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Big fuss about a hard maths question


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This came up on an exam paper this week.  I didn't teach my boys to this level, so I don't know if it's out of the ordinary.

 

For reference, the GCSE exam is taken by most 16 years olds in England.  GCSE maths and English are the very basic markers of education for most white collar jobs.

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It is on my Facebook threads and it is not out of the ordinary for 15/16 year olds :) the consensus from my friends was that they would be happy if the questions for GCE for their 15/16 year old kids were only this "hard" :lol:

 

Besides how would the exam board norm results if there are no "hard" questions as differentiators between an A or B (or between A1 and A2 for other countries like India and Singapore)

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GSCE is only for college-headed kids, right? So that's normal. Algebra II probability, correct? That sounds right.

 

No - GCSE is for everyone.  It stands for General Certificate of Secondary Education.  Here's a receptionist job that says 'preferably GCSE maths' grade B or above, as an example.  Grade C or above is essential for this receptionist job.

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No, GCSE is what everybody takes ages 16ish. A C in GCSE maths is a requirement for pretty much all jobs nowadays.

This question was terribly mean though considering the normal curriculum taught. Probability is scarcely taught in any major detail. Questions meant to sort the A*/A students are usually harder algebra/Geometry questions and I doubt most pupils would have ever seen anything like this.

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I don't know what the test is like, so I may have gotten it wrong. It was easy to see that n = 10 so I knew that there were 4 other candies. I would have written 6/10 * 5/9 = 30/90 = 1/3. I would not have written an equation with n. The question itself wasn't hard.

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One of my friend's long explanation. He is a lawyer anyway :)

 

"

a. There are 6 orange sweets and n sweets in total Probability of getting an orange sweet on first pick is therefore 6/n.

 

b. After drawing one orange sweet, there would 5 orange sweets and n-1 sweets. Probability of getting an orange sweet on first pick is therefore 5/(n-1).

 

c. To get two orange sweets in a row, you multiply the two probabilities together i.e. 6/n x 5/(n-1). That is specified to be 1/3.

 

d. Therefore 6/n x 5/(n-1) = 1/3

30 / n(n-1) = 1/3

30 / (n^2-n) = 1/3

90 / (n^2-n) = 1

90 = n^2 - n

0 = n^2 -n - 90

 

Note: This is ALL the student needs to prove. Usually we do it in less steps.

 

But you can solve it further

 

(n+9)(n-10)=0

n = -9 or 10

But n must be positiive so n=10"

 

His original working

 

"6/n x 5/(n-1) = 1/3

90 = n(n-1)

n^2 - n - 90 = 0"

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Wait up... are you saying most kids in the UK aren't taught basic probability concepts by the time they are 16?

 

I don't think anyone said that.  Probability is definitely taught, but this question seems to have tripped people up.  I've no idea why - haven't seen the curriculum.

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Probability is definitely taught, but this question seems to have tripped people up. I've no idea why - haven't seen the curriculum.

My guess is that the phrasing of the question is not similar to past year papers. For Singapore's version of GCE 'O' levels we have "Ten Years Series" so students who just get by in maths would drill on past year papers using trends their tutors spotted.

 

Looking at the question, I would have thought immediately of solving for n rather than thinking of it as a multi-step problem. The part (b) not shown could have been to solve for n.

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I don't know what the test is like, so I may have gotten it wrong. It was easy to see that n = 10 so I knew that there were 4 other candies. I would have written 6/10 * 5/9 = 30/90 = 1/3. I would not have written an equation with n. The question itself wasn't hard.

 

How is it easy to see that n = 10 straight from reading the problem?

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My guess is that it's a question meant to differentiate between students with different levels of math ability. Questions like this are a common practice in many testing situations. Yes, a strong math student at that age should be able to figure it out if they've had the basic instruction without needing to have had explicit instruction in how to solve that particular kind of problem, but I doubt an average math student would get it. That's sort of the point.

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I had no problem.  I solved it in less than 2 minutes on an index card.  I did not know n=10 before I started. I did it like Arcadia.  I have a math degree though.  ;)  I'm pretty sure I could have solved it from 9th grade on. 

 

I can see it's a problem where either you know/remember probability stuff or you don't. 

 

 

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I'm sorry. Not from reading the problem. From reading the equation. The only number squared that would be that number plus 90 is 10.

 

But the problem asks the student to show that the equation is true, given that the probability is 1/3. If you begin by assuming the equation is true (and therefore n = 10), and then showing that the probability is 1/3, you are doing a different problem altogether.

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