Recently I saw this from brilliant.org on Facebook and it struck me as an interesting problem:
the first solution is trivial and obvious:
But the Facebook post said there was two, so I set out in search of the next one. As there were exponents I thought I’d take logs of both sides:
Then realised I could take logs to base X and make things a whole lot simpler….
So x = 9/4
As you can see it reduces to an easily solvable problem, and all that was left was to check the answer:
A lovely little problem that gives a good work out to algebra and log skills.
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:
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.
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.
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.
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.
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:
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.
I’ve written before about the app “Brilliant“, which is well worth getting, and I also follow their Facebook page which provides me with a regular stream questions. Occasionally I have to think about how to tackle them, and they’re excellent. More often, a question comes up that I look at and think would be awesome to use in a lesson.
Earlier this week this question popped up:
What a lovely question that combines algebra and angle reasoning! I can’t wait to teach this next time, and I am planning on using this as a starter with my y11 class after the break.
The initial question looks simple, it appears you sum the angles and set it equal to 360 degrees, this is what I expect my class to do. If you do this you get:
7x + 2y + 6z – 20 = 360
7x + 2y + 6z = 380 (1)
I anticipate some will try to give up at this point, but hopefully the resilience I’ve been trying to build will kick in and they’ll see they need more equations. If any need a hint I will tell them to consider vertically opposite angles. They should then get:
2x – 20 = 2y + 2z (2)
3x = 2x + 4z (3)
I’m hoping they will now see that 3 equations and 3 unknowns is enough to solve. There are obviously a number of ways to go from here. I would rearrange equation 3 to get:
x = 4z (4)
Subbing into 2 we get:
8z – 20 = 2y + 2z
6z = 2y + 20 (5)
Subbing into 1
28z + 2y + 6z = 380
34z = 380 – 2y (6)
Add equation (5) to (6)
40z = 400
z = 10 (7)
Then equation 4 gives:
x = 40
And equation 2 gives:
60 = 2y + 20
40 = 2y
y = 20.
From here you can find the solution x + y + z = 40 + 20 + 10 = 70.
A lovely puzzle that combines a few areas and needs some resilience and perseverance to complete. I enjoyed working through it and I’m looking forward to testing it out on some students.
Cross-posted to Betterqs here.
Today one of my Y12s was looking through a C3 paper he found on my desk. (For those unaware A level maths, studied in Y12 and Y13, ie from 16-18, is currently modular. There are 4 Core Pure modules known as C1, C2, C3 and C4. The first two are studied in Y12 and the second two in Y13.) He came across this question:
While looking at it he said, “are you sure this is a C3 question?” I told him it was and he then said “But I can answer it.”
I looked at the question, all the main skills it tests are taught at C1 and C2, but the chain rule for differentiation isn’t taught til C3. I thought about it and realised that yes, with the application of the binomial expansion (a C2 skill), or indeed a long winded brackets expansion, it would give him a polynomial he could differentiate.
Then it occurred to me that it was in fact a brilliant question to set my Y12s as revision. It allows them to see links between the things they’ve learned, allows them to practice important skills from C1 and C2, namely the differentiation, the coordinate geometry involved finding te equation of a tangent and the binomial expansion, and to solve a problem using those skills.
It took them longer than it would have taken someone who knew about the chain rule, but it was time we’ll spent and I got some perfect answers from them. I didn’t tell any of them how to do it, they managed to talk each other through it, and I only had to pick up on one slight error when one of them had a slight hiccup with a power. I think I need to have a good look through some more higher level papers to see if I can find any other gens to test the earlier skills.
This post has been cross-posted to Betterqs here.
This is a guest post written by my brother Andy (@andycav_25). Andy is a primary teacher in West Yorkshire and currently teachers Year 5.
Fractions Jigsaw from NRICH
In the 2014 Primary National Curriculum, there’s much more emphasis on problem solving in maths. Henceforth, we’ve had a few staff meetings and twilights on this recently. We’ve been encouraged to use ‘rich mathematical tasks’, and some colleagues (myself included) expressed sheer horror at some of the ideas. However, I did embrace it and took a bit of a risk with my Year 5 class last Friday. If an OFSTED inspector had walked into the first 30 minutes of that session, I would’ve been mortified. If they’d walked in at the end, I would’ve been ecstatic.
Definition from Simon Borget
We’d been doing fractions, decimals and percentages all week and, in my timetable, I have a 2 hour session every Friday morning. I normally do 2 lessons in this time unless it can be used for a science experiment, art or D&T depending on the long term plans. Last Friday, maths was the winner. The NRICH website is great for providing these rich mathematical tasks and I found this one, which I really liked the look of. Children had to complete a jigsaw on a 5×5 grid. Each square is cut into 4 triangles that were either blank (these would eventually form the outside of the jigsaw but there were a few extras thrown in) or they had a calculation using fractions. Obviously, the children had to match up the triangles to create a jigsaw.
