D07: Three quarters divided by two ninths

 Here we look at a range of ways we can model and solve ¾ ÷ ²⁄₉. Some of these tasks should work with some pupils, but the section is primarily aimed at us as teachers. How accessible do the approaches seem to be? Might we use them in class?

If you do decide to use some of the tasks in class, you might start by simply asking pupils how they might solve this and roughly what size of answer they expect. What range of ideas do the pupils come up with?

TASK 07A: We can estimate the value of the division by thinking of division as quotition or measurement. How many times does the brown segment fit into the yellow segment? We can fit in 3 of the brown segments with perhaps another 3/8 or so of the brown segment.

[We know, using formal methods that the actual answer is 3/4 × 9/2 = 27/8 which is indeed 3⅜!]

TASK 07B: Here we have added a line to the diagram in Task 07A to produce a double number line (DNL). We can see (and know from the previous task) that the number represented by '?' on the top scale is (about) 3⅜.

We can calculate the value by finding the number that scales ²⁄₉ to 1 and applying this scale factor to ¾. The scale factor is, of course (?!), ⁹⁄₂, the reciprocal of ²⁄₉, since ²⁄₉ × ⁹⁄₂ = 1.
And ¾ × ⁹⁄₂ = 27/8 = 3⅜.

TASK 07C: Here is an alternative DNL interpretation of the division. It can be thought of as partitive division or sharing: if ²⁄₉ of a share is ¾, what is 1 share?

Analytically, we can find the number that scales ²⁄₉ onto ¾ and apply this to 1. In practice, this takes us round in circles, unfortunately, since it is this very factor that the given division represents. However, we can solve the division by 'skipping along the number lines' using 'rated addition':
2/9 corresponds to 3/4, so 4/9 corresponds to 6/4, so 6/9 corresponds to 8/4, so 8/9 corresponds 12/4, and so 8/9 + 1/9 = 1 corresponds to 12/4 + 3/8 = 3⅜.

TASK 07D: A ratio table can be a useful way of organising multiplicative information, though it has lost the (geo)metric information carried by a double number line. Also, putting information into such a table is not always a trivial task. For example, in the given table, how do we know to put the '1' under 2/9 and not under 3/4?

We can solve the task by finding the 'horizontal' multiplier that maps 2/9 onto onto 3/4 and applying it to 1; or by finding the 'vertical' multiplier that maps 2/9 onto 1 and applying it to 3/4. Here, the vertical multiplier is perhaps easier to find as it is simply the reciprocal of 2/9. [Note that the latter provides us with a formal derivation of the turn it upside down and multiply rule: multiplying both terms in our given expression by 9/2 transforms it into (¾ × ⁹⁄₂) ÷ 1, or simply ¾ × ⁹⁄₂.]

TASK 07E: This task is asking pupils (or us) to work in quite an abstract way. However, it might be more accessible to some people than the pictorial models in the tasks above. [Though we could of course use the models to help us solve some of the steps.]

The result of the first line is 9. (Why?) In the second step we halve that (4½). We then divide by 4 (1⅛). Finally we multiply by 3 (3⅜).

TASK 07F: Here we work our way to the desired expression by leaving ¾ unchanged, in contrast to the previous task where 2/9 was left unchanged. Which progression is easier?

The result of step 1 leaves ¾ unchanged. In step 2 we divide by something 9 times as small so the result will be 9 times as large (27/4 or 6¾). In step 3 we halve the current result (3⅜).

TASK 07G: Here we start with the expression we want to evaluate and transform it into simpler but equivalent expressions. What is going on in each case?

TASK 07H: This is the first of three tasks where we use two dimensions to model our given expression, rather like a Cartesian graph with its orthogonal axes. This task's diagram bears a close resemblance to the DNL in Task 02B, where we can think of the lines as parallel axes.

There are lots of (closely related) ways of looking at this task, based on the fact that we have two similar triangles. If we use a 'measurement' interpretation like the one in Task 07B, we can say that the question "How many times does the brown segment fit into the yellow segment?" is equivalent to "How many times does the shorter green segment (of length 1) fit into the longer green segment (of length 3⅜)?"

TASK 07I: This relates quite closely to the DNL in Task 07C, which we interpreted in terms of 'sharing'. Here we could ask something like this: "If a triangle whose base is 2/9 of the base of a similar 'whole' triangle, has a height of 3/4, what is the height of the whole triangle?"

TASK 07J: Finally, we model the situation using a proper Cartesian graph, though whether it Fosters greater insight is perhaps a moot point!

We could, if we so wished, interpret the graph algebraically. For example, we could think of the slanting line as having the equation y = ²⁄₉x, or x = ⁹⁄₂y.
So when y = ¾, the value of our expression is given by the value of x at that point.