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Posts Tagged ‘GCSE’

Late tiering decisions

March 20, 2019 1 comment

Last week year 11 sat their mocks. Some did really well, others did really poorly. It’s the latter group that has me purplexed. Students sitting the higher tier paper but only scoring single digits per paper, or even earlt teens per paper. What to do with them?

Some of them asked if they could move to foundation, I think its best for them. 1 student got 32 marks over 3 higher papers, did the 3 foundation and was well over 100. 1 student got 40 marks over 3 higher papers spent 30 mins in a foundation paper and got 60 marks. The grade 5s they want seem more achievable on foundation.

My issue lies with a few students desperate to do higher and try for 6s. Scoring around 50 marks over 3 higher papers it seems a risk. But having taught them both i feel that it’s within their capabilities. But from November to march they have made only tiny gains in marks. On the ine hand, foundation means they cant get a 6 and for at least one of them means rethinking post 16 choices, but on the other hand sitting higher means they might end up with only a 4 or less and thst would mean rethinking post 16 again. It’s tricky, any thoughts are welcomed.

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Categories: #MTBoS, Curriculum, GCSE, KS3 Tags: , , ,

Proof by markscheme

March 16, 2019 2 comments

While marking my Y11 mocks this week I came across this nice algebraic proof question:

The first student had not attempted it. While looking at it I ran through it quickly in my head. Here is the method i used jotted down:

I thought, “what a nice simple proof”. Then I looked at the markscheme:

There seemed no provision made in the markscheme for what I had done. (Edit: It is there, my brain obviously just skipped past it) How did you approach this question? Please let me know via the comments or social media.

Anyway, some of my students gave some great answers. None of them took my approach, but some used the same as the markscheme:

And one daredevil even attempted a geometric proof…….

Cereal Percentages

March 13, 2019 Leave a comment

This week my Y11s are sitting mock exams. One of the questions that came up on paper 1 stumped a lot them.

They came out if the exam on monday, and said the paper was very difficult. One of them asked me one of the questions:

“Sir, if you have a box of cereal and increase it by 25% but keep the price the same, what percentage would you need to decrease the price of the original box by to get the same value?”

I immediately said “20%”, an answer which flummoxed the student and the others stood around. They couldn’t work out how I had got that answer, never mind so quickly.

I tried to explain it to them, but in that moment, on the corridor, I didn’t do a very good job. For me, it was intuitive. A 25% increase and a 20% decrease would yield the same value as in one you are changing the top of a fraction and the other the bottom of a fraction so you need to use the reciprocal, 4/5 is the reciprocal of 5/4 and 4/5 is 80% hence it needs to be a 20% decrease. Cue blank looks and pained expressions. I was seeing the students again later in an intervention session so I promised to go through it in more detail then.

I talked about the idea of value, how you could consider mass/price and get grams per penny – how many grams for each penny you spend – or you could consider price/mass and get penny per grams – how much you pay per gram. I said either of these would give an idea of value and you can use either in a best value problem.

I showed them the idea of the fraction, said you could call the price x and the size y.

The starting scenario is:

y/x

The posed scenario is:

1.25y/x

but we know 1.25 is 5/4 so that becomes:

(5/4)y / x

which in turn is:

5y/4x

I then showed that the second scenario meant getting to the same value but altering x. To do this you would need to mutiply x by 4/5:

y/(x(4/5))

(y/x)÷(4/5)

(y/x) × (5/4)

5y/4x

This managed to show some of them what was going on, but others still massively struggled. I tried showing them with numbers. 100 grams for £1. This again had an effect for some but still left others blank.

I’m now racking my brains for another way to explain it. If you have a better explanation, please let me know in the comments of via social media!

Categories: #MTBoS, GCSE, Maths Tags: , , ,

Proving Products

June 26, 2017 1 comment

Just now one of the great maths based pages I follow shared this:

So naturally I figured I would have a go. I thoughts just get stuck in with the algebra and see what happens, normally a good approach to these things.

My first thought was that if I use 2n – 3, 2n -1, 2n + 1 and 2n +3 then tgere would be less to simplify later. I know that (2n + 1)(2n – 1) = 4n^2 – 1 and (2n – 3)(2n + 3) = 4n^2 – 9 so I multiples these together.

(4n^2 – 1)(4n^2 -9) = 16n^4 – 40n^2 + 9

I thought the best next move would be to complete the square:

(4n^2 – 5)^2 – 16

This shows me that the product of 4 consecutive odd numbers is always 16 less than a perfect square and as such that the product of 4 consecutive odd numbers plus 16 is always a square.
(4n^2 – 5)^2 – 16 + 16 = (4n^2 – 5)^2 
A nice little proof to try next time you teach it to your year 11s.

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.

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