Week 8-alt: Non-symbolic ratios

There are researchers* (see bottom of the page), particularly in the field of neuroscience, who posit the notion that humans (and some other species, perhaps) possess an innate, non-symbolic ratio processing system (RPS). Whether or not such a thing exists, there clearly are ratio situations that we cope with successfully at an early age. Here we consider ratio tasks that lend themselves to an intuitive approach - but which might also prompt children to supplement this with an analytic method to provide certainty.

Task 08A: Here is a sketch of an everyday situation where we compare the height of objects that are different distances from the viewer. It involves two bus stops, which we can assume are the same height in reality (though not on the page), which allows us to compare the heights of the people (Heather and Tom) standing next to them.

It is pretty clear that Tom is taller than Heather. We can confirm this in various ways. For example, we can see that the height of two Toms is much closer to the height of a bus stop than the height of two Heathers:

We can express this numerically, for example by saying that a bus stop looks to be about 3 times as tall as Heather and about 2½ times as tall as Tom - or that Heather's height is about ⅓ and Tom's about ⅖ of the height of a bus stop. We can get more precise estimates by measuring the various lengths: it turns out that Heather and Tom's heights are roughly 0.31 and 0.38 of the height of a bus stop respectively.

Another approach would be to use perspective (or mathematical enlargement); the red ray touches the top of Tom's head but goes well above Heather's, so Tom is taller:

Task 08B: Here we can plainly see that Page 2 is more red (or less blue) than Page 1 and so contains more examples of the letter a.

We can confirm that Page 2 contains more as than Page 1 in various ways. For example, we could count them all! Or, more sensibly, we could compare some individual lines. Or we could analyse the individual words adaptable and adapt. The letter a occurs in 3 out of 9 or 1 out of 3 letters in adaptable, and in 2 out of 5 letters in adapt. Which fraction is larger, ⅓ or ⅖? [Similarly, we could compare the occurrence of a in adapt and in able: ⅖ and ¼ - which is larger?]

Task 08C: Here we can see that Merle has far fewer red grapes than Reece, but also fewer white grapes. It turns out that Merle has relatively more red grapes than Reece. This might not be obvious to everyone, especially as it is not easy to count the various numbers of grapes, especially for Reece. Whatever conclusion children come to, most will realise that they have to consider the numbers of white grapes as well as red. But how? - additively or multiplicatively?

It turns out that Merle has 10 red and 4 white grapes, and that Reece has 20 red and 14 white grapes. Additively, the relations are the same: they both have 6 more red grapes than white grapes. On the other hand, Reece has twice as many red grapes than Merle but over three times as many white grapes. What do these facts mean for the colour of the resulting juice?

Another approach would be to put Reece's grapes into two groups of '10 red and 4 white grapes', to match Merle's group. This would produce twice as much juice for Reece than for Merle, but the same colour - until we add the remaining 6 white grapes to Reece's juice.

Task 08D: Here it is fairly easy to see that Merle's plate is fuller than Reece's, without needing to quantify the 'fullness' or density of the sets of gooseberries.

It turns out that Reece's plate is about 1.41 (or √2) times as wide as Merle's and so has twice the area (we wouldn't necessarily expect children to determine this). However, Reece has less than twice as many gooseberries as Merle - only 18, compared to Merle's 12. So Merle's plate is indeed 'fuller'.

In the variant below, the numbers of gooseberries are unchanged as are the areas of the plates: Reece's plate again has twice the area of Merle's.

Task 08E: Here we have three rectangles which we can think of as part of a family, where the width and height of successive rectangles both change by the same amount (5 units) each time. This will lead some children to conclude that the rectangles are all the 'same shape' (in the sense of being mathematically similar) even though they might not look it!

The task can be made considerably more challenging by removing the yellow rectangle. It is quite difficult to tell, just by eye, that the red and blue rectangles are not similar. One way to try to do so is to visualise the diagonal, from bottom-left to top-right, for each rectangle. Do these diagonals overlap?

