ALG 20

Week 20: Here we look at some figurative patterns where there is (or could be) a rule between different elements of the pattern (eg between the number of blue tiles and yellow tiles).
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Monday: Can we invent a simple (linear) rule that allows us to determine the number of yellow tiles when we know the number of blue tiles?

From the limited information that we are given, there is no unique rule connecting the numbers of blue and yellow tiles. However we can invent rules that fit our two examples (Terry and Pete) and assume they apply to Nona.
The simplest rule can be expressed like this:
every Dogtile has 4 yellow tiles for its feet, plus the same number of yellow tiles for its neck as it has blue tiles, and one fewer yellow tile for its tail.
For Terry and Pete, this would give 4 + 4 + 3 and 4 + 2 + 1 yellow tiles respectively.
For Nona, it would give 4 + 9 + 8 = 21 yellow tiles.
We can express this rule in slightly different but equivalent ways, and also in ways that show the structure more explicity - as we do in Tuesday's task.
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Tuesday: Can pupils see what the various numbers in Ella's expressions refer to? Nona has 9 blue tiles. Where would 9 appear in Ella's expression?

The two 4s in Ella's expression for Terry, and the two 2s in her expression for Pete, play very different roles. In each case one of the 4s and one of the 2s is a 'quasi variable' and would be replaced by 9 in the expression for Nona, as here: 2×9 – 1 + 4.
We can show this more clearly by putting the quasi variable in blue, say:
Terry: 2×4 – 1 + 4.  Pete: 2×2 – 1 + 4.  Nona: 2×9 – 1 + 4.
Notice that we are quite close here to having a general formula for the number of yellow tiles when there are, say, B blue tiles: 2B – 1 + 4.
Can pupils explain Ella's rule? You might want to consider equivalent rules, such as this:
Terry: 4 + (4 – 1) + 4.  Pete: 2 + (2 – 1) + 4.  Nona: 9 + (9 – 1) + 4.
Here the first term refers to the neck, the second to the tail and the third to the feet.
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Wednesday: Meet the Blocks... Can pupils find a simple rule?

As before, there is no unique rule connecting the numbers of blue and yellow tiles. However we can invent rules that fit our two examples (Tertius and Juno) and assume they apply to Octavia.
The simplest rule can be expressed in this (perhaps rather long winded) way:
Each person has 2 yellow tiles for the ears and then the same number of yellow tiles for each hand and each foot as there are blue tiles for the body.
For Octavia, who is made of 8 blue tiles, we could express the rule in ways such as these:
2 + (8–1) + (8–1) + (8–1) + (8–1), or
2+ 4×(8–1).
We could write these expressions more simply as 4×8–2. Does this still make sense in terms of our tile pattern? 
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Thursday: The stick pattern here is reminiscent of the matchstick pattern in tasks 10C and 10D.

Here it is reasonable to assume that the sticks needed for the head, neck, arms and legs don't change but that adding an extra blue body tile requires 3 extra sticks (as in task 10D). So Octavia, who has 5 more body tiles than Tertious, will need 5 lots of 3 extra sticks, making 31 sticks in all.
This is quite challenging. It would be more challenging still to show the full structure of the relationship between the number of blue tiles and sticks, but we could do it with expressions like this for Octavia's sticks:
4 + 1 + 2 + 2 + 4 + (8–2)×3; or
9 + 4 + (8–2)×3, where the head, neck, arms and legs require 9 sticks, one of the body tiles requires 4 sticks and the remaining (8–2) body tiles require 3 sticks; or
9 + (8–1)×3 + 1, where we are now saying the body tiles each require 3 sticks, plus 1 extra stick.
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Friday: "Quincy, how many yellow parts do you have?"
"Excuse me, I will ask Alexa."

Here, the default rule would seem to be that each robot has 2 yellow parts for each foot, twice the number of arms as blue parts, and this number of yellow parts for each hand.
So Quincy, who has 5 blue parts, would have this number of yellow parts:
2 + 2 + 5×2×5, or
4 + 2×5²,
which makes a total of 54 yellow parts.
If we accept this rule, then Quincy will look something like this:

Bye, bye, Quincy. Goodbye dear reader.

ALG 19

Week 19: Having bitten the bullet in Week 18, we continue working with algebraic expression in this week's tasks. The focus again is on evaluating expressions, though we do also compare how their values change and we introduce two informal ways of determining when two expressions are equal.
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Monday: As with the expressions in Week 18, you might have to remind pupils of the conventions we use in writing expression, for example that 10e+3 means (10×e)+3. However, it would be worth first finding out what meaning pupils give to the expressions.
Before embarking on parts a), b) and c), it could be interesting to ask pupils, "Which expression is the larger?" It is quite likely that some pupils with think that the first expression is always larger (because of the 10 and the 2), while others think the second is always larger (because of the 3 and the 47). It might come as a surprise that the relative size of the expressions changes.
It is not expected that pupils will (or should) solve part c) using a formal method (such as taking 2e from both expressions). Rather, we expect pupils to use some form of trial and improvement, although this is quite challenging both because both expressions change on each trial and because the solution is not a whole number. For some pupils, this process will be fairly hit and miss, while others might get a sense that a change in e has a greater effect on the first expression than on the second (Why?). However, we anticipate that both sets of pupils will gain satisfaction from (eventually?) finding the solution!

When e = 2, the value of the first expression is considerably less than the value of the second (23 compared to 51). However, as e increases, the first expression rapidly catches up.... The two expressions have the same value when e = 5½.
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Tuesday: Here we compare our two expressions in a systematic way. Some pupils might already have done this, so this is an opportunity for drawing the class together.

Starting at e = 2, and increasing e by 1 each time, we see that the first expression rapidly approaches the value of the second and overtakes somewhere between e = 5 and e = 6.
Some pupils might notice that in the table, the difference in the value of the two expressions decreases by 8 each time (the differences are 28, 20, 12, 4, -4). Why is this happening?
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Wednesday: Here we compare two new expressions. We again use a systematic table of values and this time explicitly focus on the difference between the values and how this changes.

As u increases by 1, the difference in the value of the expressions decreased by 5 each time. When u = 3, the difference is 35, so it will take seven more steps of 5 to reduce this to 0. So this happens when u = 3+7 = 10.
This is an elegant (and quite powerful) method which might inspire some pupils. However, for others it might be taking things too far, so don't push this too hard. The solution u = 10 is relatively easy to spot using trial and improvement, and if some pupils are content with this, that's fine at this stage!
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Thursday: Here we introduce another method for equating the two expressions that involves matching elements in the expressions.

We are adding 3u+1 to 5u in one expression and to 50 in the other. So the two expresions will have the same value when 5u is equal to 50, ie when u = 10.
[Note: We are using the same expressions here as in Wednesday's task, so we already know the solution. It is worth making a virtue of this: we are not so much interested in the solution per se, but in gaining insight into methods (and into the nature of expressions) that we can make use of elsewhere.]
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Friday: These expressions may look daunting.... at first.

We have contrived to have two expressions that appear challenging but where the matching method used in Thursday's task might suddenly click for pupils, leading to a relatively simple solution.
Thus we can write the expressions as 23m + 7m + 33 and 7m + 33 + 2300, so they will have the same value when 23m and 2300 are equal, ie when m = 100.

