MUL 12 new and old

[Note: GERMAN versions of the Week 12 tasks are available HERE.]

 

 Note: We have replaced the original Week 12 tasks by a completely different set, which is shown here. The original tasks can be seen at the end of this sequence.

 

Overview: Here we revisit the notion of mixtures, which we met in Week 3 in the context of mixing black paint and white paint. This time the focus is on the strength, or taste, of liquids such as orange squash after they have been mixed with water.
In a paint mixture, the colour of each paint contributes to the colour overall. In this week’s mixtures, water plays a subtly different role: rather than contributing its own taste to the mixture, we treat it simply as a medium that ‘distributes’ or dilutes the other liquid.

Mathematically the tasks are subtle too, especially when it comes to using ratio to compare mixtures. Comparing the ratio of ‘orange’ to water works perfectly well when verifying that two mixtures have the same strength, or when the amount of one of the components (‘orange’ or water) is the same. As we shall see, ratio doesn’t work so well when the strengths and amounts are different, as in tasks 12C, 12D and 12E.

Monday: At first sight, it might seem that the two mixtures in this task will taste the same, as they contain the same amount of ‘orange’.  However, students might notice that the mixtures consist of different amounts of liquid overall (1 litre, 50 ml in the first case; 1 litre, 100 ml in the second case). This should give students pause for thought.

We can think of the 100 ml of squash from Bottle B as equivalent to 50 ml from Bottle A plus 50 ml of water. This means the second mixture contains the same amount of ‘orange’ as the first, but an extra 50 ml of water - so it is slightly weaker.

Tuesday: As with Monday’s task, this is more tricky than it might at first appear. A useful first step, and one that students are quite likely to take, is to consider adding 400 ml of squash from Bottle B to the 1 litre of water. This produces a situation similar to Monday’s task, in that the amounts of ‘orange’ in the two mixtures are the same, but the total amounts are not.

The first mixture contains 200 ml of’orange’ and 1000 ml of water. If, as above, we add 400 ml from Bottle B to 1 litre of water, this is equivalent to having 200 ml of ‘orange’ and 1200 ml of water. So this second mixture would be weaker. We therefore need to add more squash from Bottle B.

It is likely that students who have taken this path will continue with a trial and improvement approach. It turns out that adding 500 ml from Bottle B to 1 litre of water produces a mixture with the same strength as the first:

500 ml from B is equivalent to 250 ml from A plus 250 ml of water, thus making 1250 ml of water overall. 250:1250 = 1:5 = 200:1000.
[Note: Comparing the ratio of ‘orange’ to water in the two mixtures works perfectly well here, where we are verifying that the strengths are the same. It doesn’t work so well in the next three tasks, where we are guaging mixtures with different strengths.]

Wednesday: At first sight, it might seem that mixture (a) has half the strength of the original mixture as it contains half as much cordial, and as the ratio of cordial to water is 100: 1000, or 1:10, compared to 200:1000 or 1:5 in the original mixture. However, the picture is complicated by the fact that the mixtures have different total amounts: if we used the mixtures to fill two identical cups, would one cup have half the amount of cordial as the other?

In the case of mixture (b), we add the same amount of liquid to 1 litre of water as with the original mixture, but what we add contains only half the amount of cordial. So the resulting mixture has half the strength. If we express the amount of cordial as a fraction of the total amount of mixture, we get this clear picture:
original mixture: 200/1200 = 2/12
mixture (a): 100/1100 = 1/11 > 1/12
mixture (b): 100/1200 = 1/12 = half of 2/12.
[Note: Expressing mixture (b) as a ratio of cordial to water is of limited help: how does 100:1100 = 1:11 relate to the earlier ratios, 1:10 and 1:5 ?]

Thursday: This task is closely related to Wednesday’s task, though this time one of the components is not water but olive oil. This would complicate matters if we wanted to compare the combined vinegar-oil taste of the two given mixtures, but our interest is only in the strength of the vinegar component.

The wording of the task is likely to prompt a ratio perspective at first: the ratio of vinegar to oil for the two mixture is 1:3 and 2:3, which might suggest that the second mixture is twice as ‘vinegary’. However, the task then asks students to imagine a spoonful of each mixture, which should prompt some students to switch to seeing the mixtures in terms of fractions:

1/4 of Ruby’s spoonful is vinegar, compared to 2/5 of Stan’s. So Stan’s spoonful contains more vinegar than Ruby’s but not twice as much: ¹⁄₄ < ²⁄₅ < ²⁄₄.

Friday: As with Wednesday’s and Thursday’s task, we again contrast the effect of using ratio and fraction relations to compare relative quantities. The task is essentially a visual version of Thursday’s task, with the same numerical relations and with identical rectangles substituted for Ruby’s and Sam’s spoonful of dressing.