The task can be made less challenging by including a more extreme member of the family of rectangles, as in the diagram below. This might well nudge some children into seeing that the larger rectangles get progressively more square! We can quantify this in various ways. For example, the smallest of these rectangles (10 units by 5 units) is twice as wide as it is high, whereas the largest (25 by 20) is only 1¼ times as wide.

 Comments welcome! Send emails to mathsuntangle@gmail.com

 

Week 12: Hidden fractions

This is essentially about fractions, though we don't mention them by name! We use the number line to focus on the measurement and ratio aspect of fraction rather than part of a whole.

Task 12A: This gives us an opportunity to see what strategies children use to estimate parts of a line. We use a line with a large number of divisions to discourage children from simply counting (though they can do so to check their estimates) and to convey the idea that we don't expect the estimates to be spot on. We have chosen numbers where the use of benchmarks can be helpful. For example, the 18 mark will be just under halfway along the line segment, 12 will be just over a quarter (or under one third), 25 is over one half but less than three quarters, say. Some children might spot that 25 is 5/8 of the way along and, similarly, that 30 is three quarters along the 0 to 40 segment.

Task 12B: We don't require anything precise here, beyond the fact that the 13 mark is more than halfway along its line segment (actually quite close to 2/3) and that 23 is slightly less than halfway along its segment - so 13 is nearer the right-hand end of its segment. Item 2 attaches a story to the same numbers - do children appreciate that it makes sense to use ratios to compare the two types of bean rather than the differences in the number of germinating and non-germinating beans? In other words, do they appreciate that it is more appropriate to use multiplicative rather than additive reasoning?

Some children might express the relations using fractions or, more likely perhaps, decimals or percentages. If it seems appropriate, you might want to discuss the idea that comparing 13-out-of-20 and 23-out-of-50 is essentially about division: 13÷20 is greater than 23÷50.

Task 12C: Again, we don't expect a very precise answer here (the red arrow is actually pointing at 35.2), but do children at least appreciate that the red arrow is pointing more than half way along the line segment?

Some children might argue that since 11 is 1 more than half of 20, the red arrow will be pointing at 1 more than half of 64 (that is, 33). This is quite a sophisticated answer, though it involves additive rather than multiplicative reasoning. The gap between the red arrow and the halfway mark, 32, will actually be just over 3 times 1 (since 64 is just over 3 times 20).

Task 12D: The 3 mark is further along the 0 to 19 line segment than along the 0 to 20 line segment: 3 out of 19 is greater than 3 out of 20. However, some children will think the opposite, especially if they are focussed on 'parts of a whole': 3 out of 19 involves fewer parts so will be smaller.

 

Task 12E: The red arrow is pointing at the 15 mark. How close do children get to this? Do they at least chose a number below 20 (half of 40)?

Some children will recognise that we are in effect dealing with the equivalent fractions 3/8 and 15/40. And/or they might see that numbers on the bottom line are 5 times the corresponding numbers on the top line. Some children might think of splitting the top line into 8 equal parts, perhaps by halving, halving and halving again. Doing the same on the bottom line reveals the numbers 20, 10 and then 15.

Week 11: Another SAT interlude

We look at some more SAT items taken from Paper 1 of the Key Stage 2 National Curriculum Mathematics tests from 2025. As before, we ask children to solve a given SAT item (in any way they choose) and then look at some variants that explore some of the mathematical relationships engendered by the item.

Task 11A: The SAT item here is Question 8 from the 2025 Paper 1 test. The most obvious way to solve it is to evaluate 12×3 and then multiply the result by 10, which we can express like this:  (12×3)×10. Variant 1 makes this explicit. Variant 2 brings out the fact that we could also solve the item by evaluating 3×10 and then multiplying by 12: 12×(3×10). This works because multiplication obeys the associative law: (A✯B)✯C = A✯(B✯C). Addition obeys this law too.

