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Posts Tagged ‘Maths’

AS Levels

April 3, 2019 Leave a comment

We are now fully into “Exam season”, Year 11 have their GCSE exams, and Year 13 have their A Levels. Then Year 12 have AS Levels.

AS levels are a weird thing. They are no longer a component part of the A Level, they are very early in the exam session and it seems to me an unnecessary added pressure.

Last year we took a decision as an academy not to enter pur maths students for the AS exams. We did this to maximise pur teaching time and avoid unnecessary stress. This year the decision was taken at trust level to enter them in all subjects.

I can now see two sides of the argument. Last year our students focussed heavily on their other subjects and not maths as they had external exams for those subjects. This meant we lost teaching time and their homework suffered during exam season. This year we have not finished all the content early enough to really focus the revision. I really dont know whats best. I do think, however, that it is important to have a decision made for all subjects.

Are you entering your students for AS Levels? I’d love to know if you are or not and why you made that decision. You can answer in thw comments or on social media.

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Categories: #MTBoS, A Level, Teaching Tags: , , ,

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.

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: , , ,

Simultaneous Equations

March 10, 2019 3 comments

It’s been a while since i last wrote anything here. Which says more about how busy I’ve been than my desire to write, but I hope to start writing more regularly.

This week I was teaching simultaneous equations and a student asked a question that made me think about things so I thought i would share.

I was teaching elimination method and I had done some examples with the coefficients of y having different signs and I put one on the board with the same signs and asked the class to think how we may go about solving. One of the students in the class put uo his hand after a while and said he thought he had solved it.

5x + 4y = 13

2x + 2y = 6

I asked hime to talk us through his thinking and he said “first I multipled the bottom equation by -2”

5x + 4y = 13

-4x – 4y = -12

“then I added the equations as before”

x = 1

“Then I subbed in and solved.”

2 + 2y = 6

2y = 4

y = 2

“so the point of intersection is (1,2)”.

This wasn’t what I was expecting. I was expecting him to have spotted we could subtract instead, but this method was clearly just as correct. It wasn’t something I had considered as a method before this, but I actually really liked it as a method and it led to a good discussion with the class after another student interjected with her solution which was what I expected, to multiply by 2 and subtract.

It was a great start point to a discussion where the students were looking at the two methods, and understanding why they both worked, the link between addition of a negative and subtracting a positive and many more.

I was wondering, does anyone teach this as a method? Have you had similar discussions in your lessons? What do you think of it?

A lovely circle problem – two ways

December 7, 2017 5 comments

So, I was working with some year 12s on a few problems around circles out of the new Pearson A Level textbook. (Incidentally, it’s this book, and I think it’s probably the best textbook I’ve come across. I would certainly recommend it.)

This question appears in a mixed exercise on circles:

 

 
It’s a lovely question. Before reading on, have a go at it – or at least have a think about what approach you’d take –  as I’m going to discuss a couple of methods and I’d be interested to know how everyone else approached it.

Method 1:

I looked at this problem and saw right angled triangles with the hypotenuse root 52. I knew the gradient of the radii must be -2/3 as each radius met a tangent of radius gradient 3/2. From there it followed logically that the ratio of vertical side : horizontal side is 2 : 3.

Using this I could call the vertical side 2k and the horizontal side 3k. Pythagoras’s Theorem  then gives 13k^2 = 52, which leads to k^2 is 4 and then k is 2 (or -2).

So the magnitude of the vertical side is 4 and of the horizontal side is 6.

From here it follows nicely that p is (-3,1) and q is (9, -7).

Finally there was just the case working out the equation given a gradient and a point.

L1:          y – 1 = (3/2)(x +3)

2y – 2 = 3x + 9

3x – 2y + 11 = 0

 

L2:          y + 7 = (3/2)(x -9)

2y + 14 = 3x – 27

3x – 2y – 41 = 0

I thought this was a lovely solution, but it seemed like a rather small amount of work for an 8 mark question. This made me wonder what the marks would be for, and then it occurred to me that perhaps this wasn’t the method the question writer had planned. Perhaps they had anticipated a more algebraic approach.

Method 2:

I had the equation of a circle: (x – 3)^2 + (y + 3)^2 = 52. I also knew that each tangent had the equation y = (3/2)x + c. It follows that if I solve these simultaneously I will end up with a quadratic that has coefficients and constants in terms of c. As the lines are tangents, I need the solution to be equal roots, so by setting the discriminant equal to zero I should get a quadratic in c which will solve to give me my 2 y intercepts. Here are the photos of my workings.

 

As you can see, this leads to the same answer, but took a lot more work.

 

I’d love to know how you, or your students, would tackle this problem.

Angle problem

December 5, 2017 2 comments

Today has been quite a geometric based day for me. I spent a couple of hours solving non-RAT trigonometry problems with year 10 and then a while with year 11 looking at various algebra angle problems. Then I went on Twitter and saw this from Ed Southall (@solvemymaths):

A couple of nice parallel lines questions that I might grow at y11 tomorrow.

Both are fairly straight forward to solve. I looked at the first one, imagines a third parallel line through the join if x and saw x must be the sum of 40 and 60 hence 100.

The second I saw an alternate angle to the 50 in the top triangle and used angle sum of a triangle is 180 to spot that x is a right angle. I glanced down at the responses and saw the vast majority had the same answers as me. That would probably have been the end of it but then I noticed this response:

The same thought process for the first one, but a significantly different approach to the second.

It made me wonder what approach others would take, and which approach my students would take. I wondered if the first problem had led this respondent into this solutions the second, and if so why it hasn’t had the same effect as me.

I don’t know if either approach is better, I just thought the differences were interesting. I’d love to hear your thoughts on it and how you would approach it.

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