UKC

http://www.bbc.co.uk/news/education-33017299

GCSE - the hardest that maths gets for everyone. It's not beyond reason to expect that they can multiply fractions, do simple probability, and use algebraic notation. This question doesn't even require them to solve the simple (it doesn't require the algebraic quadratic formula) you can see the answer.

This is the question causing the fuss:

Hannah had a bag containing a total of n sweets of which 6 were orange. It said the chances of Hannah picking two orange sweets one after the other was one third and then said use that to prove that n²-n-90=0.

Simple probability :

6/n multiplied by 5 / (n-1) = 1/3

so 6x5 / n x (n - 1) = 1 /3

multiply through

(6x5) x 3 = 1x n x (n-1)
90 = n^2 - n
0 = n^2 - n - 90

If you want the simple next step, then 90 being 10 x 9 gives the clue that the quadratic factorises into:
(n -10)(n+9)

So n = 10 or -9, you can'r have -9 sweets, so n =10


Just overly frustrated at the moment about the poor maths and IT skills of young people.

In reply to Philip:

As I understand it the question was set by 1 exam board and there was not a similar question in other GCSE exam board papers for Maths.

So have a level playing field across all boards.

That is why the topic has trended ( well according to my 15 year old daughter).

In reply to Philip:

As it happens, I'm in the middle of editing a bunch of GCSE maths books for Edexcel exams. The question is harder than anything in the suite of books I've been working on - students are used to working out the probabilities when the number of sweets is given as a number, rather than as an variable n, and they aren't used to being asked to solve equations arising from the probabilities.
I suspect it's this step up that has caused the angst, in that the exam question doesn't match anything previously encountered. Good students will be able to work through it, but average or weak students will be thrown completely by the unfamiliar presentation.

In reply to neilh:



This is always going to be a problem though isn't it? I really don't understand why there are different exam boards...

As for the question, years since I've done GCSE type stuff, but does sound quite hard. But the fuss seems to reflect the 'teach to the test' cutlure. Surely exams are supposed to be hard and push kids a bit?

In reply to Philip:

I'm not so much bothered by the question, the thing that is, is how fast it was being talked about.
Am I right in saying (this is how it was in my day) is that everyone across the country sits the same exam at the same time that's fair. How ever its not always the case I remember one of my GCSE's they sent some bit wrong for the higher level one and a bunch of us sitting that paper had to be isolated and chaperoned by a teacher until the right bit had been faxed thru.
Technology is far more advanced, faster and accessible these days, if a situation had occurred where for some reason a group of students hadn't sat that exam at the right time but had seen twitter this gives them an unfair advantage. (I know its only one question but still the principle)

As to the person that commented about discrepancies between different papers from different boards, there's always been that that's why RE GCSE is with some random welsh board and at my sisters school the music department put people through with trinity rather than associated board because it was easier to get higher grades with them.

just my 2p.

In reply to tony:


Is that not the point of an exam? To distinguish differing levels of ability. As long as it's taught that algebra is simply used to replace specific numbers with generic numbers then it should be fairly obvious. I feel quite confident a large number of the smarter kids will have had no trouble here.

In reply to Philip:

If everyone can get 100% then there is no point in having an exam. If virtually nobody can do that question then that will be taken into account during marking anyway.

Bitching about it on twitter shows a misunderstanding of what exams are for.

In reply to Philip:


It is easy when you know how. I did ok in GCSE maths and A-level maths, then went onto do Engineering maths as part of my degree at uni. Alas it has all flown my nest and I didn't get this one.

In reply to Philip:

What were the other colours and are there any left?

In reply to Philip:
The initial twittering was a piss take of the weird wording of the question, starting with "Hannah has n sweets. Hannah eats a sweet." and suddenly jumping to a quadratic equation. And similar funny conversationalist questions on the rest of the paper... having read it, it almost read as if the examiners were taking the p*ss.

As you say, there's nothing wrong with the question though. It's amongst the harder type of question you get on the exam paper, but it will be answerable by the more capable students, and grade boundaries are variable on each paper anyway.

My main problem with the question is that we don't now how many sweets Hannah has, but we do know that the probability of eating two orange ones in a row is 1/3. WTF? If you're going to set problems, they should be ones that could exist in the real world, or else don't try, and get the kids to solve an equation in the abstract instead.

In reply to Philip:

Hannah is a very good friend of mine, and I know that she's allergic to orange, so the probability of her picking 2 orange ones is zero.

In reply to Philip:

Did she replace the first sweet in the bag?

In reply to hang_about:

No, she eats it. The exact wording is:

There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow.

Hannah takes a sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet.

The probability that Hannah eats two orange sweets is 1/3. Show that n²-n-90=0.

In reply to climbwhenready:

does the question ask for n as well ?

In reply to lithos:
Ten, obviously...

ps. it ought to ask that question, yes, as it's an obvious follow-on, and a good way to demonstrate 'cross-topic' application of knowledge.

The hard part in the exam was probably writing the essay on what the sweet tasted like, and how difficult it was to choose between the orange and yellow sweets (GCSE maths seeming to have turned into a Humanities subject, if my experience of my niece's homework is anything to go by...)

In reply to bigbobbyking:

Interestingly my daughter says that the question is straight forward. The trending has been done by those who cannot figure out the answer and expected it to be easy.

Go figure in this social media age amongst 15/16 year olds.

In reply to lithos:

Yup, that's part b.

They will have separated it so if you can't work out the equation but can solve quadratics, you can get the marks for part b.

In reply to climbwhenready:

thanks I was curious as I'd not seen the whole question, i'd solved the first part already (it's not very hard !)

and to cap-pan - saying "it's 10 obviously", will not get you all the marks, show your working

In reply to climbwhenready:

What do we think about solving for n first then saying 6/10 times 5/9 = 1/3?

Seems OK to me in this context

In reply to lithos:

Someone put the paper on Dropbox. Not sure if it's meant to be there, but hey

Lost the link now.

In reply to John Gillott:

seems fine (I'd like a line about 6/n * 5 /n-1) first though.

Not sue how they want the answers / allocate marks but
given they are going out of their way to couch it in applied terms ....

don't need to see the paper, just idle wondering

In reply to lithos:

I'm not sure that first line is needed for a correct answer, though it seems that the idea might have been that students had to write it.

If n^2 - n = 90 then there are only two possible answers for n, and only one that makes sense as a number of sweets in a bag (10). Now calculate 6/10 times 5/9 to see that the bag did indeed contain ten sweets because that makes the probability calculation correct. No need to generate the quadratic in order to answer the question.

In reply to John Gillott:

It's common for GCSE questions to finish with "show that...", this is basically saying "the answer is... show us how you would get to that answer". If you use this answer in your working, you wouldn't get any marks - they're not asking you to reverse engineer a solution either.

I don't think the question was that hard, but perhaps could have been worded better. For example:

"A bag of sweets contains six orange sweets, while the rest are yellow. The total number of sweets in the bag is n.

Two sweets are taken from the bag at random. The probability of both sweets being orange is 1/3.

Using probability, show that n²-n-90=0."

I'm pretty sure GCSE kids are still taught basic conditional probability and basic algebra, so why not expect them to combine the two?

In reply to TobyMakins:

I agree to an extent. But I tried it on my son and he came up with the answer I posted and he found it easier. I think it is correct, probably, as in this case reverse engineering is mathematically correct or a correct answer to the question as worded (see wording here):

http://www.theguardian.com/science/alexs-adventures-in-numberland/2015/jun/...

In reply to John Gillott:
to write it.

in a bag (10). Now calculate 6/10 times 5/9 to see that the bag did indeed contain ten sweets because that makes the
probability calculation correct. No need to generate the quadratic in order to answer the question.

well yes but its also quite 'easy' to see that n=10 without solving the quadractic (by any means other than maybe
iteration ie you never find -9 as an option or care!) then plug that in and have you really shown much understanding ?

but you have solved the problem. Hence my comment about applied - this what id call an engineering approach, we
have a valid correct answer, not a pure maths approach - whats the abstract stuff going on. Ive no idea what they
want but rather assume its the latter

disclaimer i know nothing about GCSE etc these days its 38+ years since i did mine,
Robert Durran is a maths teacher whose opinion will be more relevant

In reply to lithos:

I think that my advice to a pupil would be that finding the solution n=10 by trial and error, showing it worked (6/10 x 5/9=1/3) and then showing that n=10 was a solution of the quadratic equation, ought to be an acceptable answer and get full marks if it is explained fully and logically, but that it is certainly not what the examiners had in mind, and that, examiners and their marking schemes being potentially very arsey, full marks could probably not be relied upon. ie I would always advise against such approaches except as a last resort.

Anyway, nothing to moan about in the question; some questions should be hard!

In reply to tony:


I suspect that this was the point of the question.

I agree with your point about context. In 20 years of teaching Maths to GCSE and A level, its the first time I've encountered a problem like this (personally I think it's a fantastic question), but with the changes to the GCSE coming up, we can expect a lot more like it in the next couple of years.

In reply to hang_about:


No. She ate it. This means the question is based on conditional probability, and is why this crucial piece of information is included.

In reply to TobyMakins:


Because putting two things together, or expecting them to link two of three sequential steps together, or just asking about something in an unfamiliar way, are all sufficient to totally flummox many of the kids.

Even at university level, many of them want to more-or-less rote-learn all the necessary methods, and complain if you change things just a little in order to test who actually understands.

In reply to Philip:

What's this got to do with IT? If people think IT is relevant to this question then their skills are indeed poor. If you can't solve this then you definitely don't deserve an A at GCSE. This is indeed straight forward if you are an A grade student,

In reply to Robert Durran:


Someone replied to my post saying such a solution should get 0/3 marks, but then deleted their post. I don't know whether they changed their mind or not, but, on reflection, I concede that such an approach does not show that the solution n=10 is unique. To show this you need to explain that the probability of getting two orange sweets strictly decreases as the number of lemon and total sweets decreases, so that two different positive values of n cannot give the same probability.

In reply to Robert Durran:



You would be correct in surmising that this is not what the examiners have in mind. Not because they are being 'arsey', but because they are asking the students to prove the quadratic.

Working backwards, by trial and error or even by 'factorising the quadratic' n^2 -n - 90 = (n - 10)(n+9) (grade c gcse btw) and then subbing in only shows that 10 is the particular solution, not that n^2 - n - 90 = 0, and no amount of explaining, no matter how 'fully and logically' makes it so.

In reply to Robert Durran:

I deleted the post because you said you'd advise your students to do something which you conceded later in the post wouldn't get the marks, so I mistakenly critiqued a post which you yourself retreated from.

In reply to Le Chevalier Mal Fet:


They are being asked to show that n^2-n-90=0 when n is the number of sweets (see the OP's link). If you establish (by whatever method) that n=10 and then show that 10^2-10-90=0, you have unequivocally done this.

I am not quite sure what you mean by "prove the quadratic".

In reply to Philip:

and to the rest of us on plane earth .....the relevance to the question and the answer ( 1974 CSE maths )

In reply to Philip:

Shock horror, it's teenagers complaining about exams on the internet.

It's all part of the parcel to have a moan surely? I bet even you did, it's just perhaps you didn't do it on twitter.

It is hardly representative of the academic ability of young people as a whole, the ones I work with are exceptionally talented.

Let them have their moan, we are not all academically minded, perhaps their talent is elsewhere!

In reply to Le Chevalier Mal Fet:


It should certainly get most of the marks unless the examiners are being arsey. Do you now accept that, if n=10 (uniquely) is established, the method should get full marks?

In reply to Robert Durran:

No, you have not. Regardless of what the op posted, the actual question asks you to prove that n^2 - n - 90 = 0 from the information that Hannah has n sweets, 6 of which are orange; she takes 1, eats it, then eats another. The P(2 orange sweets)= 1/3.

n=10 being a unique solution is irrelevant. It's also a unique solution to an infinite number of other equations.

In reply to Philip:

I thought it was a lovely question. It sorts out the "monkey see monkey do" types who are moaning about it wasn't like that on the past paper, from those that can think and apply the algebra they learnt in a different context of probability.

The actual algebra isn't hard, and I would personally use a tree diagram to set out what is being multiplied (not strictly needed but it does show it in a nice visual way)

I am going to give this question to my top set year 9 class next week as a challenge. I reckon more than half of them will be able to do it without help once they remember the AND rule (which I will make them look up for themselves if they don't remember.)

In reply to Robert Durran:


No.

In reply to Robert Durran:

I find exam mark schemes odd, and unimaginative.

In reply to marsbar:

Tree diagram is exactly how I approached it when a student asked me to explain it immediately after the exam, and you're right, it is an interesting question, and in my opinion, an indication of where things are heading in the future.

In reply to marsbar:


Solving the quadratic and showing that it happens to fit the information given does not prove that the QUADRATIC is a unique solution, so there is nothing odd about the exam ms if (as I suspect) they disallow solutions based on solving the quadratic and subbing in.

In reply to Le Chevalier Mal Fet:

I thought the same, but it's only really asking you to show that the value for n satisfies the probability and the quadratic. They could equally have asked "Show that 3n+1 = 31" - The difference being that in the question they did ask there is no requirement to find n (and it's a more logical way to solve it if you don't find n)

In reply to Le Chevalier Mal Fet:


The OP posted a link which showed a photograph of the actual question in the exam paper.

If you establish that n=10 uniquely satisfies the information about about Hannah and the sweets and then show that n=10 fits n^2-n-90=0 then you have, without any doubt, answered the question. What do you see as a problem with this? So far you have failed to do so!

If you initially (and actually quite cleverly!) solve the quadratic to get n=10 rather than using trial and error then that is fine.

In reply to drysori:

I don't have the paper with me, but my recollection is that it specifically asks the student to prove that n^2 - n - 90 = 0 from the information given. I'll certainly check on Sunday when I go in for P2 revision.

In reply to drysori:


Indeed. Or what about the question: n is the number on the Prime Minister's door. Show that n^2-n-90=0.

In reply to Le Chevalier Mal Fet:

Here is the link: http://www.bbc.co.uk/news/education-33017299
"Show", not "prove", which makes no difference.

In reply to Robert Durran:


Maybe it's my internet, but the link just took me to a bbc article.


No, it doesn't and it honestly worries me that as a maths professional you think it does. As I pointed out, n = 10 is a solution to any number of equations. You have not established a unique link between the stated conditions and this unique equation (not solution)


Do you honestly think that solving a quadratic by factorising rather than brute force trial and error is a clever trick????

In reply to Robert Durran:

Showing than n = 10 does not show that n^2 - n - 90 = 0 uniquely.

In reply to Robert Durran:

The correct solution is helpfully written in the link you just posted.

In reply to John Gillott:

According to the stepdaughter that would get nill points, because 'we're not allowed to prove backwards'. Go figure.

In reply to Philip:

The interesting point is that if n = -9, then Hannah must have 6 orange sweets and a liability of 15 yellow sweets. She is in negative equity, obviously, but what happened to those yellow sweets, and is Hannah an economist?

In reply to Le Chevalier Mal Fet:



Yes it does. And it is seriously worrying that a maths teacher such as yourself is unable to see that the approach is valid. So worrying, in fact, that I suspect you have not understood the approach rather than understood it but failed to see its validity.

Just to clarify: I establish (without any mention of a quadratic equation) that n=10 is the unique solution to the problem about Hannah and the sweets. I then show that n=10 is a solution of the given quadratic equation.

