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UK diy (uk.d-i-y) For the discussion of all topics related to diy (do-it-yourself) in the UK. All levels of experience and proficency are welcome to join in to ask questions or offer solutions. |
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#1
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
GCSE's getting harder causes Paul Dacre's head to explode.
or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 ==== |
#2
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/15 15:01, Tim Watts wrote:
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 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. |
#3
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/15 15:04, Tim Watts wrote:
On 05/06/15 15:01, Tim Watts wrote: 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 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. which is odd, because as stated it is clearly 100%.. ;-) -- New Socialism consists essentially in being seen to have your heart in the right place whilst your head is in the clouds and your hand is in someone else's pocket. |
#4
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
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 -- From the Sofa of Brian Gaff Reply address is active "Tim Watts" wrote in message ... On 05/06/15 15:01, Tim Watts wrote: 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 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. |
#5
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
GCSE's getting harder causes Paul Dacre's head to explode.
or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 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? |
#6
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/15 15:23, Muddymike wrote:
GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 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 ****ty handwriting as it was on a phone screen *cough*): http://tinypic.com/m/ip86qs/3 I had a look back. For comparison, here is avery uk.d-i-y oritent Uni London Board O-Level question from 1988: http://tinypic.com/m/ip8900/3 And here's a 1957 Cambridge paper, also O-Level: http://tinypic.com/m/ip8900/3 |
#7
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/15 15:32, Tim Watts wrote:
On 05/06/15 15:23, Muddymike wrote: GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 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 ****ty handwriting as it was on a phone screen *cough*): http://tinypic.com/m/ip86qs/3 I had a look back. For comparison, here is avery uk.d-i-y oritent Uni London Board O-Level question from 1988: http://tinypic.com/m/ip8900/3 And here's a 1957 Cambridge paper, also O-Level: Paste error: http://www.cambridgeassessment.org.u...tion-paper.pdf |
#8
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
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 -- From the Sofa of Brian Gaff Reply address is active "Tim Watts" wrote in message ... On 05/06/15 15:32, Tim Watts wrote: On 05/06/15 15:23, Muddymike wrote: GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 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 ****ty handwriting as it was on a phone screen *cough*): http://tinypic.com/m/ip86qs/3 I had a look back. For comparison, here is avery uk.d-i-y oritent Uni London Board O-Level question from 1988: http://tinypic.com/m/ip8900/3 And here's a 1957 Cambridge paper, also O-Level: Paste error: http://www.cambridgeassessment.org.u...tion-paper.pdf |
#9
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
In article , Brian-Gaff
scribeth thus 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 If it were my younger daughter she'd have simply scoffed the lot and that would have been that.. Then there would have been a cat fight when the older found out she'd nicked them and she'd have her fingers down younger's throat getting her to regurgitate them;!.. Girlies, sooo very sweet;!... -- Tony Sayer |
#10
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/2015 15:32, Tim Watts wrote:
I had a look back. For comparison, here is avery uk.d-i-y oritent Uni London Board O-Level question from 1988: http://tinypic.com/m/ip8900/3 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. And here's a 1957 Cambridge paper, also O-Level: http://tinypic.com/m/ip8900/3 |
#11
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The end of the world cometh - 4 boney dudes on horseback spotted byDaily Mail readers...
'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 |
#12
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
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 -- From the Sofa of Brian Gaff Reply address is active "Tim Watts" wrote in message ... GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 ==== |
#13
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/15 16:15, Brian-Gaff wrote:
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 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! |
#14
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/2015 16:36, Tim Watts wrote:
On 05/06/15 16:15, Brian-Gaff wrote: 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 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! But who buys sand or cement in cm3 quantities? -- mailto: news {at} admac {dot] myzen {dot} co {dot} uk |
#15
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On Friday, 5 June 2015 20:43:12 UTC+1, alan_m wrote:
On 05/06/2015 16:36, Tim Watts wrote: On 05/06/15 16:15, Brian-Gaff wrote: 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 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! But who buys sand or cement in cm3 quantities? if you go to wilko NT |
#16
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/2015 16:36, Tim Watts wrote:
On 05/06/15 16:15, Brian-Gaff wrote: 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 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! 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) -- Cheers, John. /================================================== ===============\ | Internode Ltd - http://www.internode.co.uk | |-----------------------------------------------------------------| | John Rumm - john(at)internode(dot)co(dot)uk | \================================================= ================/ |
