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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 ==== |
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. |
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? |
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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 |
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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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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 |
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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. |
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'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 |
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 ==== |
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. |
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 |
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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! |
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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. |
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"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 |
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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 |
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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 |
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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 |
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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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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 | \================================================= ================/ |
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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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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 |
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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 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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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 |
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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 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 |
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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"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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"Tim Streater" wrote in message .. . In article , 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 At the time I wrote that, I hadn't seen/heard the full question. All I heard was the equation part as reported on the Today prog. Neither had/have I. but there's enough information in what you posted to tell you that working out in you head that: n=10, will not get you full marks (or even close) (unless maths exams have been really dummed down in the past 40 years since I took mine) tim |
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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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"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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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. |
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 | \================================================= ================/ |
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 |
The end of the world cometh - 4 boney dudes on horseback spotted by Daily Mail readers...
"John Rumm" wrote in message o.uk... 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 You are misunderstanding the way that "proof" in mathematics works. You have been asked for a proof, so in order to get the marks you first have to find something that you can "prove". Simply using the equation to solve n=10 isn't it, because that is not a proof. And as the equation resolved down to n=10 the thing that you have, that can be proved, is that n=10 is, in fact, the solution to the narrative part of the question. Thus there becomes a requirement to prove that n=10 solves the sweet problem, because that's all you have that you can use as the result of a mathematical proof. (and finding that n=10 would be finding a solution, not providing a proof) Correct, so it's can't be the solution to the question that gets you the marks tim |
The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 07/06/2015 11:12, tim..... wrote:
"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 That's ok, I appreciate your reply. It's a confusing issue, and I think the problem could have been put in a less confusing way. At first I simply found n - which was not the question, as you quite rightly pointed out. |
The end of the world cometh - 4 boney dudes on horseback spottedby Daily Mail readers...
On 07/06/2015 11:20, tim..... wrote:
"John Rumm" wrote in message o.uk... 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 You are misunderstanding the way that "proof" in mathematics works. Well actually that was the accusation I was levelling at you[1] ;-) Although I suspect we are both singing from the same hymn sheet - but something has been lost in translation. [1] Due to your response suggesting that one needs to prove n = 10 You have been asked for a proof, so in order to get the marks you first have to find something that you can "prove". Indeed. Farmer Giles correctly provided a proof IMHO - i.e. derived the required equation algebraically from the source equations extracted from the problem description. Simply using the equation to solve n=10 isn't it, because that is not a proof. Agreed. And as the equation resolved down to n=10 the thing that you have, that can be proved, is that n=10 is, in fact, the solution to the narrative part of the question. Not really - you can make the proof without ever finding n (which can also be -9, a valid solution to the equation, although makes no sense in the context of the problem) Thus there becomes a requirement to prove that n=10 solves the sweet problem, because that's all you have that you can use as the result of a mathematical proof. I am not sure about that - you can't prove that n = 10, since its not the only valid answer. The best you can say about 10 is that it is one possible solution for n. Finding any number of solutions is not usually an adequate mathematical proof (unless proving a negative of course! I suspect that Andrew Wiles may be a tad upset if you can find integer values for a, b, & c where a^3 = b^3 + c^3 ) (and finding that n=10 would be finding a solution, not providing a proof) Correct, so it's can't be the solution to the question that gets you the marks Which is what I was saying! -- Cheers, John. /================================================== ===============\ | Internode Ltd - http://www.internode.co.uk | |-----------------------------------------------------------------| | John Rumm - john(at)internode(dot)co(dot)uk | \================================================= ================/ |
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