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Westpac Mathematics Competition (1 Viewer)

mica

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Re: Westpac Australian Maths Comp Senior Question !!

Ok modezero, Care to share how ?
If its to complicated to do on computer, could you possibly scan it up?
 
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Re: Maths comp [westpac]

omg.. such a waste of 2 periods.. and a damn hard comp at that.. my theory is high distinction with 3 outta 30:) but yeah.. frkn missed chem and my free for it and tomorrow i miss my double chem for the damn chem comp.
4 great periods down the drain...:mad1:
 

Mark576

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^^argh!! i had the right calculation but i ended up with 872 ...godammit!
 

drynxz

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Re: Maths comp [westpac]

First ten questions was not that bad...but when i got towards the end i had no bloody idea how to do it, so i guessed like 10 questions :p
 

ugly14

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Re: Westpac Australian Maths Comp Senior Question !!

i got 46 as my answer!
dont know how though :)
 

priesty

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Re: Westpac Australian Maths Comp Senior Question !!

The maths teacher who was supervising us during it was doing the test at the same time as us calculated the answer to be 110.
 

mica

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Re: Westpac Australian Maths Comp Senior Question !!

i got 44 i think. Appears we all got different answers lol.
Keep them comming, hopefully someone else will say they got 44 :)

Still looking for the solution
 

Ennaybur

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Re: Westpac Australian Maths Comp Senior Question !!

that was a bitch of a test if ever i did see one. HORRIBLE HORRIBLE. i do extension aswell but it doesnt help coz there's no formula work :(
 

brownsound

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if u find (which i did sum wierd way)
∑α
∑αβ
∑αβγ

u could do this
x^3 - (∑α)x^2 + (∑αβ)x - (∑αβγ) = 0

then:
∑α = 4
∑αβ = 3
∑αβγ = -2

so
x^3 - 4x^2 + 3x - (-2) = 0

and since alpha satisfies the eqn

α^3 - 4α^2 + 3α +2 = 0

now times that thru alpha to get:

α^4 - 4α^3 + 3α^2 +2α = 0

α^4 = 4α^3 - 3α^2 - 2α

now do that with β and γ, then add the eqns and u get:

∑(α^4) = 4[∑(α^3)] - 3[∑(α^2)] - 2α

substitute your numbers from the answer

∑(α^4) = 4[22] - 3[10] - 2[4]

∑(α^4) = 88 - 30 - 8

∑(α^4) = 50

probly not the best way to do it in an exam.. i actually missed out cuz i had a trial for economics.. which was just as sh*t.
 

SoulSearcher

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Re: Maths comp [westpac]

If I had 10 more minutes, I would have figured out Q. 27, and maybe added 5 marks out of 50 for that last section :p It didn't help that I started nodding off halfway during the test becuase I stayed up until 2am the night before :(
 

mica

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modezero and brownsound well done. It appears you have the understanding lol.
Thanks for the solutions and all the best
 

ice ken

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math comp suked that stupid question with 3 numbers sum of 4 then squared =10 then cubed and 22 wat if they were to power 4 wtf?????????
 

short!ez

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heyyy !! =]

i think westpac owned me but.. this is what i got for question whatever..

x + y + z = 4
x^2 + y^2 + z^2 = 10
x^3 + y^3 + z^3 = 22
x^4 + y^4 + z^4 = ??

if you look at the results.. difference between 4 and 10 is 6.. difference between 10 and 22 is 12.. so.. i just doubled it again and put down 22 + 24.. wrote down 46. hope it's right lol.

i liked the last few questions. i actually got one or two right for once. lol
 

Riviet

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short!ez said:
if you look at the results.. difference between 4 and 10 is 6.. difference between 10 and 22 is 12.. so.. i just doubled it again and put down 22 + 24.. wrote down 46. hope it's right lol.
Yeah, that was something I considered when I first saw that question posted in this thread. According to a previous post in this thread, 46 is not the correct answer. :)
 

Forbidden.

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OMG I did the maths competition, when you have an answer its not there !
i am NOT doing it again, please tell me a good reason why i should do it again ... it makes the hsc look easy ..
 

Sober

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short!ez: Shamefully I too resorted to that reasoning when I ran out of time to approach it correctly. Plugging the values into mathematica confirms the aformentioned solution of 50.

For question 29 I managed to get (with trial and error) 3x3x3x3x3x4 = 976 but did anybody else find a higher value?
 

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