#17
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/2015 23:30, John Rumm wrote:
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 Drippings? WTF?? (a) the number of spadefulls of chippings needed to make 22 spadefulls of concrete. Oh, chippings. You mean gravel? {b) the amount of concrete, in cm^3 to 2 significant figures, which would be obtained by using 2850 cm^3 of clippings. .... or do you? I worked out n=10 on the way down to the shower. And I'm not very awake at that time of day... Andy |
#18
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 09/06/2015 21:04, Vir Campestris wrote:
On 05/06/2015 23:30, John Rumm wrote: No, I did not write - I scanned and OCRed the graphic copy of the paper for those who wanted a text version... 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 Drippings? WTF?? OCR remember? (a) the number of spadefulls of chippings needed to make 22 spadefulls of concrete. Oh, chippings. You mean gravel? Its the way I would read it. However it does not matter since the questioner is defining the parameters. They could say that that the ingredients for concrete are milk, flour, chocolate chips, and evo stick - it would not change the fundamentals of the aptitudes they are testing. {b) the amount of concrete, in cm^3 to 2 significant figures, which would be obtained by using 2850 cm^3 of clippings. ... or do you? I worked out n=10 on the way down to the shower. And I'm not very awake at that time of day... Which, as several have pointed out, was not an answer to the question posed (nor required for its complete answer) -- Cheers, John. /================================================== ===============\ | Internode Ltd - http://www.internode.co.uk | |-----------------------------------------------------------------| | John Rumm - john(at)internode(dot)co(dot)uk | \================================================= ================/ |
#19
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/15 16:39, Tim Streater wrote:
In article , Brian-Gaff wrote: 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. Wrong. The essence of dealing with the real world is that many problems at the outset make no sense. You have to apply thought to determine what the problem is first, before you can solve it. All this rubbish about calling her Hannah and making it a concrete problem "from the real world" removes half the point of education in the first place. Agreed. Wrapping 'real world' stories around these problems confuses people who tend to see the story and not the problem. The essential skill that needs to be learnt is to see through the sugary wrapping and abstract the essence of the problem. |
#20
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"Tim Watts" wrote in message
... GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 I still cannot do Cheryl's Birthday. http://en.wikipedia.org/wiki/Cheryl%27s_Birthday -- Adam |
#21
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/2015 15:01, Tim Watts wrote:
GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 ==== 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 |
#22
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The end of the world cometh - 4 boney dudes on horsebackspotted by Daily Mail readers...
On Fri, 05 Jun 2015 15:01:13 +0100, Tim Watts wrote:
GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 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. :-) -- Johnny B Good |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/2015 22:59, Johnny B Good wrote:
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. Yup, just call it "numerical analysis" or "stepwise refinement" ;-) 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. :-) 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) ) / 2 which gives roots of -9 and 10. -- Cheers, John. /================================================== ===============\ | Internode Ltd - http://www.internode.co.uk | |-----------------------------------------------------------------| | John Rumm - john(at)internode(dot)co(dot)uk | \================================================= ================/ |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On Friday, 5 June 2015 15:01:19 UTC+1, Tim Watts wrote:
GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... http://www.bbc.co.uk/news/education-33017299 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 Its an easy question, I'm bemused by the complaint. NT |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"The Natural Philosopher" wrote in message
... On 06/06/15 00:02, wrote: Its an easy question, I'm bemused by the complaint. The complaint was that in addition to actually regurgitating learnt by rote formulae for solving quadratics, it forced people to think. Since the modern education system discourages thinking, and doesn't teach it, the pupils felt cheated. Which of course, they have been. Armed only with a tape measure and a pencil could you work out and mark the centre of a 600x600 ceiling tile? -- Adam |
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
In message , ARW
writes "The Natural Philosopher" wrote in message ... On 06/06/15 00:02, wrote: Its an easy question, I'm bemused by the complaint. The complaint was that in addition to actually regurgitating learnt by rote formulae for solving quadratics, it forced people to think. Since the modern education system discourages thinking, and doesn't teach it, the pupils felt cheated. Which of course, they have been. Armed only with a tape measure and a pencil could you work out and mark the centre of a 600x600 ceiling tile? Yebbut, marking the centre from the diagonals only needs a pencil and a straight edge! -- Tim Lamb |
#28
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 06/06/15 08:29, ARW wrote:
