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An excellent puzzle – alternate methods

July 19, 2017 2 comments

Yesterday I wrote this post looking at a nice puzzle I’d seen and how I solved it.

The puzzle again:

Lovely, isn’t It?

After I published my previous post I wondered if I may have been better using a vector approach or a coordinate geometry approach. So I gave them a try.

Coordinate Geometry

I started by sketching the figure against an axis.

I place the origin at the centre of the circle, worked out the equation if the circle and the right leg of the triangle and solved simultaneously for x. Giving x =1 and x=1/3. These x values correspond to half the base of each triangle, which shows the scale factor from the large triangle to the small one is 1/3. As the area of the large one is rt2 this gives the area of the small as rt2/9.

I like this method, probably a little better than the one prior to it.

Vectors

First I sketched it out and reasoned I could work it out easy enough with 4 vectors.

I saw that I could write AC as a sum of two others:

I knew that the length of AC was 1 so I used Pythagoras’s Theorem to calculate mu. It left me with the exact same quadratic to solve. This time mu was the fraction of DB needed so was automatically the length scale factor. The rest falls out as it did before.

As well as this, Colin Beveridge (@icecolbeveridge), maths god and general legend, tweeted a couple of 1 tweet solutions. First he used trig identities:

Trig Identities 

I assumed this was right, but checked it through to ensure I knew why was going on:

We can see beta is 2 x alpha and as such the tan value is correct. The cos value (although it is missing a negative sign that I’m sure Colin missed to test me) follows from Pythagoras’s Theorem:

This is again the scale factor as it is half the base of the small triangle and the base if large triangle is 2.

Complex Numbers

Then Colin tweeted this:

At first I wasn’t totally sure I followed so I asked for further clarification:


I had a moment of stupidity:

And then saw where Colin was going. I tried to work it through, by way of explaining here in a better manner.

I sketched it out and reasoned the direction of lines:

Then I normalised that and equated imaginary parts to get the same scale factor:

I am happy that is is valid, and that it shows Colin is right, but I’m not entirely sure this as the exact method Colin was meaning. He has promised a blog on the subject so I will add a link when it comes.

I like all these methods. I dontvthink I would have though of Colin’s methods myself though. I’d love to hear another methods you see.

Infuriatingly impossible exam questions

February 9, 2017 4 comments

Today I was working on some Vectors exam questions with my Y13 mechanics class and I came across this question:

A student had answered it and had gotten part d wrong. What he had done was this:

I have recreated is incorrect working.

Obviously he had found out when the ship was at the lighthouse, instead of 10km away. I explained this to him and started to explain how he should have tackled this when a sudden realisation angered me.

Now for those if you that didn’t work through the question, here is the actual answer:


Can you see what had me infuriated?

This is an impossible answer! If the lighthouse is on the trajectory of the ship and it will hit said lighthouse at t=3 then that would stop the ship! At the very least it would slow it down!!!! In reality it would have to avoid the lighthouse and change trajectory. Meaning the second answer, T=5, would not happen under any circumstances!

My initial thought was: “are they expecting students to spot this and discount the second answer? That’s a bit harsh.”

So I checked the markscheme:


Nope, they are looking for both answers. Argh! I can understand using a real life context in mechanics, I really can. But why not check for this sort of thing!

What do you guys think? Is this infuriating or am I just getting get up over nothing? I’d love to hear your views in the comments or via social media.

A lovely old problem

July 11, 2016 Leave a comment

Recently Ed Southall shared this problem from 1976:

I’m not entirely sure if it is from an A level or and O level paper. It covers topics that currently sit on the A level, but I think calculus was on the O level at some point. Edit: it’s O level I saw the question and couldn’t help but have a try at it.

First, I drew the diagram – of course:

I have the coordinates of P, and hence N so I needed to work out the coordinates of Q.  To do this I differentiated to get the gradient of a tangent and followed to get the gradient of a tangent at P.

Next I found the equation, and hence the X intercept.

And then, because I’m am idiot, I decided to work out the Y coordinate I already knew and had used!

The word in brackets is duh…..

Now I had all three point.

It was a simple division to find the tangent ratio of the angle.

The next 2 parts were trivial:

And then I misread the question and assumed I’d been asked to find the shaded region (actually part d). 

Because I decided calculators were probably not widely available in 1976 I did it without one:

I thought it was quite a lot of complicated simplifying, but then I saw part c and the nice answer it gives:

Which makes the simplifying in part d simpler:

I thought this was a lovely question and I found it enjoyable to do. It tests a number of skills together and although it is scaffolded it still requires a little bit of thinking. I hope to see some nice big questions like this on the new specification.

Edit: The front cover of the paper:

Consultation time again

June 13, 2016 Leave a comment

Is it cynical of me to question the DoE’s repeated tactic of releasing consultations either just before the summer, when most teachers are in the midst of high stakes exam testing, or over the summer when a lot of teachers are either away or spending time catching up with their families who they haven’t seen through the heavy term time?