I thought that it had to be a pair or small group activity, so do you group them as high/low ability or assign a high ability with a low ability? I kept the high achievers together this time, as an experiment in itself as much as anything.
What unfolded in front of me was just amazing. Every single kid in my class engaged. I don’t think I’ve ever achieved that before in 5 years as a teacher (and another few before that as a TA/HLTA).
One child in my class is ridiculously talented in maths (earlier in the term, it had taken him about 5 minutes, without a calculator, to find all of the factors of 256. He’s 9!). He absolutely loved this jigsaw task. He thrived on the challenge but was also made to reassess a few times when things didn’t appear to be working. It took his group about 45 minutes to complete the jigsaw, but when I spoke to them about how they did it, it was one of the lesser but still more able kids who had spotted the pattern in the jigsaw. They had taken on different roles in the group with child A quickly calculating in his head and child B spotting patterns to assist child A in finding the correct pieces.
That group also complained to me at one point that they had 6 ‘top’ pieces, which doesn’t work on a 25 square. They told me that it was wrong. ‘Look again,’ I said, numerous times. Eventually, they did realise that there were two blank areas inside the grid too.
The middle ability groups really struggled at first. As had been the plan at the offset, I shared a few answers and gave them a starting point after about 20 minutes. Bang. They started working through that grid as fast as I’ve ever seen them work in the 4 weeks they’ve been in my class. Not everybody finished (in fact probably only about half did, if that) but that didn’t matter either to me or them. They had all achieved something and we all knew it.
Once the HA group solved it, I gave them a template of the solution to make their own – they had fun testing each other. At that point, I also shared that template with the rest of the class as an extra piece of scaffolding. That made some of them look and immediately say ‘we need that shape here so which piece fits?’ Another great use of mathematical thinking. Key question: what’s the same and what’s different about the pieces and what does this tell us?
At the end of the session, I did two plenaries. First, what skills have we used this morning? Answers: ‘fraction skills’, ‘teamwork’, ‘spotting patterns’. It took a little bit of guidance but that’s fine.
I would’ve liked ‘problem solving’ to be an answer but alas…
We’ll get there in the end!
The second plenary was simple: ‘who has been completely confused and felt like your head is about to explode at least once this morning?’ (62 thumbs up – there’s 31 children in the class).
Who thinks you’ve made progress with your maths this morning? (again, 62 thumbs up). I think this risk definitely paid off! I will be trying to do at least one of this type of task or investigation for every topic in maths from now on. The skills and achievement shown by every child at their own level was amazing and they really enjoyed it too! Plus, there’s no marking to do… Bonus!
I enjoyed reading this post. It’s nice to get a fuller view of what maths is taught at primary and how it is taught. I love the resources on nrich, I’ve not used this specific one, but I certainly would if I have a class that it fits too. Have you used anything similar? What are your views on this task/these tasks?
I came across the following area puzzle on Ed’s (@solvemymaths) site.
I found it quite an interesting idea and had a little go at solving it.
First I considered an equilateral triangle, all the angles are 60 degrees so we can see the area would be (x^2 sin (60))/2. As sin 60 = 3^(1/2)/2 it was easy enough to solve.
I then realised I could have used Heron’s Formula, so did it that way too. Luckily I got the same answer:
The square was a tough one:
Then I considered a pentagon, I wasn’t sure how to approach it at first, so I reverted to my favourite shape, the triangle. Split the pentagon into 5 congruent isosceles triangles and solved with trig.
Wolfram alpha gave me the lovely exact answer:
I then considered the hexagonal case, which is really just 6 equilateral triangles.
I approached the decagon in the same manner I had the Pentagon.
And got an equally lovely exact answer:
I enjoyed working through these, and thought it would make a nice lesson to build resilience and cognitive activation. I also thought about what else could be done. What does the sequence of side lengths generated look like? Could an equation be formed to describe it? What if we were looking at the areas of regular polygons with side length six, what would the sequence look like then? All these would make nice investigations.
This week’s puzzle from Chris Smith is a nice contextualised simultaneous equations puzzle I intend to use next week.
It boils down to:
2X + B + T = 135
2X + 2B + 3T = 269
X + B + T = 118
Via elimination using 1 and 3 we can see that X (number of tandems) is 17.
34 + B + T = 135
34 + 2B + 3T = 269
17 + B + T = 118
1 and 3 rearrange to the same leaving:
B + T = 101
2B + 3T = 235
Elimination gives T (number of Tricycles) to be 33 which leave B (number of Bicycles) to be 68.
A nice little problem.