"The Natural Philosopher" wrote in message ... On 06/06/15 00:02, wrote: Its an easy question, I'm bemused by the complaint. The complaint was that in addition to actually regurgitating learnt by rote formulae for solving quadratics, it forced people to think. Since the modern education system discourages thinking, and doesn't teach it, the pupils felt cheated. Which of course, they have been. Armed only with a tape measure and a pencil could you work out and mark the centre of a 600x600 ceiling tile? Tape measure? A straight edge... |
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"ARW" wrote in message ... "The Natural Philosopher" wrote in message ... On 06/06/15 00:02, wrote: Its an easy question, I'm bemused by the complaint. The complaint was that in addition to actually regurgitating learnt by rote formulae for solving quadratics, it forced people to think. Since the modern education system discourages thinking, and doesn't teach it, the pupils felt cheated. Which of course, they have been. Armed only with a tape measure and a pencil could you work out and mark the centre of a 600x600 ceiling tile? Only if you present me with the appropriate risk assessment for my standing on the chair and working with my hands above my head tim |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 06/06/15 10:26, Tim Streater wrote:
In article , Tim Watts wrote: I have taken ones from people who's profiles or other tweets suggest whose. Never sed I cud do inglish |
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"Tim Watts" wrote in message
... GCSE's getting harder causes Paul Dacre's head to explode. or "Slightly hard" GCSE Maths Question causes outrage... I wonder if there is a GCSE Religious Education question that asks you to name the 4 boney dudes on horseback (ISTR that only two of them are boney). -- Adam |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 05/06/15 15:19, Tim Streater wrote:
In article , Tim Watts wrote: Prove that: n^2-n-90=0 For n = -9 or n = 10, easy enough to do in one's head. https://twitter.com/hashtag/edexcelmaths makes depressing reading... |
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"Tim Streater" wrote in message .. . In article , Tim Watts wrote: Prove that: n^2-n-90=0 For n = -9 or n = 10, easy enough to do in one's head. That's a solution not a proof tim |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 06/06/2015 11:15, tim..... wrote:
"Tim Streater" wrote in message .. . In article , Tim Watts wrote: Prove that: n^2-n-90=0 For n = -9 or n = 10, easy enough to do in one's head. That's a solution not a proof tim Correct, it is not a proof - neither was my answer earlier. This is, though: 6/n x 5/n-1 = 1/3 -- 30/n^2-n = 1/3 -- 90 = n^2-n -- 0 = n^2-n-90 |
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"Farmer Giles" wrote in message ... On 06/06/2015 11:15, tim..... wrote: "Tim Streater" wrote in message .. . In article , Tim Watts wrote: Prove that: n^2-n-90=0 For n = -9 or n = 10, easy enough to do in one's head. That's a solution not a proof tim Correct, it is not a proof - neither was my answer earlier. This is, though: 6/n x 5/n-1 = 1/3 -- 30/n^2-n = 1/3 -- 90 = n^2-n -- 0 = n^2-n-90 but it's still not a proof in the context of a maths question, which is prove that n = 10 (from the information in the question, not from the equation) tim |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 06/06/2015 19:07, tim..... wrote:
"Farmer Giles" wrote in message ... On 06/06/2015 11:15, tim..... wrote: "Tim Streater" wrote in message .. . In article , Tim Watts wrote: Prove that: n^2-n-90=0 For n = -9 or n = 10, easy enough to do in one's head. That's a solution not a proof tim Correct, it is not a proof - neither was my answer earlier. This is, though: 6/n x 5/n-1 = 1/3 -- 30/n^2-n = 1/3 -- 90 = n^2-n -- 0 = n^2-n-90 but it's still not a proof in the context of a maths question, which is prove that n = 10 (from the information in the question, not from the equation) It didn't ask me to solve for n - but I have done. I formed the equation from the information in the question - thereby proving it. And, in my previous reply, I derived that n=10 from the equation. |
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The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"Farmer Giles" wrote in message ... On 06/06/2015 19:07, tim..... wrote: "Farmer Giles" wrote in message ... On 06/06/2015 11:15, tim..... wrote: "Tim Streater" wrote in message .. . In article , Tim Watts wrote: Prove that: n^2-n-90=0 For n = -9 or n = 10, easy enough to do in one's head. That's a solution not a proof tim Correct, it is not a proof - neither was my answer earlier. This is, though: 6/n x 5/n-1 = 1/3 -- 30/n^2-n = 1/3 -- 90 = n^2-n -- 0 = n^2-n-90 but it's still not a proof in the context of a maths question, which is prove that n = 10 (from the information in the question, not from the equation) It didn't ask me to solve for n - but I have done. I formed the equation from the information in the question - thereby proving it. And, in my previous reply, I derived that n=10 from the equation. you're right sorry |
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The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 06/06/2015 19:07, tim..... wrote:
"Farmer Giles" wrote in message ... On 06/06/2015 11:15, tim..... wrote: "Tim Streater" wrote in message .. . In article , Tim Watts wrote: Prove that: n^2-n-90=0 For n = -9 or n = 10, easy enough to do in one's head. That's a solution not a proof tim Correct, it is not a proof - neither was my answer earlier. This is, though: 6/n x 5/n-1 = 1/3 -- 30/n^2-n = 1/3 -- 90 = n^2-n -- 0 = n^2-n-90 but it's still not a proof in the context of a maths question, which is prove that n = 10 (from the information in the question, not from the equation) There is no requirement to "prove" n=10 (and finding that n=10 would be finding a solution, not providing a proof) -- Cheers, John. /================================================== ===============\ | Internode Ltd - http://www.internode.co.uk | |-----------------------------------------------------------------| | John Rumm - john(at)internode(dot)co(dot)uk | \================================================= ================/ |
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