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).