In reply to Philip:

However, if she is only in possession of 6 orange sweets, the probability would be 1, not 1/3, therefore the quadratic has to be solved with 10.

But, if her ledger book accounts for the owing 15, and she runs it through CERN a few times, then Gordon Brown caused the global crash.

In reply to Le Chevalier Mal Fet:


Not sure what you mean by this. Certainly, if n=10 implies that n^2-n-90=0 (Though n^2-n-90=0 does not imply that n=10)

In reply to Le Chevalier Mal Fet:
http://ichef.bbci.co.uk/news/936/media/images/83449000/jpg/_83449984_tussfo...

So, does the word show, rather than the word prove, mean you can get away with trial and error?

Personally I thought it more straightforward to write the equation, and rearrange.

I wouldn't solve until the next step.

In reply to Le Chevalier Mal Fet:


A correct solution is written (the one you and I and all mathematically literate people would immediately have come up with and which the examiners were expecting), but not the only correct solution.

In reply to Robert Durran:

How did you establish n= 10 without the equation?

In reply to Robert Durran:

We had the 'show' vs 'prove' argument earlier. The gcse student didn't get the difference, and from what she was saying, neither do her teachers.

In reply to Robert Durran:



Yeah, I got thatn=10 is a solution to the given quadratic, thanks and I understand what you think you've done.

The point is that n=10 is also a solution to other equations too. What n=10 doesn't do is establish a link between the information given and that unique solution.

I think for the time being we're going to have to agree to disagree on this point, until the markscheme is published.

In reply to All:

Just finished reading the thread. Apologies for repeating points made earlier.

In reply to David Alcock:


I don't think there is a difference. What do you think the difference is?

In reply to Le Chevalier Mal Fet:


But there is no need to do so. The question would have still been valid (though decidedly odd and misleading) had it said "show that n^2+n-110=0".


It wouldn't surprise me if the markscheme did not accept my approach, However, this would not mean it is invalid.

In reply to Robert Durran:


This is why your argument fails. n=10 doesn't neccesarily imply n^2 - n - 90 = 0 as n=10 is also a solution to n + 1 = 11 for example. Your reasoning is backwards.

In reply to marsbar:


Argue that the probability of getting two orange sweets decreases the more lemon sweets there are (ie that no two values of n can give the same probability, so that any valid value of n is unique). The show that n=10 is valid by showing that 6/10 x5/9 = 1/3.

In reply to marsbar:


by iteration if you like. n=1 - nope, n=2 nope ...n=10 yep

im not arguing in any direction, other than my initial thought about couching the question in applied terms,
which IMHO leads to a different a valid approach (ie show that 10 is a valid answer), they should ask a less applied
more abstract question if thats what they want to test - does the question as set test what you want it to

http://www.theguardian.com/science/alexs-adventures-in-numberland/2015/jun/...

has pic of question and alexs take on it.

In reply to Robert Durran:

Showing in this context that 10 or -9 derived from the quadratic fulfils the terms of the probability. Proving being expressing the probability in terms (n, n-1) that can be rearranged to be the quadratic. This is the gcse kid's POV. I personally think if you can logically justify a proof it doesn't matter which end you start at. However, I did O levels, and she (gcse kid) is more familiar / indoctrinated with the examiners' mark schemes.

In reply to Robert Durran:



I'm afraid that's exactly what it would mean. Examiners are not there to piss on your innovative approach to proof. If a solution is disallowed it's because it's invalid.

In reply to Le Chevalier Mal Fet:


Yes it does. Now I really am worried!

In reply to Le Chevalier Mal Fet:


So you think that markschemes are infallible?! I would prefer to judge a solution on it's logical merit rather than on the judgement of those who make up markschemes.

In reply to Philip:

No-one's answered my question about there being -9 sweets in the bag. (Think of the bag being an overdrawn bank-account). I know this is tongue-in-cheek.

-9^2 --9 -90 = 0

6/-9 x 5/(-9-1) = 1/3

In reply to lithos:

That link is how I did it. I get the iteration bit, but the bit about it can only have one value that RD said is the important bit.

In reply to David Alcock:
That falls in the "stupid answer" category. For year 7 its normally how many buses do we need. You can't hire 2.3 buses, so we hire 3.
2 is a stupid answer as well because we would leave people behind.

If you want a better reason, probability is defined using a 0 to 1 scale. 0 is impossible, 1 is certain.
You cannot have negative probabilities. So you can't have -9 sweets.

In reply to David Alcock:


Oh dear...............

In reply to Robert Durran:

I'm with you on this. It's quite an interesting one.

The line in the question that says 'show that n^2 - n - 90 = 0', can be replaced with 'show that n = 10', because the former is true if and only if the latter is true when n is a positive number, which it must be in this case.

So, we can re-write the question with 'show that n = 10' instead of the quadratic.

The two statements are logically equivalent but the question now feels a bit different. It could seem to be a more advanced question - the child is expected to formulate then solve the quadratic without being given the quadratic as a stepping stone. The child is expected to go straight to answering (b) via (a) but without being given (a) to prove / formulate.

But, the reformulation also allows a more direct solution to the validity of the quadratic because we can then use your second argument about the probability calculation: n = 10 works and as you say increasing or decreasing the number of sweets makes the probability of two oranges less likely or more likely than, so there is only one value of n which makes it work, so it is 10. Basically, the child goes from answering (b) to justifying (a) by reference back to the 1/3 chance.

Of course those familiar with algebra already knew this because the quadratic only has one positive solution. The various ways of looking at it all interlink. So we can go round in circles. But essentially, there is more than one watertight way to answer the question.

In reply to marsbar:

I know that is a "stupid answer". But my point was about the rancid heap of bollocks that is modern economics. I am aware that even a bag with 6 & -15 sweets does not fulfill the probability. Hence 'tongue in cheek'...

In reply to Robert Durran:


That's what I felt.

In reply to Robert Durran:


Ok. Good luck. Don't let me or the examination boards stand in the way of your unique insights into inverse deductive logic.


>Yes it does. Now I really am worried!

Just because n=10 is a solution to that quadratic, it does not mean that this quadratic is the only possible solution to the stated conditions (note my use of the qualifier necessarily).

An analogous argument would be
'Tonight I drank some beer. Stella is beer. Therefore I drank Stella'
'n=10 solves the probability problem. n=10 also solves n^2 -n-90=0. Therefore n^2-n-90 is the solution.
Goodnight.

In reply to John Gillott:





I'm gonna have to let this one go 'cos I'm climbing in the morning, but this is the exact point; the two statements are not logically equivalent. It's possible for n=10 without n^2-n-90=0. The logic only works the other way. So n^2-n-90=0 implies n=10, but not the other way round (Just like 3^2 implies 9 but sqrt9 doesn't neccesarily imply 3?)

In reply to Robert Durran:

To really boil it down:

the quadratic if and only if n=10;

question becomes show n=10;

n=10 works (6/10 times 5/9 = 1/3);

every other value for n is bound to fail as the chance always goes down as n goes up.

In reply to David Alcock:

Woosh over my head it went...

Sorry.

In reply to Le Chevalier Mal Fet:


You might be saying that the quadratic implies n=10 if n is positive, but we can just play around with the idea of n having a value without that value being the solution of a particular quadratic.

Sure, we can.

But the question as posed does not demand of the child that they generate that quadratic by multiplying two probabilities together. Rather, it states that there are n sweets in the bag and asks the child to show that the value of n satisfies the quadratic. So, it is perfectly reasonable for the child to note that the quadratic is the same as saying that n=10 and to find a way of showing that there are 10 sweets in the bag without generating the quadratic.

In reply to Le Chevalier Mal Fet:

I think the qualifing statements make it more like tonight I drank beer from my fridge. In my fridge the only beer is Stella, so I drank Stella.

It has to be positive because its probability. Once you have worked out it is 10, again it links back to the probability by saying it must be 10 and can't be any other answer.

Logically it seems right to me. I don't think the exam board will agree, but that doesn't make them right. Are we training kids to pass exams, or to think?

(To pass exams. Sad but true.)

In reply to Le Chevalier Mal Fet:


How? In my world 10^2 = 100. 100 - 10 = 90. QED as my Dad would say.

It doesn't mean its the only quadratic or equation that has a solution of 10. But that wasn't the question. Nor is that what you said, although I'm guessing it is what you meant.

In reply to marsbar:

we training kids to pass exams, or to think?


Yes. That's the impression I got from posing this one to the stepdaughter whose just done her AS. She got the question wrong btw, and blamed not having revised 'that sort of question recently'.

In reply to Le Chevalier Mal Fet:


It doesn't ask for a unique solution. n=10 inherently implies that n^2 - n - 90 = 0. It also implies that 2n = 20, or that n^3 = 1000 etc ad infinitum. It doesn't work the other way round, because the quadratic has more than one solution.

In reply to Le Chevalier Mal Fet:

I am sure that if you thought about it for a moment you would realise that that is nonsense. You are, after all, a maths teacher! You seem to be confusing "It's possible for n=10 without n^2-n-90=0" (False) with "It's possible for n^2-n-90=0 without n=10" (True).

In reply to Le Chevalier Mal Fet:


You've actually contradicted yourself here.

In reply to neilh:




Either way, this is probably a mistake for the school choosing that exam board, , as if their pupils are sitting the 'hardest ' exam in that year , then arguably their results will ( on paper) be worse than other schools, putting them down the league tables.

As I understand it, schools can choose which exam board their pupils to sit ?

In reply to Jim C:


The pass mark isn't the same for each exam board. So harder questions make lower pass marks. As for being unhappy with it, I would be pleased that we are stretching the bright students. Its not a grade C question so its not disadvantaging the majority.

In reply to marsbar:
I've lost touch a bit but was under the impression they gave up norm referencing ages ago and that (alongside competition) was part of the key to the rot. I'm pretty cynical about exam boards, especially in maths, having experienced first hand with new 1st year engineers the very real decline in standards since I did my exams at O , A and S level from 78 to 80. It was hard initially to evidence the decline was more than strong anecdote but the research at Coventry Polytechnic/Univeristy and Engineering Council work that built out of it scotched that. On standardised Engineering entry tests (designed to work out who needed extra help and in which areas) it was apparent the standards of the top end of the E grade when the tests started (early 80s? ) had drifted to the equivalent to the bottom of the A grade by the late 1990's, for A level Maths.

In reply to Offwidth:

I get the impression that they base it on how many marks are allocated to topics at different levels of difficulty and then they discuss it and decide.
http://www.bbc.co.uk/news/education-33025378

In reply to marsbar:
Thats not the same thing as the old norm referencing. Edexel in my humble opionion have always been vandals for what the did to the HN and ON qualifications and worse still happened in overseas dodgy sales tactics (that I witnessed in Malaysia).

In reply to Jim C:


Not necessarily. The grade boundaries will be different for different exams, even within the same exam board's papers, so a hard paper doesn't automatically mean it's harder to get a given grade.

Also, schools may want to choose the board which has questions of this sort, which forces a better approach to problem solving, rather than focus on learn-by-rote methods which pupils then cannot apply to any real scenario. Schools which want to be successful at A-level, for example, might prefer this approach to prepare their pupils.

In reply to Offwidth:

The problem with norm referencing is that is only works if the intake are near enough identical each year. In a year where there are more clever pupils it would be harder to get the top grade. So grades from year to year are not comparable. With this method it should be about the question level. In theory. In practice I don't know that it makes much difference.

In reply to drysori:

Totally agree. We want to test maths, not how many past papers someone has learnt the answers to.

In reply to marsbar:
Well if you want understanding , the current system seems (certainly up to the changes a couple of years back) to me to have delivered the opposite. Time and time again our new students tell me they were told to rote learn techniques.

I still think norm referencing delivers the most accurate comparative results, possibly at the expense of those at boundaries who are amongst those who need least to worry (I was never convinced of the quality of the research at the time, given the huge sample sizes involved in math.... maybe you can link me something modern to help me understand) .

School math was too 'academic focussed' in my day but the old ON and HN qualifications taught maths in context of the engineering and led to a back door route for a significant proportion of the successful graduate engineers. When Edexel took over we had to drop the HND in my University (like quite a few) as the syllabus lacked rigour and the ability to link properly even to our year 2 accredited course (which should have, been covering some of the same ground).

I forgot to add I agree with those who think, although the derivation isn't difficult, the wording is very sloppy. Proof means something very specific and such sloppy wording is a bad idea and can waste time and increase anxiety for the good students who know this and might be looking for something they have missed.

In reply to Offwidth:

I agree, in practice the norm thing was probably OK. And I also agree about the rote learning, hence why I am please to see a backlash now.
The wording says show, not prove. I am not sure if this was sloppy or deliberate, but it could be clearer.

In reply to Philip:
So. I looked at this equation for a minute or two and worked out that n must equal 10.
If id put n equals 10, would I have got the marks?
CSE grade 1 maths me

In reply to Philip:


Im doing a science PhD and i havent got a clue, so im pretty sure i wouldnt have a clue at 16...everythings simple when you know how, if you dont, well that question may as well be written in latin!


not being able to answer that question hasnt caused me too many issue in life so far...

In reply to Webster:

At first sight it floored me then I thought about it hard and I think the problem is the wording rather than the work required in the question. But we didn't do any probability iirc in O level so would have been flummoxed, but syllabus did include a bit of calculus.

In reply to marsbar:

My apologies... muddled with the OPs version... my concern with possible confusion was more than just the "prove" point.. it's also the formulation of the quadratic (the second root should be stated in the question as non-available).

In reply to Webster:

Could you work out that n=10... if not you possibly should be ashamed of the gap in your knowledge but you are far from uncommon as a PhD student. Some of the worst moments I've experienced in PhD vivas is students struggling with fairly relevant basic math.

In reply to Offwidth:

*maths

In reply to Offwidth:
The second root should not be stated as non available, I see that as a key part of the understanding of the question, the negative root does not fit the constraints of the problem and so the student should work out which answer is sensible and which may be dismissed.


>probability is defined using a 0 to 1 scale. 0 is impossible, 1 is certain.
You cannot have negative probabilities. So you can't have -9 sweets.

In reply to Webster:

http://gizmodo.com/is-it-math-or-maths-514561142

In reply to marsbar:
The equation does have two roots and is bugger all to do with probablility and that sort of thing IS potentially confusing in my teaching experience (especially to those say with autistic tendancies).

In reply to Offwidth:

I could have qualitatively worked out that n=10, but i wouldnt know how to calculate it...

if you have 10 swwets and 6 are orange, the first time you pick a sweet you have a 60% chance of an orange, you are then left with 9 sweets, 5 of which are orange so a 55% chance independantly...

ok im struggling, but i (hopefully) never have to sit an exam again. but i know where to look to find out how to do things, and to critically assess a sources reliability. how many people actually remember how to do everything they were taught at school? thats what google is for!

In reply to Offwidth:


that video proves or disproves nothing! the only conclusion in it is that vocabulary is illogical and therfore the only reasoning i need for correcting you is that this is a UKc forum and we are conversing in ENGLISH therfore it is maths

In reply to Webster:

English is not a fixed language, it is based on usage, many people use math so both versions are fine (math is more commonly used in N America and maths more so in the UK). You should move to move to France, they love being inflexible with language.

In reply to Offwidth:

I have autistic tendencies and then some. Its no excuse.

It has everything to do with probability in the context of this question.

If the student can't put the maths into the given context then that is part of what is being tested.