ALG 18

Week 18: In this set of task we demonstrate some of the power, intrigue and utility of algebra, though perhaps not its serious purpose. We use algebraic expressions (involving a variable, u) to describe the position of the vertices of a quadrilateral, and then look at what happens to the quadrilateral as the value of the variable changes.
This week's tasks involve a rare excursion into the use of algebraic symbols. Pupils might need some gentle support to help them read the expressions correctly, though we are not asking pupils to do much with the expressions, other than to evaluate them for different values of the single variable involved.
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Monday: This task sets up the work for the rest of the week. You might need to work through it quite carefully, to remind pupils of various conventions and ideas, for example that A, B and C, etc, refer to points on the given grid, and that they form the 'vertices' or corners of a 4-sided shape - a 'quadrilateral'. Also, when we write a term like 4u, we mean '4 times u' and that if we know the value of u we can find the specific value of the term, or of an expression like 4u–8.
Other, more subtle ideas are also at play here, for example that an expression like 4u–8 can be thought of as a number in its own right, not just a set of instructions for finding a number; moreover, this number varies as u varies - and different expressions may vary at different rates. These ideas need time to develop over many years.
There is another issue here which may puzzle pupils even if they don't raise it explicitly. We may decide to let the issue stay dormant at this stage, but nonetheless it is important that we ourselves are aware of the issue, which concerns the 'status' of the algebraic expressions. In a 'real life', practical situation, the expressions would have a 'purpose'. They would arise out of the nature of the situation and the relationships involved, and they might help us to solve a worthwhile problem. That is not the case here! Rather, the only 'purpose' of our current expressions is to gain experience of working with such expressions and of getting a sense of how the behaviours of different expressions compare. Our hope is to do this in an intriguing and mathematically illuminating way, despite the absence of a vital problem that algebraic expressions could help us to solve. That is for another time (hopefully!).

In part b), we can tell from the diagram that
4u – 8 = 20, and
4u – 6 = 22, and
u + 7 =14.
We can use any one of these sentences to determine that for this diagram, u = 7.
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Tuesday: Part a) refers back to Monday's task. In part b) the value of u is increased by 1. Before embarking on this part, you might want to ask pupils to predict what might happen.

Points A, B and C all move to the right, but not all by the same amount. Why?
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Wednesday: Here we increase the value of u by 1 again. What will happen?

Do any pupils predict that the points move to the right again, and by the same amount as each moved when we previously increased the value of u by 1?  How do they explain this? Can they relate it to the numbers in each expression?
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Thursday: This is quite challenging, though it can obviously be solved by trial and improvement. Given what has happened to the shape for the values of u that we have considered so far, some pupils might realise that value required here will be smaller.

Some pupils might be able to solve the task analytically, and it might be worth discussing such an approach, even if the class first solves the task by trial and improvement.
Pupils might have noticed that side AB is always the same length and at the same angle, regardless of the value of u (Why?). This means that point A is the 'bottom' vertex of the 'diamond' and the fixed point D is the left-most vertex of the 'diamond'. Also, D will be 2 units to the left of B. We know that D is always 10 units from the red line, so A will be 12 units from the red line. So 4u–8 = 12 when the quadrilateral is this 'diamond' shape. What is the value of u when 4u–8 = 12 ?
We get the 'diamond' when u = 5, as shown in the bottom diagram:

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Friday: Again, we can find the values of u by some form of trial and improvement, or in a more analytical way.

When A and E coincide, the distance of A from the red line is 0 units, ie 4u – 8 = 0. What value of u fits this sentence?
When B and D coincide, the distance of B from the red line will be 10, ie 4u – 6 = 10. What value of u fits this sentence?
The resulting shapes of the quadrilateral (ie when u = 2 and when u = 4) are shown below.

 It is useful to think of what happens to the quadrilateral as u increases over the complete range that we have considered (ie from u = 2 to u = 9). You might want to ask pupils to describe what happens.
When u = 2, the side AB is way over to the left, while point C is just to the left of point D (see the above slide). As the value of u increases, side AB and point C move to the right, but AB moves faster (How much faster?). When u = 4, B has reached D, so the quadrilateral has 'collapsed', and C is just to the right of D (see the above slide). When u = 5, the quadrilateral has opened out again, to form a 'diamond', with C being caught by A and ovetaken by B. As u continues to increase, the quadrilatral becomes more and more stretched, with AB moving further and further to the right and away from C as well as D.
How does this geometric description relate to the algebraic expressions? (A question more for us than the pupils!)