Multiplication is also commutative, so (12×10)×3 would be another way to solve the item, which is a variant that children might well come up with.

Variant 3 is perhaps more challenging. Where did the 24 come from?! One explanation would be to say we have doubled 12 and halved 10. Here is a more formal way of explaining it:
12×3×10 = 12×3×2×5= 12×2×3×5 = (12×2)×(3×5) = 24×15.

A systematic way to generate more variants like Variant 3 is to write the original expression as a product of prime factors: 12×3×10 = 2×2×3×3×2×5.
The factors can then be reordered and combined in a variety of ways.

Task 11B: The SAT item here is Q9 from the 2025 Paper 1 test. The missing term in the given equation is 54. Variants 1 and 2 have each been formed by adding or subtracting a given amount to/from the right hand side of the equation and to/from the first term on the left hand side. So in each case the second term on the left hand side of the equation remains the same: 54.

Variant 3 is rather different: 326 and 380 have both been halved. What happens to 54?

Task 11C: There is a very simple rule that can be used to solve this SAT item (Q11): 'multiply the numerators, multiply the denominators', or 'multiply tops and bottoms'! The aim of the task is to give some meaning to what is going on, by using a number line to model the fractions. This requires some quite careful thought:
we can think of 5/6 as being 5/6 of the distance from 0 to 1; however, the fraction 2/5 is operating on this 5/6 distance, not on the 0 to 1 distance. The diagram shows that there are 5 equal intervals from 0 to 5/6. 2/5 of these 5 intervals is 2 intervals, which is 2/6 of the way from 0 to 1.

The distance that is half of the distance from 0 to 1, can be though of as 3 of the 6 given intervals from 0 to 1; this is also 3 of the 5 intervals from 0 to 5/6 which is 3/5 of the distance from 0 to 5/6:
3/5 of 5/6 = 1/2.

Task 11D: The SAT item here (Q14) is testing children's knowledge of the long multiplication algorithm. Michael Gove and Dominic Cummings had to know this algorithm when they were at school; in 2014 they declared that our young children should have to know it too. The algorithm rests on the fact that multiplication is distributive over addition. So instead of trying to multiply 614 by 32 in one go, we split 32 into 30+2 and multiply by 30 and by 2 separately. Formally, we write this: 614×32 = 614×(30 + 2) = 614×30 + 614×2.

Variant 1 is a slightly longer form of the standard algorithm: it shows explicitly that we are multiplying by 30 and by 2. How aware are children that this is what the algorithm is actually doing?!

Variant 2 underlines the fact that we are splitting the multiplier 32 into smaller parts. We normally do this by splitting the number into its 10s component and 1s component, but we can spit it in other ways, for example as here: 32 = 20+12. We should get the same answer!

Variant 3 uses a different idea which takes us back to the associative law:
32 = 4×8 and so 614×32 = 614×4×8 = (614×4)×8.

Task 11E: This SAT item (Q15) is rather neat, though I don't know the rationale for including it in the Paper 1 test. Are children expected to use the 'bus stop' algorithm? (4 into 2 doesn't go; 4 into 20 goes 5, remainder 0; 4 into 0 is 0; 4 into 0 is 0; answer 500.) Or do they think of 2000 as 100×20 as in Variant 2; or do they visualise 2000 split into 4 equal parts as in Variant 3?

Week 10: A ÷ B, how many times does B go into A?

Here we adopt a particular interpretation of division - quotition or measurement. For a division like 100÷20, we ask, How many times does 20 go into 100? We can also think of division in terms of partition, or sharing. This is probably the interpretation that children meet first, but, strangely perhaps, when we perform division, for example by using chunking or the long division algorithm, we tend to use quotition. Division by a rational number is also more easily interpretative as quotition than partition. Consider, for example, 10 ÷ ⅓. It seems more natural to think of this as 'How many ⅓s are there in 10?' than 'What is 1 share if ⅓ of a share is 10?' [Or, compare 'How many ⅓ pint bottles can I fill with 10 pints of milk?' with 'If 10 pints is ⅓ of my milk supply, how much milk do I have?' We might be able to solve the latter fairly easily, but it seems an odd way to view a situation.]