Putting the equation into context, in this context while the algebra gives 2 answers but either common sense or the probability constraint from the question tells us that 10 is the correct answer of the 2 possible answers given by the equation.

In reply to Offwidth:

Damn ye Sire, Maths it were and Maths it shalle be evermore! Tis writted in stoane.

In reply to marsbar:

It's not about listening to excuses, its about aiming to be as clear and as fair as possible in assessment. There are numerous forms of clearer wording or math expressions that could have been used.

In reply to wercat:
Times change but the pilgrimage of language is sweete breeth inspired....

"Whan that Aprill with his shoures soote the droghte of March hath perced to the roote, and bathed every veyne in swich licour of which vertu engendred is the flour; whan Zephirus eek with his sweete breeth inspired hath in every holt and heeth the tendre croppes, and the yonge sonne hath in the Ram his halve cours yronne, and smale fowles maken melodye, that slepen al the nyght with open ye (so priketh hem nature in hir corages), thanne longen folk to goon on pilgrimages...."

In reply to marsbar:


Agreed, however, it would be interesting thought to know if a very similar question was set say back in the 1950's or 60's . And if it was , at what level was it aimed at back then. ( it can't be an entirely unique question surely ?)

In reply to Offwidth:
But has language changed, or is it perhaps just a lot of the spelling, and the spoken words remain similar or the same?

For an absurd example , take 'Ye Olde Chippe Shoppe' , although a made up phrase to infer age, should it really just be spoken as 'The old chip shop' ( although you often hear people pronounce it as they imagine it was spoken )

In the same way, when we look at old ( or Olde) writings, with the spelling of the day, was the spoken word actually so different as the spelling would suggest?
We have no recordings of the spoken word from the 16th or 17th century, so how do we know how it was spoken?


Edit - When

In reply to Robert Durran and Le Chevalier:

Robert's initial method (solve the quadratic and take the negative solution to obtain n = 10; then confirm that ten sweets would give a probability of 1/3), as he conceded early on, was incorrect and did not answer the question, which was to show that the quadratic equation holds. But Robert's subsequent method (start from the info about sweets, and show in some way that n = 10; then it follows immediately that n^2 - n - 90 = 0) is perfectly valid.

I think there's a meta-theorem that one can apply here: if two intelligent people are disagreeing over something "simple", then at least one of them holds a misconception that is probably quite subtle and interesting.

In the case at hand, Le Chevalier Mal Fet is arguing that if polynomial has a root at n = 10 it does not necessarily follow that the polynomial has to be n^2 - n - 90. He's quite right, but it's not relevant -- you are not asked to show that there exists some special unique polynomial connected to the sweets problem; we're just asked to show that the polynomial n^2 - n - 90 happens to vanish at the particular value of n that solves the sweets problem. Thus it is enough to establish, by any means at all, that n = 10 (although obviously, you can't use the fact that n^2 - n - 90 = 0, as that is what we set out to prove).

Some ponderings that this interesting thread have thrown up:
I'd be interested if Robert, or anyone else, could prove that n = 10 without as an intermediary step showing that n^2 - n - 90 = 0, or some simple re-arranging of that equation.

I have always understood "show" and "prove" to be synonyms, but perhaps that is not conventional in school mathematics?

As to the wording, spelling out that each sweet gets eaten really hammers home the fact the there is no replacement going on. I think this is helpful to state explicitly; I think more students would get confused if the question were condensed as suggested above. Does anyone disagree?

Don't mark schemes list common methods that students might come up with, but then allow for any other valid method in proof questions?

In reply to Offwidth:
Thank you for the reply in kind! We lucky enough to have those words read to us in class by JRR Tolkien's grandson MGR Tolkien, though I must admit to finding middle English quite hard and obscure.

In reply to crossdressingrodney:

I thought RD did show that n=10 because 6/10 x 5/9 = 1/3

In reply to wercat:

wercat - do you understand about combing the probabilities (ie know that you have to multiple 6/n x 5 /n-1)
if not i worry about your understanding of statistical analysis of your exps etc

i teach at uni and have had undergrads unable to square -3 and the reach for calculator

if i get a chance im going to ask my 2nd years if they can do this (I predict most will be able to, >10% will not
- all have B or above in GCSE maths)

In reply to lithos:

Be interesting to compare your undergrads to my year 9 class!

In reply to Jim C:


A fascinating question that I believe has partial answers in the rhyme and rhythm of period writing, particularly poetry, and in the accents of various groups who left our shores in that time and them remained comparatively isolated from continentally influenced shifts in speech, such as Tangier, Virginia.

Oh, and with a large dose of speculation!

In reply to marsbar:

That shows that n=10 is one possible solution but does not rule out others. If you want to show that n=10 is the only solution you need one more line of reasoning, which Robert supplied in his next post.

In reply to crossdressingrodney:

Yes I agree. I didn't repeat him, but I agree it is needed.

In reply to crossdressingrodney:



That is not what I said! It doesn't matter how one comes up with n=10 (though solving the quadratic would be cunning exam technique!) as long as it is checked that 6/10 x 5/9 = 1/3. However I always said that you then have to check that n=10 satisfies n^2-n-10=0. All I later corrected was that it was also necessary to show that n=10 is a unique solution to the probability problem (by arguing that the probability strictly decreases as n increases - pretty obvious, but it can be explained by observing that while the number of pairs of orange sweets remains constant, the number of pairs of any colours increases)


I doubt it (except by trial and error)!


It is perfectly conventional; they are synonymous.
There is absolutely nothing wrong with the wording of the question; I really don't think it could be clearer. Anyone who complains about it lacks understanding of the problem and the mathematics being tested and deserves to lose marks in my opinion.


Yes they do, but markschemes and individual markers are not infallible and it wouldn't surprise me if valid solutions to this questions (such as outlined above) are marked down. I hope not. It will be interesting to see.

In reply to Offwidth:


There are many potentially confusing things in mathematics. Good candidates with good understanding will not be confused and get more marks. There is nothing wrong with the question.

In reply to Jim C:


I've got one of the practice papers we used in the run-up to my 1971 "O" Level. Here are a couple of questions from it. Not quadratic-related, but give an idea of how the questions were structured. If I have time I'll transcribe some more later. NB: There were 2 syllabuses (syllabi?) for the JMB examinations. The "A" syllabus included some calculus, the "B" syllabus didn't. Also, the practice paper was from the previous year (pre-decimalisation so prices are in "old" money)

_______________________________________

A smooth uniform rod AB is 3 m long and has a weight of 30 N. The rod is supported in a horizontal position by a trestle at Q, 70 cm from B and by a rope passing under the rod at P, 80 cm from A. Both parts of the rope supporting the rod are vertical.

(a) Calculate the values of the vertical reaction at Q and the tension in the rope, assuming that this tension is constant throughout that part of the rope in contact with the rod.

A body of weight 50N is now suspended from the rod AB at a point X.

(b) Calculate the least value of the distance AX if the rod is not to lose contact with the trestle at Q.

(c) If X is between P and Q, calculate the length of PX when the tension in the rope is equal to the reaction at Q.


___________________________


(i) Find the area enclosed by the curve y = x + 1/x^3, the x axis and the lines x=2 and x=5.

(ii) An open rectangular box with a square base of side x cm is made of thin metal sheeting. The cost of the sheeting used for the base is 3d per cm^2 and that used for the sides 1d per cm^2. The total cost of the sheeting is £0 18s 9d. Find the maximum volume of the box (neglecting the thickness of the sheeting).

In reply to Robert Durran:


Another way of presenting the different approaches, and a possible tweak:

if A is the statement about the bag of sweets, n, and the probability calculation;

and B is the quadratic equation,

Then the question asks the student to prove that A implies B.

To show A implies B by going from the truth of A to the truth of B, I'm struggling to come up with a method other than the generation of the quadratic.

