts and multipliers are sometimes non-integer, e.g. 'for every 2 apples there are 5 oranges; with 20 oranges, how many apples?' (×4 → 8 apples).
   2. For each, state the multiplier and show it applied to both parts.
   3. Compare a tempting additive answer with the correct scaled answer and explain which relationship each uses, e.g. for 2 apples : 5 oranges with 20 oranges, the additive move gives 2+15 = 17 while the correct scaled answer is 2×4 = 8.
   4. As an optional extension, tackle one non-integer case using measurable quantities (e.g. 2 cups juice : 5 cups water with 12 cups water → ×2.4 → 4.8 cups juice) to test that the method still works when whole-group build-up cannot reach the number.