We have a movie to go with this, which you might want to show the pupils at this stage. Stills from the movie are shown below.

ALG 17

Week 17: This week we look more closely at the Cartesian graph. We plot points that fit a story about numbers of 2p and 5p coins. It turns out that the points lie on a straight line, and we use, and try to make sense of, this fact.
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Monday: Here we introduce the story that will feature throughout the week, and find some values (of numbers of 2p and 5p coins) that we will subsequently plot on a graph.

This task is straightforward [apart from the fact, perhaps, that in part c) the given number refers to Ella's coins rather than Dan's]. We end up with the ordered pairs 50,60 and 100,40 and 125,30 for the numbers of Dan's and Ella's coins respectively.
It is possible that some pupils will start looking for patterns, for example that if we increase the number of Dan's coins by 5, we decrease the number of Ella's coins by 2. Is this always true? Why?! The existence of this pattern could be said to be the reason why we get a straight line when we represent these values as points on a graph (as we do in Tuesday's task). This is a nice insight - and useful for us as teachers! However, we don't take things to this depth in the current set of tasks.
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Tuesday: Here we provide a reminder of how we can represent pairs of values as points on a Cartesian graph. Pupils might notice that the points seem to form a pattern - namely that they lie on a straight line...

We have deliberately asked pupils merely to sketch the graph and we suggest that it is worth getting them to do this on plain paper (you could follow this up by asking pupils to use graph paper if you wish!). It is then interesting to see

• whether pupils place the points in a straight line in their sketches, or 
• whether, if they don't place them in a straight line, they nonetheless expect that they would lie on a straight line on a 'proper' graph, or 
• whether they have no expectations about this at all.

Wednesday: It seems fairly clear that the line cuts the 2p axis at the 200 mark, but the cut on the other axis is less clear. How do we find its value?

A primary aim of this task is to help pupils experience the fact that graphs usually make sense! So, here we have a set of points that fit a particular story. However, for this to happen, pupils will need to realise (or be made aware) that the 200 mark actually represents the pair of numbers 200,0 (ie 200 2p coins, 0 5p coins). Similarly, the mark on the 5p axis also represents a pair of numbers. But this mark represents more of a challenge. Do pupils estimate its value by eye and then check it fits the story (assuming they appreciate that it should fit the story!)? Or do they derive its value from the story, namely that it is the number of 5p coins that make 400p, ie 400÷5 = 80.
[Note: We have used informal notation for coordinate pairs, to convey the fact that the notation is arbitrary. However, you might prefer to use the conventional notation, eg (200, 0) rather than 200,0, especially if pupils are already familiar with this convention.]
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Thursday: It is likely that pupils will focus on the number of 2p coins first, when looking for the 4th point - so it will represent 75 coins, as 75 is midway between 50 and 100. How will they go about finding the second number (of 5p coins)? Use the story or use the numbers (60 and 40) shown on the graph?

Again, the task can help pupils appreciate that graphs can make sense! Here the 4th point fits the pattern shown on the graph (in that the numbers 75,50 lie midway between 50,60 and 100,40) and it fits the story: 75×2p + 50×5p = 150p + 250p = 400p.
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Friday: Would a 45˚ line through the origin help?!

There is no exact solution to this task, since the number that fits the desired condition (namely, 400÷7 = 571⁄7) is not a whole number.
The coordinates 55,55 provide quite a good visual estimate for a point on the graph's line. However 55×2 + 55×5 = 385, so we have not quite reached 400 and so 55,55 is not quite on the line. We can get the extra 15 needed to reach 400 by having three more 5p coins, giving us the ordered pair 55,58. From here we can get other pairs by exchanging two 5p coins for five 2p coins. Notice that this is equivalent to hopping in small steps along the line! So two other solutions would be 60,56 and 50,60. However, neither has a gap between the numbers as small as the gap for 55,58. It turns out that this is infact the closest pair.

