With Burns Night fast approaching, I was reminded of one of his more memorable couplets last week:
"The best laid schemes o' mice an' men
Gang aft a-gley."
I was in a Grade 5/6 class who had been learning about mean, median and mode. THey were able to tell me what each of these meant: the mode is the most common value; the median represents the middle value; the mean is the sum of all the values divided by the number of values. Some were unsure what the mean actually represented so I showed them concretely how it is an evening out of all the values as shown in this post.
I then gave each pair of students ten random playing cards and asked them to find the mean, median and mode for their set of ten values (we agreed that jacks, queens and kings count as 10). By giving them cards, it encouraged them to put them in order first (a step sometimes overlooked when calculating the median).
With ten cards, it allowed them to pull out the middle pair and use these to work out the median. Also, to work out the mean involved them practising their mental arithmetic (adding numbers and dividing by ten). When each pair finished, I gave them a new set of ten cards. It was a neat activity that was easy to organise and kept them engaged.
As they were doing so well, I decided to give a more challenging question:
Find a set of ten cards that has a mode of 5, a median of 7 and a mean of 8.
Before you read any more, try this one for yourself.
I have given questions like this before, and although I hadn't actually tried this particularly combination of numbers, I figured that there would be a few solutions. The students got stuck into it right away and as I moved around the room, I heard lots of great reasoning:
"The cards have to add up to 80!"
"Make sure we have more 5s than any other card."
"If 6 and 8 are the middle pair, then the median MUST be 7!"
When necessary, students would come to me to exchange some of their existing ten cards for ones that they needed. If a pair thought they had a solution they would show me and had to verify it worked:
In this case, whilst they have the correct mode and median, the mean is not 8. Back to the drawing board in this case.
There was lots of great thinking happening but after a while I was surprised that not one pair had found a solution. This had me wondering: had given them an impossible problem? I had picked those values on the spur of the moment thinking that there would be a solution but now I was not so sure.
So with just a few minutes of the class left, I asked them to stop and pointed out that whilst there was great math happening, we still didn't have a solution: why was this? A few students asked if it was impossible. So as class we decided to see if we could reason our way through this. We agreed that we needed to get ten cards:
...and that at least two of these cards need to be 5s:
...and that the middle pair had to be something like 6 and 8 (i.e. have a sum of 14) to give a median of 7. Some students argued that you could have two 7s as the middle pair but only if you add an extra 5 to make this the unique mode:
So far we have 4 cards totalling 24. The remaining 5 cards must add to 56 if we are to get a mean of 8. Four of these cards must be (in the case above) greater than 8 and two must be (in this case) less than or equal to 6. Now the students were quick to reason that this could not be done: the four cards greater than 8 can have a maximum value of 40 and the remaining two cards can have a maximum value of 11 (if 5 is to be the unique mode). So as we cannot get a total of 80 with these cards, we cannot get a mean of 8.
The question was impossible to solve.
This is not what I had planned!
And yet, it worked really well. The students learned as much (if not more) from proving that this was impossible than they did from finding the mean, median and mode of a given set of numbers.
James Tanton refers to the importance of giving learners the opportunity to experience 'funstration' in Math. In other words, making sure they get engaging problems in which they will hit roadblocks. Impossible problems (e.g draw a quadrilateral with 4 acute angles; find a number whose square is bigger than 5 but whose square root is less than 1; draw a triangle with lengths 4cm, 5cm and 10cm) are one way of doing this and this is why I like giving them occasionally.
Even when I hadn't planned to!
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Note: the proof outlined above does not contain all the arguments and cases that the students thought about and as such isn't necessarily complete. Can you improve upon it?