Anyway, this year they have released another one. It focusses around the new GCSEs, and more specifically the awarding of grades. The consultation states that for the first award there will be a heavier reliance on statistical methods to set the grade boundaries, allowing the same proportion of grade 4s as we currently have of grade Cs, likewise similar proportions of 1s to Gs and of 7s to As. The rest will be split arithmetically ie the boundaries in between will be equally spread. From Year 2 onwards it will revert back to examiner judgement, but use the statistical analysis as a guide as well as the national reference tests.

This immediately raises questions – how do we know that the first year to sit it should have a similar proportion of 4s as Cs? It seems that this has been decided without much thought about the prior attainment; the consultation certainly doesn’t mention it for the first year. It does going forward, but that doesn’t really explain how this prior attainment will be measured. I have been under the impression that the KS2 SATs are moving from level based assessments to assessments where the students’ scores will be reported as percentiles – surely then comparisons of prior assessment will always be the same? “This year, bizarrely, we saw exactly 10% score above the 90th percentile, what’s more bizarre is that is exactly the same proportion as last year!”

It seems strange to me to put such a heavy reliance on these prior attainment targets anyhow. We live (for now) in a society that has a fairly fluid immigration system, so the students who get to year 11 haven’t always been through year 6 in this country. There is also a question of the validity of the assumption that every year group will progress over the 5 years of secondary at the same rate.

The obvious elephant in the room is floor targets. By setting the boundaries so the same proportion of students get above a grade 4 as get above a C, but switching the threshold to a grade 5 you immediately drop the results of a whole host of schools down, what happens then remains to be seen, but I can imagine lot of departments will become under pressure and scrutiny for something that is statistically inevitable given the new grading formula.

This is all interesting, but it’s not much different to previous announcements and consultations, what is different is the formula for awarding grades 8 and 9. The formula looks to be a fair way of doing it, but it seems strange to me to use this formula just for the first year. Why then revert to examiner judgement about the grade standard? The government seem to be happy to use statistical analysis and similar grade proportions in parts of their grading system, but not in all of it, and that seems odd to me.

Have you responded yet? If not you can here (but hurry, the consultation closes June 17th). I’d love to hear other people’s views either in the comments or via social media.

Patient problem solving

June 7, 2016 5 comments

I recently read a piece by D Pearcy called “Reflections on Patient Problem Solving”, from Mathematics Teaching 247. It was an interesting article that looks at how teachers need to allow time for students to try their own ideas out while problem solving, rather than just coax them along in a “this is how I would do it” kind of way.

Pearcy’s definition of problem solving is looking at something you have never encountered before that is difficult and frustrating at times, takes a reasonable amount of time, can be solved more than one way and can be altered or extended upon easily. He then goes on to ask whether this is actually happening in classes or if teachers are just walking students through problems, rather than allowing them to problem solve.

He quotes Lockhart (2009) – “A good problem is one you don’t know how to solve” and states that it follows that if you give hints then it defeats the point of setting problems. He goes on to say that maths advocates talk of the importance of maths as a tool to problem solving – but that this isn’t actually happening if students are not being allowed to get frustrated and struggle through to a solution.

He explains how he finds it difficult not to give hints when students are struggling, both because it is in most teachers nature to help, and because of the external pressure to get through the syllabus quickly. This is something I too have encountered and something I have become increasingly aware of as I try to allow time for struggle. Other factors at play are maintaining interest, and increasing confidence. If we let students struggle too much they may lose interest and confidence in their ability – thus it is important to strike a balance between allowing the struggle but not letting it go too far. This is certainly something I keep in mind during lessons, and I feel it is something that we all should be aware of when planning and teaching.

This is an interesting article that looks at a specific problem and allowing students time to struggle and persist. This importance of this is paramount, in my view, and this is also the view expressed by the author of the article. I find it very hard to not offer hints and guidance when students are struggling. One way I manage to combat this at times is by setting problems I haven’t solved yet, thus leaving me a task to complete at the same time. This can work well, particularly at A Level and Further Maths level as then I can take part in the discussion with the students almost as a peer. This is a technique I have used often with my post 16 classes this year.

I have been reading a lot about problem solving recently, and a recurring theme is that teachers can often stifle the problem solving they are hoping to encourage by not allowing it to take place. This is something we need to be aware of, we need to have the patience to allow students the time to try out their ideas and to come up with solutions or fall into misconceptions that can then be addressed.J
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Have you read this article? If so, what are your thoughts on it? Have you read anything else on problem solving recently? I’d love you to send be the links if you have and also send me your thoughts. Also, what does problem solving look like in your classroom? Do you find it a struggle not to help? I’d love to hear in the comments or via social media.

Further Reading on this topic from Cavmaths:

Dialogic teaching and problem solving

Understanding students’ ideas

References:

Pearcy.D. (2015). Reflections on patient problem solving. Mathematics Teaching. 247 pp 39-40

Lockhart, P. (2009). A Mathematician’s Lament. Retrieved from: https://www.maa.org/external_archive/devlin/LockhartsLament.pdf

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:

image

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.

image

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

image

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.

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