ALG 16

Week 16: Here we revisit the idea that we met in Week 8, of determining what a minor change to the numbers in an expression does to its value, without simply calculating the modified expression.
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Monday: This might look daunting at first, but after a brief pause for thought it should become apparent that the change made to the first expression is fairly straightforward. What does it do to the expression's value?

We have increased the value of the expression by 100. Its new value is 7336.
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Tuesday: This is a lot more demanding. A first glance might suggest the expression has just been increased by 1.

It can help if pupils try to give some kind of meaning to 4×38 (and similarly to 4×39), for example by applying a context such as '4 boxes of 38 pencils' or '38 bunches of 4 flowers'. Or, more abstractly, reading the expression as '4 lots of 36' or '4, 38 times'. Pupils are then likely to see that changing 38 to 39 increases the expression by 4. Of course, some pupils will simply calculate 4×38 and 4×39, and thus determine that the value has increaed by 4. This is perfectly valid, but encourage these pupils to try to make sense of our neater, 'incremental' approach.
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Wednesday: It is likely that many pupils will find this more challenging than Tuesday's task.

Formally, this can be said to rest on the distributive property of division. We can think of 242145÷5 as 242135÷5 + 10÷5. So we have increased the value of the expression by 2.
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Thursday: We are not asking for the actual value of the second expression, only whether it is smaller or larger than the first. Nonetheless, this is challenging!

Here again, it can help pupils to make sense of the expressions if they give them some kind of meaning. For example, we can imagine sharing £46,385 between 3 people and an amount that is £1 more between 4 people. Or we can ask, how many packs of 3 apples can we get from 46385 apples, and what would happen if each pack were to contain an extra apple and we had 1 more apple altogether.
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Friday: The simplest way to solve this task is to 'add a naught' to some of the numbers. But which ones?

There is quite a lot going on here! Multiplication is distributive over addition, so both '26×435' and '18' have to be multiplied by 10. However, multiplication is not distributive over multiplication, so when we multiply '26×435' by 10 we don't multiply both numbers in the expression by 10!
Here are two ways of making the whole expression '10 times as large':
260×435 + 180 and 26×4350 + 180.
It is likely that some pupils will 'add a naught' to just one of the three given numbers, while others might do so to all of the numbers.

ALG 15

Week 15: In this week we continue the representations theme, with a focus again on mapping diagrams. We start with a 'circular' mapping diagram but move on to the more orthodox parallel-axes form. We use the diagrams to solve a set of classic clock puzzles involving two hands rotating at different speeds - when do the hands coincide? We adopt an empirical step-by-step approach whereby we repeatedly 'plot' the position of the hands as they progress around the clock face. As pupils work through the tasks, some pupils may sense the possibility of using a more analytic approach, and you might want to extend the work in that direction, though we don't do it here.
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Monday: In later versions of this task, we focus on a particular method of solution. But it is interesting to see what spontaneous methods pupils use, so let them adopt any approach that comes to mind. What approach would you use?!

What methods did pupils use? Did some draw a diagram of the clock, or draw a table? Did they work systematically towards the answer or try to home-in on the answer using trial and improvement or perhaps even some kind of analytic* approach?
[*Here is an example of an 'extreme' analytic approach that you might find interesting, though it is unlikely that pupils will have used it: "When the minute hand moves forward by a cetain amount, the gap between the hands increases by twice that amount. Starting at 0, and with a gap of 0, the hands will overlap again when the gap has increased by 60 minutes, ie after the minute hand has moved through 30 minutes."]
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Tuesday: Here we present pupils with a step-by-step method, by treating the clock as a kind of circular mapping diagram. When does the 'mapping line' join equivalent points on the two circular clocks?

You might want to project the first of the two slides, below, or print it for pupils to use as a worksheet. The second slide shows when the arrows overlap again - namely when the minute hand has travelled through 30 minutes, and the 3-minute hand has travelled over 3×30 minutes = 60+30 minutes.

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Wednesday: The approach here is similar to Tuesday's approach, except that the circles have been transformed into infinite parallel lines.

The clock hands overlap when the mapping lines point to the same number or to numbers that differ by multiples of 60. So they overlap at 0,0 minutes and again at 30,90 minutes - and every 30 minutes thereafter!
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Thursday: Here we consolidate the work of the previous slides by using the parallel-axes mapping diagram to solve a variant of the original task, namely where the red arrow turns 4 times as fast, rather than 3 times as fast, as the blue arrow.

 You might want to print out the first of the two slides below as a worksheet for pupils. The second slide shows the solution, namely when the blue arrow has moved through 20 minutes and the blue arrow has turned through 4×20 minutes = 80 minutes = 60 + 20 minutes.

It is worth writing this sentence, showing the solution, in full on the board:
4×20 minutes = 80 minutes = 60 + 20 minutes.
Some pupils might discern that they can use a similar sentence, looking something like this, to help them find the solution:
4 × Blue = 60 + Blue
(where 'Blue' stands for the number of minutes turned through by the blue arrow).
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Friday: Here we short-circuit out step-by-step approach by recording times in a table rather than on a mapping diagram.

After 10 minutes, the red arrow has gone through a full turn (ie it is pointing at 60 minutes). After 15 minutes it has gone through 1.5 turns (ie it is pointing at 30 minutes) so it has overtaken the blue arrow. So we need to consider what happens between 10 and 15 minutes, eg by considering the position of the arrows after 11 minutes, 12 minutes, etc.
After 11 minutes the red arrow has turned through a full turn plus another 6 minutes (6×11 = 66 = 60+6), so it hasn't quite caught the blue arrow again.
After 12 minutes the red arrow has turned through a full turn plus 12 minutes (6×12 = 72 = 60+12), so the arrows overlap at this time. So here we have
6 × Blue = 72 = 60 + Blue,
using the previous notation, and where Blue = 12.
Or, if we want to go a few steps further (!),
6x = 60 + x,
where x is the number (of minutes) that we are trying to find (namely x = 12).

ALG 14

Week 14: We have used an ordered table and a Cartesian graph to represent the changing score in a football match. This week we look at yet more representations. Most of these are not commonly used (and inded some could be said to have been invented for this blog) and pupils certainly don't have to become 'fluent' in their use. The set of tasks might help pupils to see

• that mathematical representations are not handed down from on high but are invented by us; 
• that we might like some representions more than others (and different representations have different affordances), and 
• that we can learn to read and make sense of representations.

We use information from some of the matches that we met in Week 13, so pupils can refer back to these, at appropriate times, to check their work.
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Monday: Here we use arrows to show the football scores. How is the arrow diagram related to the graph?

Instead of drawing dots, we have drawn the lines that join successive dots. The arrow-heads help to emphase the direction in which the scores build up.
If we rotate the new diagram through 45˚ anticlockwise, it maps neatly onto the graph.
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Tuesday: Here we consolidate the ideas from Monday's task by asking pupils to interpret an arrow diagram.

Pupils can check their reading of the arrow diagram by looking back at the Monday and Tuesday tasks of Week 13. We also show the complete table in the next task.
You might want to ask pupils to describe how the arrow diagram conveys the fluctuating fortunes of the two teams.
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Wednesday: This new representation might look very strange! But the Cartesian graph probably did too, when pupils first met it! What exactly is going on?!

 Here we represent a score not by a single dot, but by a line segment, or rather by its two end points. This is rather like a mapping diagram (see Friday's task) excpet that the axes are at right angles ('orthogonal') rather than parallel. We could convert our new representation back into a Cartesian graph by replacing each slanting line segment with a a vertical and horizontal line segment connecting the two endpoints, and by putting a dot where the new segments join.
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Thursday: This shows the completed mapping diagram from Wednesday's task. How can we tell which lines represent a score where Rangers are in the lead?