However, there is the other way of showing A implies B, which is to show that Not (B) implies Not (A). This is in effect what the alternative method amounts to. It starts from the equivalence of the quadratic and n=10, then shows that n=10 generates the correct answer to the probability calculation, then uses n=10 as an anchor to show that any value for n other than 10 makes the probability calculation not 1/3.

Here's the tweak, to the mark scheme perhaps: what are the possible reasonable readings of the question? Can the child read it as asking them to show that n=10 makes the probability calculation true? Namely, can they get away with showing B implies A?

In reply to Robert Durran:

Saying it doesn't make it true, its just typical forceful RD UKC rhetoric.... (why not have a go at proving it ? . Of course my science and maths research colleagues who agree with me that the question could easily have been worded better could all be wrong. I dont think its a bad question, I just think its improvable.

In reply to Jim C:

I was having fun at the expense of spelling pedants and using language with poetic licence (no logic intended) !?

In reply to Offwidth:


I work in an Engineering company, I work day in day out with design engineers , we send our people to our clients to best understand the problem, and the design engineer then does the calcs and comes up with options.

We would never expect the fact finders to come back from site, and deliberately disguise the actual problem in language riddles like they sometimes do in these exams.

In reply to lithos:
You may well worry about my ability as I am neither a scientist nor very good at maths!

I did practically nothing about probability at school but we did a lot of stuff like binary numbers quite young and I can remember some calculus stuff being done at O level too. I do understand more about probability and combination now thanks to reading QED by Feynman a bit ago but I still think the wording is a problem - the request to "prove" something if that hasn't been taught regularly as an activity to the candidates might cause a block in an exam.

Of course I discovered in the process of stating the equation and then simplifying it that the proof requirement simply dropped out of the process at the bottom but I couldn't see that at first - I wasn't in an exam situation and therefore didn't need to panic about it.

Incidentally I was looking at some of my son's GCSE Physics practice papers and found quite a few questions that i thought were badly worded unless you remember what you were (incorrectly) taught by rote.

"Microwave signals can be received at the top of building A but not at the top of building B. However, radio waves can be received at the top of both buildings A and B." is just one sloppy example.

In reply to Rob Naylor:

But sizes in metric !

I was brought up with LSD all the way through primary , and I understand it, but boy am I glad we changed it for a 'better system'.

Personally I am for examining only on the basis that there is no risk of those sitting the test being misled/ confused.
I just want to know if they pass, and are employed here in finance or engineering, that they can all do the required calculations, and get the correct answers.
I don't want us to also be testing , English comprehension, which might not be easy for someone where English might be a second language.
That is a separate examination.

In reply to Rob Naylor:


Are they from the old "Additional Maths" O Level (only sat by the most able) rather than from the standard O Level?

In reply to wercat:

i am sorry i meant webster not you (said they were doing a science phd), apologies.

webster
>"Im doing a science PhD and i havent got a clue, so im pretty sure i wouldnt have a clue at 16...everythings simple when you know how, if you dont, well that question may as well be written in latin!"

In reply to Offwidth:

Proving what? I'm not sure what you are referring to.


In which case, please give an improved wording of the question which tests all the things the question it is intended to test:
(1) Conditional Probability
(2) Algebraic Formulation and Manipulation
(3) Solution of a Quadratic Equation
(4) Discarding of one root of the quadratic which is unrealistic in the context of the question.

If you can do all this, preferably worded in a way which would disqualify my trial and error then checking approach, then I shall concede that it is improvable.

In reply to Robert Durran:


Apologies if I've misunderstood you.


Yes, this is what I thought you said! I agree with all that.

In reply to marsbar:

I liked your explanation of why n = 10 implies n^2 - n - 90 = 0 by the way. One of the (many) things that has been surprising this year (I'm doing a PGCE in secondary maths) is that plenty of kids who can manipulate algebraic expressions well seem to struggle with substituting in values. It's as if they don't realise that there's any connection between the symbols they are manipulating and actual numbers.

In reply to crossdressingrodney:

" It's as if they don't realise that there's any connection between the symbols they are manipulating and actual numbers."

there's probably no better justification for teaching a bit of programming to kids as a way of showing this in action - my own mathematical understanding increased a lot when I got to play with a microcomputer in the early 80s and used BASIC, long after school and university.

In reply to John Gillott:





I'm not sure what this actually means!


And I'm certain that the question simply cannot (as worded) enforce the generation of the quadratic.

To do so, I think the question would have to ask the for the probability of getting two orange sweets in terms of n and then say "Hence show that n^2-n-90=0".

But I really don't think this is what the question is asking, even though it is clearly what the examiners intended the candidates is to do.


Yes I think they could: solve the quadratic and then show that n=10 is the unique solution to the probability question.

In reply to crossdressingrodney:


Absolutely. Many pupils will happily use "rules" to do 2a+3a =5a or 3(t+5)=3t+15 without appreciating that the rules are only there in the first place because, say, 2x9+3x9=5x9 or 3(7+5)=3x7+3x5 (and hardly any will appreciate that both are examples of the same rule!). Any good teacher will have pointed out and demonstrated that the rules are there because that is how numbers actually behave, but too often this will be forgotten under the pressure to perform ever increasingly complicated manipulations correctly (ie in such a way that the teacher marks them as correct.........).

It is depressing when an apparently algebraically competent pupil says something that demonstrates a complete lack of understanding of what algebra actually is!

In reply to Robert Durran:

I'm using set theory / truth tables.

Proof in one direction does seem to require the generation of the quadratic by the student.

But there is another method, which amounts to the contrapositive, which doesn't require the generation of the quadratic. It requires the solution of the quadratic or a trial and error method to get n=10.

The last bit was a somewhat stronger claim / question: is it reasonable to read the question as not demanding a proof than n=10 is unique? Or (same thing / related), is it reasonable to read it as saying that there is a single answer and the student must show this answer is 10

In reply to crossdressingrodney:

There is a great resource on tes for substitution top trumps. The really great thing is that it gives the equations for each animal, but you give the values. So for bottom sets you can give really easy numbers and for top sets you can use the same cards with decimals or fractional values to mix it up again.
https://www.tes.co.uk/teaching-resource/substitution-game-algebra-6027440

If you get them printed on thin card and laminated and store in plastic takeaway tubs they will last longer.

I would probably do groups of 4 for this activity.

Be prepared for some idiot to complain that you let the children play games and they didn't do any "work" if you teach in a middle class area!!! Maybe that's just my current school of precious princesses and helicopter parents.

In reply to Rob Naylor:

Thanks for posting those Rob. I'm assuming you've chosen particularly hard ones from that paper?






For the first question I think that neither moments nor Newton's laws are covered in GCSE maths now, so it's hard to make direct comparisons, but the general form of "from a wordy question, construct simultaneous equations and then solve them" is not unreasonable at the top end of GCSE today. However the fact that there are three unknowns and that the coefficients are not small integers probably means it wouldn't make it onto today's paper. This could change next year though!

It is quite similar to a simple A level mechanics questions, where the maths is top end GCSE standard, but the challenge is understanding the physics and writing down the correct equation.



The first part is straightforward integration, which is now generally not done until A level. We've only just covered this last week with year 12 -- could that group of B to A* GCSE grade students have learnt this a year ago? Some of them, probably, yes.

The second part involves turning words into simultaneous polynomial equations; solving one of them; using calculus to maximise another (unless there's an easier way I haven't spotted). Really, really tough. No way you'd ask that at GCSE.

Any thoughts from more experienced teachers?

Did you have calculators in those days BTW?

In reply to marsbar:


That's cool!

In reply to crossdressingrodney:



doubt it very much, i remember getting a very very early calculator (one of first in school) about 1974 i think
when i was 12 ish

In reply to Robert Durran:


Isn't it!

I've really enjoyed doing "lazy maths" with my classes this year: get them to answer on whiteboards things like

What is 37 + 37 + 37 + 37 + 37 + 37 + 37 + 37 + 37 + 37?
What is 37 x 51 + 63 x 51?

The looks on the faces of those who don't spot it at first are priceless while they're trying to work out how everyone else in class apparently knows their 37 times tables.

In reply to John Gillott:


I'm not convinced! At least I'm not convinced that the question could be reasonably interpreted in this way.


I don't think so.

Or (same thing / related), is it reasonable to read it as saying that there is a single answer and the student must show this answer is 10.

No, I don't think so. There is nothing in the question to say there is a single answer. It ought to be possible to devise a similar question with two possible answers.

In reply to marsbar:

Brilliant, thank you!

My colleague did actually lend me a set of those cards but I used superhero top trumps in the end, as Marvel superheroes are one of the few things I know about that lie within the cultural sphere of my year 8s! And it's hard to base a card sort on twerking and loom bands.

In reply to crossdressingrodney:


A good one (a bit more advanced) is to work out the third side of a right angled triangle whose hypotenuse is 74.5cm and one of the shorter sides is 25.5cm.

In reply to Robert Durran:

Nice.

I got fooled by a similar one earlier this year:
Which triangle has the larger area: one with sides of length (10, 10, 12) or (10, 10, 16)?

In reply to Robert Durran:


The question doesn't require the construction of the quadratic, but one solution does (the one the setters wanted the students to provide, it seems!).



Given that the quadratic is true with a positive n if and only if n=10, we can re-write the question to say 'show n=10'. I can see how it would be reasonable to read that as strongly implying that there is only one value for n / that the student is not required to show that the answer can only be 10.

In reply to John Gillott:


I can't find anywhere what part (b) of the question was. Find n? Find the number of sweets? Show that n=10?
Does anyone know?

In reply to crossdressingrodney:


Excellent!

In reply to crossdressingrodney:



Hahaha....very good!
Got the 1st one straight off, but about 5 minutes for the other (I'm blaming the hangover )

In reply to Robert Durran:

Part b was solve the quadratic I think.

In reply to Robert Durran:

'There are n sweets in a bag. Six of the sweets are orange.
The rest of the sweets are yellow.

'Hannah takes at random a sweet from the bag.
She eats the sweet.

Hannah then takes at random another sweet from the bag.
She eats the sweet.

The probability that Hannah eats two orange sweets is 1/3

(a) Show that n²-n-90=0
(b) Solve n²-n-90=0 to find the value of n.'

In reply to John Gillott:


So anyone who had simply solved the quadratic for part (a) should have immediately realised that something was amiss when they came to part (b).

In reply to Robert Durran:


Why? This is the whole point. The question setter wanted students to do a) then b), and to do a) they wanted the student to do the probability calculation with n as a part of it and thus to generate the quadratic

However, it isn't necessary to do this to answer a)

In effect the student can answer b) first then do a) without generating the quadratic, because a), as a statement, is logically equivalent to n=10 given that n is positive.

So, the student can subvert the setter's intention, can answer a) and b) together, by collapsing a) and b) into one question: 'show n=10'.

That part b) asks the student to 'find the value of n' relates to what I was getting at before: the question is telling the student there is only one value for n, so it could be argued that using the alternative method (not generating the quadratic), the student does not need to show that n can only equal 10, just that 10 works, ie 6/10 times 5/9 = 1/3

In reply to John Gillott:


Yes. If they found themselves solving the quadratic twice it should have been clear that they were not doing what the examiners intended and expected.


Agreed.


Agreed again.


I think that if the quadratic is never actually solved to find both roots, it would really be stretching it to expect to get all the marks available for part (b). Arguing that it is not necessary because the question later goes on to imply that there is only one value of n seems decidedly dodgy to me!

In reply to John Gillott:




No it doesn't say that. It says show that n^2 - n - 90 = 0

Solving the quadratic and subbing the positive solution (n=10) into the conditional probability only shows that the quadratic COULD BE generated from the information given, not that is IS for the reason that there are an infinite number of other equations in n that also give the solution n=10. You can't just assume that the right one is the one given.

In reply to Robert Durran:





Re the last bit, my point was that the student solves to get both values, then argues to ignore -9.

So yes, solve the quadratic for both values.

However, what may not be necessary is your early correction / addition to your reasoning - the bit about showing 10 is the unique answer. The question asks the student to find the value of n, so it is telling him there is only one that makes sense as an answer to the number of sweets in the bag. It could therefore be argued that the student does not need to show that any other (positive) value of n will not give 1/3 as the answer to the probability calculation.

In reply to crossdressingrodney:



Thanks. That's exactly what I was trying to say, slightly drunk and very tired on an annoying tablet keyboard. However, I disagree about the relevance. I think the question is asking exactly that. "Show that" means "show that it is", not "show that it could possibly be" which is why I don't think Robert Durran's approach works. This type of problem is fairly common at A level and the markscheme always insists on an approach based on working forwards from the initial conditions to the equation which is then solved, for the reason that, as you state, doing it the other way doesn't prove the unique solution.


They are, but given this why do you feel that my objection to finding n and subbing in is irrelevant?


Yes, in my experience they do (20 years teaching maths to GCSE and A level maths and further maths). In this case I'm confident that Robert Durran's approach will not score full marks, not because the exam board are being 'arsey' or trying to stifle creativity as others have claimed on this thread, but because, as you say, "show that" is synonymous with "prove that" so "show that n^2 - n - 90 =0" means prove that is is UNIQUELY n^ - n -90 =0".

In reply to John Gillott:


Strictly speaking this may be true, but is probably pushing exam technique to the limit!

In reply to Le Chevalier Mal Fet:


Yes, so finding n=10 as the (unique) solution to the probability problem (WITHOUT reference to the given quadratic) and then showing that, for n=10, n^2-n-90=0, does this. This has already been pointed out to you several times!


The question, as worded, does not require you to generate the quadratic from the given information.

In reply to Le Chevalier Mal Fet:

Show that n^2-n-90=0 means exactly what it says (that the number of sweets, n, fits the equation). If you want it to mean something different (that an algebraic formulation of the of the probability problem can be manipulated into this quadratic) then you have to say so. I think the examiners have probably avoided this in order not break the question up into several parts and spoon feed candidiates throgh the question (a common form of dumbing down), but have inevitably then opened the question up to the other approach (probably uninentionally).


In which case the markscheme is wrong; any correct answer to the question as posed, HAS to get full marks.
To insist on a particular logical series of steps the question would have to lead the candidates through those steps using ther word "hence".


Would you prefer "wrong" to "arsey"?.


I agree. It doesn't change the question at all. If I said to you," t=3, prove that t^2 +5t-9=20" do you disagree that it would be sufficient to substitute t=3 into t^2+5t-9 and get 20?



What is? Without first specifying that we are talking about a quadratic derived from an algebraic formulation of the probability problem, it CAN'T mean that! The question can only mean what it actually says.

In reply to Robert and Le Chevalier:

I think this is the root of the problem:


The only thing that has to be shown to be unique is that n = 10 is the unique positive solution to the problem.

n is a number that is fixed at the beginning of the question. Thus n^2 - n - 90 is merely a number that is fixed at the beginning of the question. You get the marks if you show that that number is zero.

You seem to be treating the n as a formal variable and the expression as if it were an algebraic object, namely a polynomial expression in the formal variable n; you seem to think the examiner wants you to recover this polynomial expression from somewhere, but that's not what the question's asking. It's just asking you to show that the number n^2 - n - 90 is zero. So it's enough to show that n = 10 (as long as you don't assume a priori that the polynomial expression holds).

In reply to crossdressingrodney:


Fixed by the information about Hannah and the sweets anyway.


You certainly SHOULD do, though I fear you might not!


I think we all agree that the examiner's intention was that the polynomial is recovered from somewhere. However, the question, as posed, does not require that it is.

