Forming and Solving Equations

May 16, 2016 1 comment

While checking the work of a year 11 student on Friday I came across a question that could have been a great one for the higher GCSE students to practice their skills together and also their selection of which mathematics to use.

The question was to find the area of this triangle:

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A great question. One that to you or I is straightforward but that would take GCSE level students and below a bit of thinking and let’s them hone their skills.

The way to tackle it is to use Pythagoras’s Theorem to form an equation, solve for x then find the area. I feel is beneficial as it combines Pythagoras’s Theorem with a decent amount of algebra then includes the find the area bit at the end.

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In this case though, that wasn’t the question. There was more information on offer and the question was:

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Which is still a fairly nice form and solve an equation problem.

3x + 1 + 3x + x – 1 = 56
7x = 56
x = 8
A = 0.5×7×24 = 84

There is a niceness to this question that goes beyond the question itself.  It shows us a great way of differentiating within lessons. Just be leaving out a tiny portion of the information, in this case the perimeter, we can make the question much harder. This idea is something I’ve been working on in various places. M1 questions can be made much easier by providing a diagram, for example.

Have you used questions in a similar way? If so I’d love to see them, please do get in touch.

Cross-posted to Betterqs here.

Is one solution more elegant?

May 14, 2016 13 comments

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:

image

I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

Method 1

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule,  here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right.  I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

Method 2

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

Method 3

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

Method 4

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

(6 + x)^2 = 45 + 9 + x^2

x^2 + 12x + 36 = 54 + x^2

12x = 18

x = 1.5

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:

image

All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!

Cross-posted to Betterqs here.

Mathematical Style

May 11, 2016 5 comments

Yesterday I read this post from Tom Bennison  (@DrBennison). The post was written to start a conversation for a twitter chat that I unfortunately couldn’t make. It did, however, make me think.

He was questioning Wetherby mathematical elegance and style should be assessed at A level. Suggesting that solutions with more elegance should be awarded more marks.

Bizarrely, the example he used was almost exactly the same as a discussion if had with a year 13 class not long before I read his post. His example was finding the midpoint of a quadratic. He looked at two methods – completing the square and differentiation – and suggested that as CTS is more elegant that should be worth more.

I agree immensely that CTS is a preferable method with far more elegance, but I don’t think the marks should be different depending on the  method you choose. I feel that we should be encouraging mathematical thought, trying to create young mathematicians who can apply themselves to a problem and find their own way through. I feel if we start assigning marks for elegance and style them we would be moving towards the “guess what’s in my head” style of assessment that I feel we need to be moving away from. The way to do well would be to spot from a question what the examiner wants, rather than to apply the mathematical tools at ones disposal and find a solution.

Back to that Y13 lesson I mentioned, we were looking back over some C3 functions work and one of the questions involved finding the range of a quadratic function – so obviously it was necessary to.find the minimum. A discussion ensued as to how to do this with students coming up with 3 valid methods. The two mentioned above, both of which I find quite elegant, although I do much prefer CTS. The third method was suggested by one student who said “it’s -b/a – you just do -b/a” I knew what he meant – he was saying that this was the x value where the minimum occurred and that you put that in to find y, but he didn’t really understand what it was or why. He’d come across the method online and has learned it as a trick. When I showed him it came from completing the square and looking at it as a graph transformation, I saw the light bulb come on.

It is an interesting discussion. Some methods are far more elegan, and some are just algorithmic tricks. I think that the lack of understanding with these tricks will lead to marks being lost.  So perhaps this will self regulate.

I’d love to hear your views in this, which way would you tackle finding the minimum of a quadratic?  And do you think we should assign marks to elegance and style?

A lovely simple trigonometry puzzle

May 8, 2016 4 comments

Sometimes a puzzle can look complicated,  but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

image

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms,  it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

Sin^6 + sin^4 cos^2 – sin^4

See it now? What if I rewrite it as:

Sin^4 sin^2 + sin^4 cos^2 – sin^4

I’m sure you have seen it now, but to be complete,  take the common factor of the first two terms:

Sin^4 (sin^2 + cos^2) – sin^4

Obviously sin^2 + cos^2 = 1, so we’re left with:

Sin^4 – sin^4 = 0

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

Cross-posted to Betterqs here.