The three tinted rectangles will look like this:

 

THIS, below, IS THE ORIGINAL VERSION OF WEEK 12, which we have replaced by the materials above.

NOTE: We are replacing this set of 5 tasks (below) with a new set of Week 2 tasks (above), in order to keep the total number of weeks to 20. So this set won't appear in any future printed version of the blog.

 This week, we explore an interesting multiplicative situation taken from physics by considering the properties of a balanced see-saw (or lever). The underlying notion is that of moment of a force, though we start from a more basic principle, perhaps. And we don't delve into moments here but simply refer to a Multiply rule. In a later set of tasks (Week 18) we use ratio to explore an interesting related notion from physics, centre of gravity.

Monday: Here we introduce students to the week's context of a balanced see-saw and to a 'key rule' that can be used to keep the see-saw balanced.  

You might want to explain to students that the diagrams are schematic: some of them carry a lot of information so we have tried to keep them simple, but as a result it looks as though, in real life, the weights might well roll off the see-saw.

Students will probably quite readily see how a weight has been split to form the configurations shown in diagrams A and B. In C, two weights have been split and some of the parts joined again. This suggests a better name for the Key rule might be Equal-split-or-join rule.

Tuesday: Here we suggest that there might be a 'Multiply rule' (involving weight × distance from the balance point, or fulcrum) that can be used to show that a see-saw is balanced. We examine the rule more thoroughly in Wednesday's task.

Can students explain why Abe's Add rule can't be true? What sense does it make to add a weight and a length, and what happens to any relation if we change one or both units? Imagine that Abe claims that a 6g weight 2cm from the fulcrum balances a 4g weight 4mm from the fulcrum, because 6+2=4+4. What happens if we measure in cm? Does 6 + 0.2 equal 4 + 0.4? The Add rule is nonsense!

Regarding Kay's Steps in part b), students might take a while to make sense of Step 2. Why are we allowed to discard one of the 2g weights?

Wednesday: Here we confirm that the Multiply rule works for see-saws that we know, from Monday's task, to be balanced.

Checking that a rule works isn't quite the same as showing why it works. We can certainly show that the Multiply rule is a logical consequence of our Key rule. (For example, it is not too difficult, using the Key rule, to show that a weight of 3g, 1mm from the fulcrum is equivalent to a weight of 1g, 3mm from the fulcrum.) However, that too doesn't really show why the Multiply rule is true, or not in a satisfying way! For that, perhaps we have to turn to physics, or to our experience of moving about on a see-saw!

On the other hand, if we start with a multiplicative expression like 12×5, we can derive an equivalent expression like 6×7 + 6×3 in Diagram A not just by thinking about weights and applying the Key rule, but from the laws that govern arithmetic itself, as here: 12×5 = 6×2×5 = 6×(5+5) = 6×(5+2 + 5–2) = 6×7 + 6×3.

Thursday: Here we use the Multiply rule to check that three given see-saws seem to be balanced. We then return to the Key rule and try to apply the kind of steps shown in Tuesday’s task, and shown again here, to demonstrate that the see-saws are indeed balanced.

This task is quite challenging and it is likely that students will have to take quite a few exploratory steps to solve it. It should perhaps be treated as a puzzle.
In part b), students need to
•split weights
•join weights
•use the balance point (fulcrum) to introduce an extra weight
•use the balance point (fulcrum) to discard a surplus weight.

 

Friday: Here the total weight on each side of the balance is the same.

We can use the Key rule (or Equal-split-or-join rule) to replace the 5g and 2g weights on the left with a 7g weight. This is again quite challenging and might involve several trials before students hit on a set of steps that works.

Students might notice that the point where this 7g weight is located divides the line segment joining the 5g and 2g weights in the ratio 2:5. The match between these numbers is, of course, no coincidence and is worth exploring further.
One extension task would be to explore what happens when the 5g and 2g weights are moved 1cm to the left, say. It turns out the two weights are still equivalent to a 7g weight that divides the segment joining them in the ratio 2:5.  Notice that if the distance from the fulcrum of the 2g and 5g weights is d units and d+7 units, then the laws of arithmetic allow us to write this:
2d + 5(d+7) = 2d + 5d + 35 = 7(d+5).
You might like to consider a more general case, of weights u and v at a distance d and d+e to the left of the fulcrum!
Another extension task would be to replace the 5g and 2g weights by 4g and 3g weights, say. The two weights can again be replaced by a 7g weight, this time dividing the distance between them in the ratio 3:4. We return to this ratio property in Week 18.