Task 10A: In these items we mostly divide by 20 and provide children with a list of multiples of 20 to help then visualise the situation. However, we also try to get children to think beyond what they see in front of them by asking about multiples not in the list, or about multiples of other numbers than 20. Each of the items involves a new cognitive leap, and this might be going much too fast for many children. Thus, here and on later tasks, you might want to dwell longer on a particular idea by devising variants that are similar to a previous item (and for some children, you might want to devise variants that are even more challenging!).

Item 3 can be solved using the fact that 280 is twice 140 and will thus contain twice as many multiples of 20. This might be quite an abstract idea for some children, who might decide to list all the multiples instead. 

Item 4 can be solved using the idea that 280 will contain half as many multiples of 40 than of 20. Again, some children might solve the task in a more grounded way, by listing multiples of 40.

Item 5 is intended to bring out the idea that asking 'How many...' is a way of interpreting division. Do children see the connection, or, indeed, have they already done so? Some children might solve this or subsequent items using the idea of sharing. You might want to spend some time comparing the two interpretations of division. In the case of 100÷20, for example, one could compare 'How many 20cm lengths can be cut from a 100cm rod' and 'How long is each piece if a 100cm rod is cut into 20 equal pieces?'. In one case the answer is 5; in the other it is 5cm.

Task 10B: This is similar to the previous task but involves multiples of 15 which are probable slightly less familiar to children than multiples of 20. We also push things a bit further by involving a fractional divisor (7.5 in Item 6) and a fractional answer (1½ in Item 7).

Many of these items can be solved in several ways, which can provide plenty of scope for discussion. In Item 5, children might notice that the multiples of 75 (namely 75, 150, 225 and 300) lie on a slanting line in the table - why? Some might notice that Items 4 and 6 have the same answer, since the numbers in Item 6 are both one tenth of the corresponding numbers in Item 4.

Task 10C: Here we move into the realm of decimals and you might need to decide how far to delve into the more challenging ideas and how much to linger on simpler ones by devising straight-forward variants of the earlier items.

Task 10D: This is similar to the previous task, but children might find it to be more challenging as there is a tendency for children to be more comfortable with simple decimals than simple vulgar fractions.

Many children will have been taught a procedure for converting mixed numbers into improper fractions. It will be interesting to see whether they see any connection between the procedure and what is being asked in Items 1 and 3. (And Item 2, for that matter!)

Task 10E: This task comes from the CSMS Fractions (3 and 4) Test. It was given to a representative sample of over 200 Year 9 students in 1976/7 and to a similar sample in 2008/9. It was answered correctly by 57% of the sample in 1976/7, but by only 29% in 2008/9.
 

It always surprises me that the item turned out to be so difficult, according to these data. Is it because children reach for a taught procedure which many of them then get wrong, and which diverts their attention from thinking about the story, of runners on a running track? My hunch is, that if children paused and paid attention to the story, rather than reaching for a procedure, then many of them would be successful at mentally marking-off 1/8 km lengths until one gets to 3/4 km. (Admittedly, this grounded approach involves some knowledge of equivalent fractions, but then eighths, quarters and halves are part of the most familiar family of equivalent fractions that there is.)

If you have any comments about the blog, or would like to share experiences of using the tasks, please email me here: mathsuntangle25@gmail.com

Week 09b: Sum decimals and fractions

Here we embark on a gentle excursion into adding fractions and compare the effect of expressing numbers as common fractions or decimals.

Task 09A: This is a fairly straightforward task, but it is interesting to see what strategy children use to choose values for the numbers A, B and C, especially at the point where they are asked for another ten examples (a request that might seem rather excessive!).