In reply to Robert Durran:


Yep.

In reply to Robert Durran:

4". Or 101.7mm. Ish.

In reply to Robert Durran:

Oops!

In reply to Philip:

Perhaps it was worded more clearly on the paper than in your post? It's not clear Hannah removes the sweets she picks though it's clearly the case if you work it through and the 'it said' bit is distracting. Seems like a pretty tough unstructured question to me but then it's a long time since I saw or sat a GCSE maths paper.

jk

In reply to Robert Durran:


Yes, but that's not what you did in your post. You solved the quadratic then subbed it into the given probability information to show that it produced the result 1/3


It does. I've dug out an example of a similar question along with the ms and examiners report to illustrate this point.

In reply to Philip:

What's the probability that a simple maths question published on UKC will continue to run for more than three days?

And what's the probability of it ending in a Godwin?


Show your working.

In reply to Robert Durran:
This is a similar question from a few years back. It uses compound shapes instead of conditional probability but the principles of deriving a quadratic from initial conditions are the same. Sorry some of the formatting isn't right and the diagram wouldn't copy, but it's basically a compound shape with sides in x. The important thing is the way the question is structured and the examiners comments.

1. The diagram below shows a 6-sided shape.
All the corners are right angles.
All measurements are given in centimetres.

The area of the shape is 25 cm2.

(a) Show that 6x2 + 17x ¡V 39 = 0
(3)

(b) (i) Solve the equation
6x2 + 17x ¡V 39 = 0

x = ¡K¡K¡K¡K¡K or x = ¡K¡K¡K¡K¡K
(ii) Hence work out the length of the longest side of the shape.
¡K¡K¡K¡K¡K¡K..cm
(4)
(Total 7 marks)

MS. (a) 6x2 + 11x ¡V 10 + 6x ¡V4 = 25
6x2 + 17x ¡V 39 = 0 3
M1 for an expression for the area involving either
(3x ¡V 2)(2x + 5) +2(3x ¡V 2)
or 3x(3x ¡V 2) + (3x ¡V 2)(7 ¡V x)
or 3x(2x + 5) ¡V 2(7 ¡V x)
or (3x ¡V 2)2 + 2(3x ¡V 2) + (3x ¡V 2)(7 ¡V x)
where in each case at least one of 2 or 3 product terms
must be correct
M1 (indep) for one correct expansion involving x2
A1 for simplification to final answer


3. This proved to be the first really challenging question for the candidates. "There is still a minority of candidates who do not understand that in part (a) they are required to derive the quadratic equation from given information. They give themselves away by trying to solve the equation as their answer to part (a)." The most commonly successful approach was to identify two rectangles of areas (3x ¡V 2)(2x +5) and 2(3x ¡V 2) respectively and then set the algebraic sum equal to 25. Further marks were then gained by using correct algebra to get to the given equation. Splitting the shape horizontally proved to be less successful as often the top rectangle was given the measurements (3x ¡V 2) and (2x + 5). Other methods involved splitting into the sum of three parts and working on the difference between the area of the full rectangle (2x + 5) by 3x and the small rectangle 2 by (2x + 5) ¡V (3x ¡V 2) although in many cases the second term was not worked out correctly.
Part (b) was generally tackled by using the formula. The usual error of not spotting that 172 ¡V 4 ¡Ñ 6 ¡Ñ (¡V 39) = 289 ¡V 936 is incorrect was often seen. Other errors included a faulty evaluation of as ¡V 17 „b 35 ¡Ò 12 and 2 in the denominator rather than 12. Sometimes the negative sign was omitted from the second solution.
Some candidates realised that they could factorise the left hand side and often did so successfully. A minority once they had found the solutions reversed the signs.

The relevance to this thread is the way the question asks students to "show that" in exactly the same way as this question, and the examiners initial comments which clearly indicate that an approach based on solving the quadratic and subbing in is not acceptable for reasons I've given several times previously in this thread.

In reply to Robert Durran:


nice.

i worked this out but it took quite a few steps after the initial 'jump', i wonder if i missed something?
i needed 1 subtraction and a whole bunch of factoring and rearranging

In reply to Le Chevalier Mal Fet:

Three points:

1. In your post before this one you said that Robert solved the quadratic then subbed in 10. He did more than that: he solved it, argued that -9 is not relevant and then, more than, that he showed via a logical argument that no positive value other than 10 can satisfy the probability calculation.

2. Without seeing the image it is hard to be sure how well your analogy works: is there a valid solution in this case that does not rely on generating the quadratic?

3. Robert is of the view that the examiners would be wrong if they marked the Hannah's Sweets puzzle in the way you think they would. So, in some ways, showing that they have marked another question similarly, if indeed they have, is not all that relevant.

In reply to Le Chevalier Mal Fet:


It doesn't matter how you come up with n=10 (trial and error, solving the quadratic or any other method) as long as you check that it is indeed a correct and unique solution to the probability problem. I originally suggested trial and error and later added that solving the quadratic would be cunning exam technique (of course either could be done in rough and not handed in, so the examiners need not know how you came up with it).



Well all I can say is that the examiners do not understand their own question as worded; they are wrong. As I said earlier, I would not encourage pupils to use my alternative method because I would not trust the examiners to accept it as being a correct solution to the question as worded. You have proved me right to be cautious.

In reply to Le Chevalier Mal Fet:

Obviously just solving the quadratic and assuming this is the answer to the area problem is not adequate (just as in the Hannah and sweets problem), but again, it would be perfectly adequate to come up with the unknown by any means (including solving the quadratic) and checking that it is a correct and unique solution to the original problem, and then checking that it fits the quadratic.

Again, if the examiners disallow any mathematically and logically correct solution to the question as worded, then they are sadly at fault.

You seem to be assuming that the examiners are almost by definition correct; you just repeatedly appeal to what is on a markscheme rather than to what the question actually says. Surely mathematics and logic are much better guides to what is correct.

In reply to lithos:

>


x^2 = 74.5^2 - 25.5^2 (Pythagoras)
= (74.5+25.5)(74.5-25.5) (Difference of two squares)
= 100 x 49
x = 10 x 7
= 70

In reply to Robert Durran:
ah bugger forgot my diff of squares - i noticed it was 49 difference that was the clue then

74.5 ^2 = (25.5 + 49)^2
= 25.5 ^ 2 + 49 ^2 + 2 x 25.5 x 49
= 25.5 ^ 2 + 49 (49 + 51)
= 25.5 ^2 + 4900

x^2 = 4900, x = root 4900 = 70
long winded but made sense esp when i saw 2 * 25.5 = 51 !

In reply to Robert Durran:

Ah that's even more elegant than I thought. I'd enlarged by scale factor 2 to make everything an integer.

In reply to Robert Durran:

Or, more generally, note that the sides are a, 100-a, and x. So in the Pythagoras formula a^2 drops out to leave x^2 = 10000 - 200a. The specified value of a then gives x^2 = 4900 =70^2.

In reply to lithos:


Pythagoras' a^2=b^2+c^2 can be rewritten as c^2=a^2-b^2, so c^2=(a+b)(a-b), so (a+b)/c=c/(a-b).
The fact that Pythagoras can be written as two equal ratios, suggests that there should be a very natural similar triangles proof (there is!).

In reply to marsbar:

ok so the test is done and the results in....

"Hannah .... "

Q: can you do this

a) yes 30%
b) maybe with some help 40%
c) no 30%

n=100, approx %s

In reply to lithos:

Oh I forgot. Will do and let you know.

In reply to Philip:

The back-page of the Saturday Guardian has a puzzle section that usually includes a simple probability calculation, like this one (dressed up with the Hindu Jar brothers and balls in urns, rather than Hannah and sweets). It's much, much easier to solve than trying to do the Cryptic Crossword next door!