It is interesting to compare this mapping diagram to the kind of graph-with-added-red-line that we had in last Thursday's task. The diagrams below carry the same information about which team is in the lead or whether the scores are level, but how effectively does each one convey the information to us?
In the second slide we have changed the colour of some of the mapping lines.

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Friday: Here is another mapping diagram, but this time with parallel axes. Is this form easier to read?

The final score was Valencia 2:4 Arsenal (see Friday's task of Week 13). The vertical lines show scores that are level: 0,0 and 1,1 and 2,2. Down-Left lines show Valencia in the lead, of which there is only one, when the score is 1,0. Down-Right lines show Arsenal in the lead: 2,1 and 3,2 and 4,3.
What are the strengths and weaknesses of the different representations, ie the score card, ordered table, Cartesian graph, mapping diagram with orthogonal axes, mapping diagram wih parallel axes?

ALG 13

Week 13: Here we look at how information (about the times at which goals were scored in a football match) can be represented in an ordered way in a table, and as points (or 'dots') on a Cartesian graph. The tasks give pupils the opportunity to make sense of such graphs by thinking about how the dots are linked and how this tells the story of how the score changed during the match.
You can of course choose your own football matches, perhaps involving a local team, or women's teams - or choose a different sport. But try to include 'interesting' scorelines, eg where the lead fluctuates.
You could also simulate scores by tossing a coin several times - with Heads representing a home goal, Tails an away goal - or forget scores and simply plot Heads and Tails cumulatively, ie as they arise. What sort of path will be formed by the dots?
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Monday: This task introduces pupils to the context of football 'score cards' and how we can use the information about when goals were scored to record how the score changed. Pupils need to work in a methodical way and it might be helpful for some pupils to have a copy of the score card so that they can cross through the times as they record each change in score.

You might want to ask pupils to describe the information contained in the completed table. How does it show the fluctuating fortunes of the two teams?
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Tuesday: Here we are using the standard coordinate system to represent the information in the table on a Cartesian graph. Where will the next dot be? - one unit up or one unit to the right?

It might help some pupils to join each new dot to its predecessor. Notice that each new dot is one unit above or to the right of the previous dot (why?) and that we start, inevitably, at the origin (0, 0) (why?).
This is what the final graph looks like:

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Wednesday: Here pupils have a chance to consolidate what they have met so far. Some pupils might want to draw an ordered table of the scores before completing the graph.

You might want to ask pupils to describe the characteristics of their completed graph and how this tells us about the changing scoreline. We look at the graph in more detail in Thursday's task.
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Thursday: Here we look at the completed graph from Wednesday's task. Note that we have added a red line to the graph - what is special about any of our dots that lie on the line?

Part b) is quite challenging, but also quite fun. The effect of the extra goal for the home team would be to move all the 'later' dots (goals) one unit to the right, as shown below.
You might want to ask other questions that 'transform' the graph in some way. For example, what if Ngadeu-Ngadjui hadn't scored at all? Or what if we'd had the same goals, at the same times, but that Sevilla had been the home team?

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Friday: This task should help pupils get a better feel for what the table and graph are telling us. It turns out that there is only one possible answer for some of the missing scores, but for others there are two answers that would fit the run of scores.

There is only one way to get from 3,1 to 5,1, namely 4,1. But we can get from 5,1 to 6,2 via 6,1 or 5,2.
The graph is illuminating here. We know that if we join consecutive scores (dots), we get a path composed only of horizontal and vertical segments and where the 'journey' from the origin only ever moves upwards or to the right. This means there is only one way to fill the 'gap' between the 2nd and 4th dots, but two possible ways to get from the 4th to the 6th dot, ie two possible positions for the 5th dot:

The actual times of the goals in these two matches are listed below. If they so wish, pupils can use this information to find the actual order in which the scores changed.

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We take a further look at ways of representing football scores in Week 14.