A simple way of approaching this task is to choose fairly random values for A and B (with A and B positive and A+B<10), and then to derive C by subtracting their sum from 10. However, there are many subtle features that children could employ. For example, they might limit their choice to numbers with one place of decimals; they might allow 0 in that decimal place for some or all of the three numbers; they might try to limit the amount of 'carrying' by choosing decimal parts that sum to 1 rather than to 2; they might form new sets of numbers by re-ordering an existing set, or by modifying the set in a simple and systematic way. Where children have used restrictions that massively simplify the task, you might want to challenge children to work without them.

Task 09B: This is also fairly straightforward, and is of interest mainly for how it compares to later tasks where the decimals are replaced by common fractions.

It is likely that children will initially choose values for A and B that have only one decimal place, as in 0.1 + 0.4. However, they will soon run out of such numbers when asked to 'find ten more'. How readily do they see that they could use numbers with two decimal places, and whose sum is 0.50, as in 0.01 + 0.49?

Task 09C: Some children might find this task a good deal more challenging than the previous task. A neat way to solve it is to find a fraction equivalent to ½ (for example, ⁴⁄₈) and then to split this into two (for example, ⅛ and ⅜). An alternative approach is to choose a value for A and then derive the value for B. Children might discover that this is simpler when the denominator of A is even (as with 7/20) than when it is odd (as with 4/9). Note that the notion of equivalent fractions (and, implicitly, common denominators) is not needed in Task 09B, beyond the idea that 0.5 is equivalent to 0.50, etc.
 

Some children might solve the last part of the task, where A = 4/9, by thinking of ½ as being equivalent to 4½/9, so that B = ½/9 which is 1/18.

Task 09D: This task reinforces the idea that it is easier to find B when the denominator of A is even. It is also worth pointing out that this task would not work well if the fractions were written as decimals, as in Task 09B: most unit fractions would turn out to be recurring decimals (and when A is recurring, B is too).

As children work their way down the table, they might become more fluent in finding the values of B. If so, they might enjoy continuing the table. Children might notice that for a unit fraction A with an odd denominator, the denominator is 2 more than the numerator of B.
 

Task 09E: We open things out slightly by choosing a less familiar value for A+B, namely ⅓. Where previously it was relatively easy to find B when the denominator of A was even, now this only applies when the denominator is a multiple of 3.

Again, it can be fun to continue the table. This time children might notice that for a unit fraction A with a denominator that is not a multiple of 3, the denominator is 3 more than the numerator of B.
 

Week 09: Transforming expressions

Here we take a gentle stroll through the art of transforming expressions into equivalent expressions, including the method of 'constant difference' used in the previous week's tasks.

Task 09A: Here we have the classic situation, which might be quite familiar to some children, where an addend or subtrahend is close to a 10s, 100s or 1000s boundary, etc. In the case of a simple addition, we use the 'compensation' principle: if we increase one component we have to decrease the other by the same amount to keep the expression's value unchanged. In the case of a simple subtraction, we apply the same change to both components.

There is plenty of scope here for devising similar tasks if you feel children would benefit from further exploration or practice.

Task 09B: Here we look more closely at the 'constant difference' principle. Changing the component 27 to 30 or to 20 is equally beneficial in that they result in similar, more simple subtraction expressions. However, it could be argued that the change made to the other component, 642, is (slightly?) more demanding in the latter case since it involves
• a larger number (7 rather than 3),
• subtraction rather than addition and
• bridging a tens boundary (at 40).

Task 09C: Here we are using 'compensation' to transform an addition expression and this time it can be argued that the transformation of the first expression is more demanding since it is here that the change made to the other component, 642,  involves
• a larger number (7 rather than 3),
• subtraction rather than addition and
• bridging a tens boundary (at 40).