May 8, 2016 Leave a comment

Sometimes a puzzle can look complicated,  but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

image

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms,  it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

Sin^6 + sin^4 cos^2 – sin^4

See it now? What if I rewrite it as:

Sin^4 sin^2 + sin^4 cos^2 – sin^4

I’m sure you have seen it now, but to be complete,  take the common factor of the first two terms:

Sin^4 (sin^2 + cos^2) – sin^4

Obviously sin^2 + cos^2 = 1, so we’re left with:

Sin^4 – sin^4 = 0

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

A tale of two Graphs

May 3, 2016 2 comments

For the last two years I’ve collected some amazingly bad graphs from election material, both that has come through my door, and that other people have sent me (see this and this.) My own MP provided so many gems that Colin Beveridge  (@icecolbeveridge) started an Internet campaign to have people refer to these misleading election graphs as “Mulhollands” – after the man himself. This led to Colin and others tweeting him questions about his misleading graphs and one teacher, Adam Creen (@adamcreen) tweeting him with corrected “Mulhollands” that his Y9 class had completed.

This was obviously effective, as each time a piece of campaign literature has arrived since I have scoured it and there has only been one chart on any of them, and that was correctly drawn! A success! Bizarrely, as I have had many other folks looking out for them, this seems a success that has been widespread as I’ve not come across any hideously inaccurate graphs this year.

I did think that this election season would pass without any mention on this blog but today I came across two interesting graphs from a neighbouring ward. Both are accurate, but they tell very different stories, and reminded me a little of Simpson’s Paradox, without actually being directly related to it.

Exhibit A

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This graph is from the Lib Dems in Horsforth ward for the Leeds City Council elections, it’s not wholly accurate in terms of the bar charts,  but it’s near enough to not irk me too much. It shows that of the 5 previous local elections in the area the Lib Dems have won 3 and the Conservatives have won 2. They are using this to sell the idea that it’s only them or the Conservatives who can with the seat. However…….

Exhibit B

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This is from a Labour party leaflet in the very same ward and it shows that in the last Local Election the Labour party candidate came second to the Conservative candidate and that the Lib Dem ended in last place. The inference here is that Labour are more likely to beat the Conservatives as they came second last time.

Both leaflets are presenting true facts selected to further their narrative, and both are presenting them accurately, although one could argue they are both a little misleading.

I’ve looked at the stats from the last few years, it seems that on general election years the tories win by a fair way, but that in local election years it is tight between all three parties, but the lib dem vote has been steadily dropping. It could be an interesting ward to judge the national feeling on on Friday when the results come in, as it is really a 3 way marginal in non GE years.

The anomaly that is the general election year is interesting, more people do vote nationally when there is a GE, but the massive swing to the tories is fairly unusual, as they tend to be good at mobilising the vote. I do know that the Lib Dems in that constituency didn’t really campaign during the GE and the seat was a Tory Labour marginal, and that in a neighbouring Lib Dem Labour marginal the Conservatives didn’t campaign, so perhaps this had an effect.

If you have found any terrible election graphs, please send me them!

Examinations, Examinations, Examinations

May 3, 2016 4 comments

This post was first published on the 3rd May 2016 here, on Labour Teachers.

Sometimes it feels like the government’s main three priorities are examinations, examinations and examinations, and this fact has certainly led to many people involved in education to express their disagreement and disappointment with the system.

Most recently, a large number of people with children of a primary school age have chosen to keep their children off school in protest against the new SATs test their children will sit. This has caused me to spend some time thinking about this, and try to put together some views.

Exam factories

One of the leading criticism of these tests is that it drives schools to shrink their curriculum and focus heavily on the content which will be examined – meaning subjects like art, music, history etc get widely ignored and children miss out on an important part of their education. I can certainly find agreement with this, however I think this is already an issue with the SATs as they stood, so it doesn’t seem to warrant the furore of the new tests, which can only compound an already prevalent problem.

What are they for?

This is a key question,  and I think that a different answer to it would lead to a different outcome. The tests as a marker for informing future teachers of a students ability are very helpful. The tear that SATs were boycotted we saw real problems with the grades reported by primary schools as there were massive inconsistencies from school to school. However, this argument alone seems to be silly, as what we see often is that students primed and drilled from the test from September to May achieve well, but then do no more maths from May to September and often regress. If this was to be the sole reason then surely they could be abolished totally and secondary schools could complete diagnostic tests on entry?

The other answer to this question is to measure school performance, and this is a real can of worms. It is this exact fact that leads to the exam factory conditions and the gaming the system and as such causes a load of problems. The other side of it is, however, that there needs to be some way of ensuring that schools are doing what we expect them to do. I don’t know what the answer is, but I tend to think high stakes testing is not the answer.

Is it just a problem with SATS?

No, all the issues outlined above are transferable to GCSE and A level exams. Again, I don’t have an answer, but I think that there must be a better way to treat 16 and 18 year olds than to make them sit high pressure, high stakes, examinations at a time of increased hormones knowing that if they go wrong that could seriously affect their life chances.

I don’t have the answers, but I do feel that there are answers and our job in opposition is to find them and present them to the public, showing that if they vote differently in 2020 we can give them a better way.

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