2008 Maths Competition (1 Viewer)

[rambam92]

Re: Maths Competition

lol last year i got a distinction but this year i screwd it big time

 

[shaon0]

Re: Maths Competition

you didnt explain anything after saying to complete the square...

No one let me explain...someone came in the middle of it and started giving me another question. Its not my fault..i still got the method right.

 

[Finx]

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?

 

[Aerath]

Re: Maths Competition

Nope, didn't even understand that question.

 

[-tal-]

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?

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.

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]

Re: Maths Competition

were any of the last 5 = 328 by any chance?

 

[ahhliss]

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?

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:

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]

Re: Maths Competition

Also, were any of the questions 21-25 mulitple choice option (B)?

 

[lyounamu]

Re: Maths Competition

Also, were any of the questions 21-25 mulitple choice option (B)?

5C1 . (3/4)^4 . (1/4) = approx. 40% chance of B in those 5 questions (once). You better narrow it down.

 

[ahhliss]

Re: Maths Competition

5C1 . (3/4)^4 . (1/4) = approx. 40% chance of B in those 5 questions (once). You better narrow it down.

Is that combinations? Lol haven't learnt it yet but we did Binomial Theorem and the teacher said it combinations also use nCr

 

[lolokay]

Re: Maths Competition

doesn't the C in nCr mean combinations? I think it would make more sense to do combinations before binomial theorem

 

[lyounamu]

Re: Maths Competition

doesn't the C in nCr mean combinations? I think it would make more sense to do combinations before binomial theorem

Yeah. That's what I meant.

I learnt combination though.

 

[ahhliss]

Re: Maths Competition

doesn't the C in nCr mean combinations? I think it would make more sense to do combinations before binomial theorem

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]

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 :'(

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 though

 

[lyounamu]

Re: Maths Competition

Knowledge on combinations wouldn't help much in the binomial theorem chapter. But it would certainly benefit in some ways.

 

[ahhliss]

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 though

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.

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

 

[lolokay]

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

 

[ahhliss]

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

Oh yeah, I forgot nCr = n!/r!(n-r)! xD I haven't done my homework for greatest coefficient yet :S

 

[foram]

Re: Maths Competition

Oh yeah, I forgot nCr = n!/r!(n-r)! xD I haven't done my homework for greatest coefficient yet :S

I thought binomial was yr 12.

 

[lolokay]

Re: Maths Competition

I thought binomial was yr 12.

notice how in her sig it says "accelerated"?