Some children might feel that the difference in demand of the two methods is minimal. That is a perfectly valid stance to take! The aim of the task is not to suggest that the difference is so vital that children should learn to adopt one method and reject the other. The aim is more modest and diffuse, of helping children develop a greater sensitivity about the various elements that form part of a mathematical method.
Task 09D: Here the constant difference principle is enacted in stages.

When the given method is applied to 416 – 27, we get this:
416 – 27 = 419 – 30 = 489 – 100 = 389.
There are of course other methods for performing the calculation and you might want to ask children for some. One that works for these particular numbers is to 'decompose' 27 into 16+11, which gives this:
416 – 27 = 416 – 16 – 11 = 400 – 11.
And one way to continue would be this: 400 – 11 = 399 – 10 = 389.

Task 09E: Some children will attempt to solve this task by simply adding together all the terms of the expression. However, it is hoped that the length of the expression, and the clue about its approximate value, might prompt them to look for an alternative approach.

We are told the value of the expression is close to 300, which might encourage children to consider 50 + 50 + 50 + 50 + 50 + 50 as a way of checking whether this is true. How easily can they see that this is indeed equal to 300? By using 6×50 = 300, or perhaps by thinking of (50+50) + (50+50) + (50+50)? In turn, can they spot that the original expression will differ from 300 by this amount:
1 + 4 – 2 – 6 – 3 + 8 = 13 – 11 = 2.
This is greater than 0 and so the expression is greater than 300.

 

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Week 08: Subtracting on the number line

In this week's task we use the number line to model (and perform) subtraction, A – B. (We restrict ourselves to natural numbers, with A greater than B, and focus on the specific subtraction 44 – 28). 

We use the number line representation to model subtraction in two ways:
1. We find the difference between two positions (A and B, or 44 and 28) on the number line, in other words the displacement that takes us from one position (B, or 28) to the other (A, or 44); we can think of this as 'adding on':
B + ? = A;
2. We start at one position (A, or 44) on the number line and see where we end up after a displacement (B, or 28).

Further, we see what happens when we use the 'constant difference' principle, by, in particular, changing 44 – 28 into 46 – 30.

[Note: an interesting tension can arise when one uses a model based on a representation like the number line. Such a model can help one think more clearly about the mathematical elements being modelled; however, until one has become familiar with the model, its detailed features can get in the way of such thinking. This might well happen with some children using this week's tasks, in which case it might be best to concentrate for a while on just one of the tasks, for example 08A or 08C, and apply the particular model to a range of subtraction expressions.]

Task 08A: Here we are thinking of the subtraction 44 – 28 as finding a difference, which we model as a displacement that takes us from the location 28 on the number line to the location 44. 

Andy has found the displacement that takes us from location 28 to 44 as three skips, of 2 units, then 10, then 4. Children might think of it in other equally valid ways, for example as a skip of 2 and then 14, or of 10 and then 6, or even as a single skip of 16.

Whatever particular set of skips one might use to get from 28 to 44, it is worth emphasising that they involve number sense. If we had been given a fully marked number line as in the diagram below, we could simply have counted. To underline the point, you might want to show children this diagram.

When it comes to the last part of the task, children should be encouraged to use a semi or fully 'empty' number line as in the diagrams below.

 

Task 08B: Here we are again thinking of the subtraction 44 – 28 as finding a difference between the locations 28 and 44 on the number line. However, Bella uses the 'constant difference' principle to change 44 – 28 into an equivalent expression 46 – 30, where the difference is easier to see.

Bella has found the displacement that takes us from location 30 to 46 as two skips, of 10 units, then 6, in contrast to Andy's 3 skips in the previous task. Children might think of it in other equally valid ways, and more might see it as a single skip of 16, given the nicer numbers.

Task 08C: In this task, 44 is interpreted as a location on the number line, and 28 as a displacement that takes us to a new location.

Carl interprets the displacement of 28 units as 3 skips, of 20 units, then 4, then another 4, which lands on the location 16. Children might come up with other, equally valid skips that land on 16, for example 4, then 20, then 4.

