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2008 Maths Competition (1 Viewer)
- Thread starter Aerath
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Finx
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Re: Maths Competition
My god, EVERYONE got screwed over from behind in this test. Even the 4-unit accelerants in my year said it was ridiculous.
I'm proud I got the Triangle:Semicirlce one right - I subbed the radius for 1, thus length of one side = 2. I can't remember what I put down. I'd work it out, but it's like 2AM here -.-
I got raped by Q10 - submerging the prism into the tank of water. For some reason I kept thinking of the laws of displacement from physics, and I was all like "Wtf?!" - nothing worked out for me in that part =[
I thought I had the last question, but after reading some of your responses, I don't think I nailed it >_>.
Also, the X{1,2,3,4,5,6} w/e was freaking stupid. Did anyone understand that at all?
My god, EVERYONE got screwed over from behind in this test. Even the 4-unit accelerants in my year said it was ridiculous.
I'm proud I got the Triangle:Semicirlce one right - I subbed the radius for 1, thus length of one side = 2. I can't remember what I put down. I'd work it out, but it's like 2AM here -.-
I got raped by Q10 - submerging the prism into the tank of water. For some reason I kept thinking of the laws of displacement from physics, and I was all like "Wtf?!" - nothing worked out for me in that part =[
I thought I had the last question, but after reading some of your responses, I don't think I nailed it >_>.
Also, the X{1,2,3,4,5,6} w/e was freaking stupid. Did anyone understand that at all?
Re: Maths Competition
Maximum number of numbers you can have in a group is 3, since more than that requires consecutive numbers.
So singularly, there's 6. Then you add in groups of 2, which is (4 + 3 + 2). Then in groups of 3, which is 4.
Answer: 6 + 9 + 4 = 19
I'm pretty sure that's how you do it. But I'm not too sure coz the way it was asked made go
Just saw that question @ tutor today. I think the question asked how many groups of numbers you can make without having consecutive numbers next together.Finx said:
Maximum number of numbers you can have in a group is 3, since more than that requires consecutive numbers.
So singularly, there's 6. Then you add in groups of 2, which is (4 + 3 + 2). Then in groups of 3, which is 4.
Answer: 6 + 9 + 4 = 19
I'm pretty sure that's how you do it. But I'm not too sure coz the way it was asked made go
bored of sc
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Re: Maths Competition
were any of the last 5 = 328 by any chance?
were any of the last 5 = 328 by any chance?
Re: Maths Competition
Yeah, Q10 is a serial rapist D: I did some strange things and picked (b) or something :S
That X thing was weird. Q20 or something. Normally, the first 20 shouldn't be too bad but this year's really sucked for me D:
Ah I hope I got the semicircle one correct, but knowing me, I probably made a stupid mistake somewhere. For example, the question about sharing $3 between two people with one of them having 50c more than the other. I kept reading it as $5 and couldn't get the answer because instead of $1.25 and $1.75 I had $2.25 and $2.75. I only got it the second time doing it D:Finx said:
Yeah, Q10 is a serial rapist D: I did some strange things and picked (b) or something :S
That X thing was weird. Q20 or something. Normally, the first 20 shouldn't be too bad but this year's really sucked for me D:
bored of sc
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Re: Maths Competition
Also, were any of the questions 21-25 mulitple choice option (B)?
Also, were any of the questions 21-25 mulitple choice option (B)?
Re: Maths Competition
But Binomial Theorem didn't need any preexisting knowledge of combinations. It wasn't even mentioned until she said nCr is used in combinations. We're doing combinations after the trials and my teacher hates it :'(lolokay said:
Re: Maths Competition
but calculating nCr means that you're working out the number of combinations of r elements from set n (did I get the terms right?), so it might be a bit more clear if you did the combinations topic first. I'm sure it wouldn't make that much of a difference which order you learn them thoughahhliss said:
Re: Maths Competition
Elements from set n? That sounds like the X{1, 2, 3, 4, 5, 6} or whatever question in the Maths Comp. They were talking about elements, sets and subsets. I was just like O_O
I don't understand half of what you said lol. Are you saying that by learning combinations first, we learn how nCr works? Because I just pressed nCr into the calculator for Binomial Theorem but never considered how it worked.lolokay said:
Elements from set n? That sounds like the X{1, 2, 3, 4, 5, 6} or whatever question in the Maths Comp. They were talking about elements, sets and subsets. I was just like O_O
Re: Maths Competition
lol don't worry about the elements/sets bit
I thought you did learn what nCr was in binomial theorem, because I see questions on here where you have to calculate the greatest coefficient of a particular expansions which requires you write nCr in terms of its mathematical operation
nCr is calculate by n!/r!(n-r)!. in combinations this formula could be derived in the following way:
say you have a set of n things. now to order these (permutation) there can be n in the first place, n-1... and 1 in the last place -> giving n! ways.
but if you only want to consider the first r places, you 'unorder' the remaining (n-r) places by dividing by (n-r)!
but you don't care what order the first r places are in either, so you 'unorder' them by dividing by r!
giving nCr = n!/r!(n-r)!
(you probably didn't understand that at all since I'm not too great at explaining things)
why can this be used in binomial theorem? Take the example (x+1)^n = (x+1)(x+1)...(x+1) [n times]. You can see that the coefficient of x^r is the number of ways of choosing r x's from the n pairs = nCr
lol don't worry about the elements/sets bit
I thought you did learn what nCr was in binomial theorem, because I see questions on here where you have to calculate the greatest coefficient of a particular expansions which requires you write nCr in terms of its mathematical operation
nCr is calculate by n!/r!(n-r)!. in combinations this formula could be derived in the following way:
say you have a set of n things. now to order these (permutation) there can be n in the first place, n-1... and 1 in the last place -> giving n! ways.
but if you only want to consider the first r places, you 'unorder' the remaining (n-r) places by dividing by (n-r)!
but you don't care what order the first r places are in either, so you 'unorder' them by dividing by r!
giving nCr = n!/r!(n-r)!
(you probably didn't understand that at all since I'm not too great at explaining things)
why can this be used in binomial theorem? Take the example (x+1)^n = (x+1)(x+1)...(x+1) [n times]. You can see that the coefficient of x^r is the number of ways of choosing r x's from the n pairs = nCr
foram
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Re: Maths Competition
I thought binomial was yr 12.ahhliss said: