GCSE's getting harder causes Paul Dacre's head to explode.
or
"Slightly hard" GCSE Maths Question causes outrage...
The actual question was:
Hannah has 6 orange sweets and some yellow sweets. Overall, she has n sweets. The probability of her taking 2 orange sweets is 1/3.
Prove that: n^2-n-90=0
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I should add, that is a paraphrased version (as are all of them as they've been typed from memory).
For the avoidence of doubt, Hannah takes a sweet and it is orange. She eats it and takes another, which is also orange.
The probability of her managing to do this is stated as being 1/3.
For n = -9 or n = 10, easy enough to do in one's head.
GCSE's getting harder causes Paul Dacre's head to explode.
There are six orange sweets and n sweets overall. If she takes one, there is a 6/n chance of getting and orange sweet. When she takes one, there is one less orange sweet and one less sweet overall.
If she took another orange sweet, the probability would be (6-1)/(n-1)=5/n-1. Now, you have to find the probability if she gets two orange sweets so you simply times the two fractions: 6/n * 5/n-1 = 30/n^2-n.
It tells us the probability of two orange sweets is 1/3 which means
1/3=30/n^2-n.We need to make the denominators the same so simply times 1/3 by 30/30 which would equal 30/90. if 30/90 = 30/n^2-n, then n^2-n=90. if n^2-n=90 then n^2-n-90 will equal zero.
Mike (with a little help from Google)
Which being in the real 2015 world is exactly what I would do if this were a real problem, so who need to pass the exam?
And this was my effort (complete with shitty handwriting as it was on a phone screen *cough*):
I had a look back.
For comparison, here is avery uk.d-i-y oritent Uni London Board O-Level question from 1988:
And here's a 1957 Cambridge paper, also O-Level:
makes depressing reading...
Paste error:
which is odd, because as stated it is clearly 100%.. ;-)
'It is understood that Edexcel has only received one formal complaint about the paper so far, and this came form a teacher.'
Proofread by prior passes?
Jim K
It does not seem a very useful form of the problem. OK I know its trying to get people to not think in a regimented way and find the missing bits of the puzzle from what is given, but my main annoyance at most exams, which I always did badly at, was the meaningless unuseful problems they put in them. Real world problems should be used in my view, as that is the situation these people are going to use their talents in, not some ambiguously worded meaningless problem. Brian
Ah but which has the most calories and can she see the colours? Also if she can, she could be colour blind in which case.. Brian
But hang on, what might happen is the first time yes she gets two oraange ones if she cannot see the colour, howeve she might get 1 of each, so the next bit is then changed. its not a real world problem, this is what always made me annoyed. Brian
It is a pity my 1988 past question is in image format and not text - you would have liked it - it was calculating concrete mixes!
I still cannot do Cheryl's Birthday.
But who buys sand or cement in cm3 quantities?
If you assume there is no change in volume then you can also assume they aren't making any concrete so the answer to a will be anything you like unless they are only counting the volume of clippings.
It's quadratic - and it factorises.
n^2-n-90=0
(n+9)(n-10) = 0
so n = 10
1st selection: = 6/10 probability of selecting an orange sweet (there are 10 sweets overall, and 6 of them are orange)2nd selection: = 5/9 probability of selecting an orange sweet (there are now 9 sweets remaining and 5 of them are orange.
overall probability = 6/10 x 5/9 = 30/90 --> 1/3
That's actually quite a cunning stinker. What makes it such a stinker is the reliance on inferring the true nature of the probability statement which needs very close scrutiny to determine that what is actually meant is defining the probability of blindly extracting the first *two* sweets from the bag. It matters not as to whether both sweets are collected in a single grab or one by one, the resulting probability remains the same.
The probability of randomly picking orange sweets at each successive trial drops each time until there are no more orange sweets left where it becomes zero in the extremely improbable event that all picks succeeding in selecting an orange sweet until no more were left to pick.
My initial guess for n was 18 (6 orange sweets outnumbered 2 to 1 by 12 yellow sweets to give the initial one in three chance I'd naively assumed). It was only the fact that 18^2 equals a whopping 324 that made me pause for thought and take another 'guess'. I tried n=12 before realising it had to 10. Only then was I able to determine that the actual probabilities of the first two picks being an orange sweet would be 60% (6 in 10 chance) for the first then 55.55555555% (5 in 9 chance) for the second giving an overall probability of 60% of 55.55555% = to 33.333333% or one third.
I'd say that 'guessing' likely values for n in this case (integer values only) would be considered an entirely legitimate approach to take in order to determine the value of n which can be the only value that also allows the stated one in three probability to be true.
I'm sure there must be a more strict formula involving the use of square roots but, since the algorithm for calculating a square root also involves the use of 'guessed' values in an iterative process to swiftly calculate a value of sufficient accuracy, I think the guessing method is still entirely legitimate nonetheless. :-)
A quick OCR:
A coarse mix of concrete is made by using 2 parts of water with 3 parts of sand, 4 parts of drippings and 2 parts of cement by volume. Assuming that there is no loss in volume on mixing, calculate (a) the number of spadefulls of chippings needed to make 22 spadefulls of concrete. {b) the amount of concrete, in cm^3 to 2 significant figures, which would be obtained by using 2850 cm^3 of clippings. A rectangular base is to be made for a garage using this coarse mix of concrete. The base measures 6 m by 3.5 m and it is to be 15 cm thick. Calculate (c) the volume of the base in cm^3, id) the minimum quantity of sand, in cm^3 to 2 significant figures, which will be needed to make this base. The cost of a bag of cement is £2.90 and it contains 40 000 cm^3; the cost of a sack of sand is ?1.50 and it contains 80 000 cm^3; the cost of a barrowfull of chippings is £1.25 and it contains 150000 cm^3. Assuming that cement, sand and chippings are only for sale in multiples of these quantities, (e) find the total cost of buying sufficient materials from which to make the base.
(12 marks)
Yup, just call it "numerical analysis" or "stepwise refinement" ;-)
It is, in the sense that the expression quoted is quadratic, and iterative methods like Newton-Raphson are valid techniques (made all the more attractive by the advent of brute force computing power).
Having said that, the normal quadratic formula would work as well
0 = 1 +/- sqrt( 1 - (4 x -90) ) / 2which gives roots of -9 and 10.
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