Task 08D: Here we again interpret 44 as a location on the number line, and 28 as a displacement that takes us to a new location. However, Dee moves shifts the location 44 two units to the right and increases the displacement by the same amount. This will still take her to the desired location but using a displacement that it is easier to apply. As with Task 08B, this makes use of the 'constant difference' principle, though it might not be as easy to construe!

Dee interprets the displacement of 30 units as a single skip that lands directly on 16. Children might come up with other, equally valid skips that land on 16, for example 6, then 20, then 4.

Task 08E: Here we use some closely related stories to represent the subtractions 44 – 28 and 44 – 16. Do some of the stories seem easier to solve than others?

If we use the number line to represent the stories, then Story W fits the model in Task 08A and Story X fits the model in Task 08B. The resulting displacements are 16 and 28 units respectively, so Story W has the smaller displacement, and I think it is for this reason that Story W seems easier to me.

Similarly, Stories Y and Z fit the models in Tasks 08A and 08B respectively, with resulting displacements of 28 of 16 units, which seems to make Z easier (for me!).

[If this doesn't seem very convincing, keep the number 44 in each story, but change 28 and 16 to, say, 38 and 6.]

Week 07: Counting dots

Here we look at quick ways of counting a collection of dots by putting dots into groups. The collections in the first two task don't have a particularly clear, overriding pattern so there can be lots of fairly efficient ways of grouping them! The aim is to encourage children to look for (or impose) structure, even though the structure might not be general. In the later tasks there is the potential to generalise, which some children might find both challenging and exciting, but the tasks can be worked on fruitfully at a non-general level.

Task 07A: The collection of dots here is not entirely random - it has rotational symmetry. However, initially, children might group the dots in fairly random ways, in which case it would help to show them groupings that contain more of a pattern. [It would also help to provide copies of the dot-collection - or squared or dotted paper for children to draw their own copies.]

Once children get the hang of it, they might enjoy the fact that one can group the dots in many, not-entirely-random, ways. Here are some examples.

 

Task 07B: Another collection with rotational symmetry. You might also want to create you own collection, or ask children to do so.

Some possible groupings:

Task 07C: This collection is already quite structured, with two lines of symmetry (and rotational symmetry). Some children might spot that we can think of the collection as showing two copies of the third triangle number, 6. If we push the triangles together horizontally we can form a  rectangle containing 4 row of 3 dots, as hinted at in the given example.

Here are some more possible groupings. The first example hints at the idea of pushing the two triangular sub-groups together vertically, to form a rectangle containing 4 columns of 3 dots. The 4th example introduces the idea of 'phantom' dots which have to be taken away.

Task 07D: Here we have moved entirely away from the idea of a semi-random collection of dots that we had in Tasks 07A and 07B. This collection is highly structured, which makes it mathematically more interesting, including in ways that are quite challenging.

A nice, but challenging, feature of these ways of structuring the dots is that they can all be generalised. Imagine a 'square' of dots with, say, 10 dots rather than 5 along each side. Then we would get these structurings corresponding to the ones in the figure below:
4×10 – 4, 4×(10–1), 4×(10–2) + 4, and 10×10 – (10–2)×(10–2).
You might want to see whether children can generalise any of their structurings in this way.

Task 07E: Here is another well-structured pattern, which can be generalised should you wish to do so.

The collection of dots might remind children of the way 5 dots are arranged on dice. One can think of the collection as three overlapping 5s, as in the first example, below: there are three 5s but with 2 lots of 2 overlapping dots. The third example includes a phantom dot.

As in Task 07D, these structurings can be generalised, although they are probably even more challenging. Consider, for example, a collection that looks like a row of 10 overlapping 5s. Then we would get these structurings correspoding to the three in the figure below:
10×5 – (10–1)×2, 5 + (10–1)×3, and 3×(10+1) – 1.
Again, you might want to see whether children can generalise any of their structurings in this way.

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