Archive

Archive for the ‘GCSE’ Category

Circles and Triangles

July 15, 2016 5 comments

Regular readers will know that I love a good puzzle. I love all maths problems, but ones which make me think and get me stuck a bit are by far my favourite. The other day Ed Southall  (@solvemymaths) shared this little beauty that did just that:

I thought “Circles and a 3 4 5 triangle – what an awesome puzzle”, I reached for a pen an paper and drew out the puzzle.

I was at a bit of a loss to start with. I did some pythag to work some things out:

Eliminated y and did some algebra:

Wrote out what I knew:

And drew a diagram that didn’t help much:

I then added some additional lines to my original diagram:

Which made me see what I needed to do!

I redrew the important bits (using the knowledge that radii meet tangents at 90 degrees and that the line was 3.2 away from c but the center of the large circle was 2.5 away):

Then considered the left bit first:

Used Pythagoras’s theorem:

Then solved for x:

Then briefly git annoyed at myself because I’d already used x for something else.

I did the same with the other side to find the final radius.


A lovely puzzle using mainly Pythagoras’s theorem, circle theorems and algebra so one that is, in theory at least, accessible to GCSE students.

I hope you enjoyed this one as much as I did!

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:

Old school Venn

July 7, 2016 Leave a comment

Recently I saw this picture from Ed Southall (@solvemymaths) and thought it interesting:

It is an O level question on Venn Diagrams from 1988. I had a go at it.

The Venn itself was easy enough to fill in and the forming and solving part followed nicely.

As did the rest.

Having gotten used to A level statistics this was relatively straightforward, it manages to test use and knowledge of Venns but doesn’t go as far as probabilities. 

I like Venn diagrams and I think questions like this are a good start point to build on, students who can do this will find A level Venns much easier. I assume that this style of question may be what we can expect from the new style GCSE, and even if it’s not its certainly something I intend to use with my classes.

Categories: GCSE, Maths

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.

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.

Accuracy with Trigonometry

April 11, 2016 Leave a comment

This post was originally posted here on Cavmaths and here on BetterQs, on 5th March 2016, however the original post somehow got deleted so I’m re posting it.

This week I was planning to cover upper and lower bounds with year 11 as on the last mock a lot of them made mistakes so I felt it would be a good topic to revise. As part of the planning process I had a look through the higher textbooks our department has bought for the new specification GCSE (we bought the Pearson ones, the full suite at KS3 and 4. Some great questions in them and the online version, activeteach, is great to take questions and place into your lessons. I’d definitely recommend it, if used correctly, but I will admit to being disappointed to see a formula triangle being advised…) to see if there were any good questions I could pilfer, and I came across the section on using upper and lower bounds in trigonometry.

My first thought was, “that’s a nice topic”, and then the full spectrum of the topic began to unfold.

Initially, I had like the idea that students would be required to think about the fraction, and how minimising the denominator actually maximises it, but thin I remembered the nature of the cosine function! This example shows what excited:

image

Not only would students be required to understand the nature of a fraction, they’d also need a deep understanding of the cosine function itself, to understand that the bigger cos x is, the smaller x is, and vice versa  (where x is between 0 and 90 of course). This could be a real deep understanding of the graph, or the unit circle, or just the geometry of a right angled triangle.

The example itself is very procedural based, which is a shame,  but it does give a teacher a good frame to start discussions. I wouldn’t use textbook example as teaching anyhow, just as an additional example to talk through one on one with students who were still struggling.

The textbook goes on to pose this awesome discussion question:

image

A real nice prompt to get an in depth discussion around the trig ratios going. I often use similar prompts when looking at maximum values for sine and cosine “what’s the biggest opp/hyp can ever be?” for example. This often gives a nice discussion focus.

I think that this topic shows how different the new specification will be. Students are going to need a much deeper relational understanding if they are to achieve the top grades with questions like this being posed.

What do you think of bounds being questioned in relation trigonometry? Have you used prompts like this before? How have you found them?

%d bloggers like this: