Author Topic: 4U Maths Question Thread (Read 665377 times) Tweet Share

lsong · « **Reply #1515 on:** October 22, 2017, 06:35:01 pm »

Hi,
For part (ii), are we allowed to use 'Similarly to argument in part (i)'? Or would we have to prove it separately? The sample answers allow the similarly argument but not sure if we should follow this.
Thanks.

RuiAce · « **Reply #1516 on:** October 22, 2017, 06:38:55 pm »

Quote from: frog1944 on October 22, 2017, 06:22:20 pm

Hi RuiAce, for the first round, what is the probability the players will meet? I understand that the number of ways to pick the 8 players is 105, but what is the number of ways that they will meet? I have the book, it says 1/7 but I'm not sure how to arrive at that using your method (I originally was attempting to do it this way)

Thanks

EDIT I'M SORRY ALL OF THIS IS WRONG

$\text{Perhaps you could try splitting the cases like this instead:}$
$\text{Case 1: A and C meet in the first round}\\ \text{We just gotta make an assortment for the others, which can be done in}\\ \binom62\binom42\binom22\frac{1}{3!} = 15\text{ ways}$
$\text{Case 2: A and C meet in the second round}\\ \text{A and C must be put with two different persons: }\binom62\text{ arrangements}\\ \text{but then the rest can be the same: }\binom42\binom22 \frac{1}{2!}\\ \text{So this case can happen in }\binom62\binom42\binom22\frac1{2!}=45\text{ ways.}$
Note that the identical groups will only occur for the other 4 people. The pairs that A and C must be with are immune from this because A and C must be in different groups, and that effectively forces all possible groups to be distinct.
$\text{Case 3: A and C meet in the third round}\\ \text{This will actually be the same as case 2 and also involve 45 people.}$
The only difference is that we don't have (A,B)(C,D) || (E,F)(G,H) but rather (A,B)(D,E) || (C,F)(G,H). This doesn't actually change the number of arrangements though.

Sanity check: If we regroup the cases, we get 15+45+45 = 105 as we expected.

That method doesn't get to the correct final answer so I need to check something

Edit #2: I've found out what the theoretical values are by working backwards. I'm gonna try again and work out how we can get to them.

RuiAce · « **Reply #1517 on:** October 22, 2017, 06:39:39 pm »

Quote from: lsong on October 22, 2017, 06:35:01 pm

Hi,
For part (ii), are we allowed to use 'Similarly to argument in part (i)'? Or would we have to prove it separately? The sample answers allow the similarly argument but not sure if we should follow this.
Thanks.

Similarly is allowed so long as if you replicate the exact same STEPS albeit with different angles/sides you will reach a very similar conclusion.

RuiAce · « **Reply #1518 on:** October 22, 2017, 07:41:34 pm »

Quote from: RuiAce on October 22, 2017, 06:38:55 pm

EDIT I'M SORRY ALL OF THIS IS WRONG
$\text{Perhaps you could try splitting the cases like this instead:}$
$\text{Case 1: A and C meet in the first round}\\ \text{We just gotta make an assortment for the others, which can be done in}\\ \binom62\binom42\binom22\frac{1}{3!} = 15\text{ ways}$
$\text{Case 2: A and C meet in the second round}\\ \text{A and C must be put with two different persons: }\binom62\text{ arrangements}\\ \text{but then the rest can be the same: }\binom42\binom22 \frac{1}{2!}\\ \text{So this case can happen in }\binom62\binom42\binom22\frac1{2!}=45\text{ ways.}$
Note that the identical groups will only occur for the other 4 people. The pairs that A and C must be with are immune from this because A and C must be in different groups, and that effectively forces all possible groups to be distinct.
$\text{Case 3: A and C meet in the third round}\\ \text{This will actually be the same as case 2 and also involve 45 people.}$
The only difference is that we don't have (A,B)(C,D) || (E,F)(G,H) but rather (A,B)(D,E) || (C,F)(G,H). This doesn't actually change the number of arrangements though.

Sanity check: If we regroup the cases, we get 15+45+45 = 105 as we expected.
That method doesn't get to the correct final answer so I need to check something

Edit #2: I've found out what the theoretical values are by working backwards. I'm gonna try again and work out how we can get to them.

Ok scrap ALMOST ALL of that - I think I've worked out a way to figure out the initial starting positions properly.
$\text{We maintain that the number of ways }A\text{ and }C\text{ can face off}\\ \text{in the very first round is }\binom62\binom42\binom22\frac{1}{3}=15!\\ \text{I am highly positive that this part at least was correct}.$
___________________________________
$\text{But everything else was wrong, so I've changed my analysis.}$
$\text{Instead, what I will do is build the initial pairings first, without worrying about}\\ \text{which round they }\textbf{actually face off in}$
$\text{To do this, I need to let }A\text{ and }C\text{ be in two distinct pairs out of the four.}\\ \text{So we have the pairs: }(A,?), (C,?), (?,?), (?,?)$
$\text{Now, we first find someone to pair up with }A: \binom61\text{ choices}\\ \text{Then, out of the five remaining, someone with }C: \binom51\text{ choices}\\ \text{then the last two run into the identical groups problem, so for them: }\binom42\binom22\frac{1}{2!}$
$\text{So the number of pairings I can have is}\\ 6\times 5\times 6\times 1 \times \frac12 = \boxed{90}$
Alternatively, I could've chosen the two extra pairs first: $ \binom62\binom42\frac{1}{2!}$ and then arranged the last two; one with A and one with C: $2\times 1$. This still gives 90.
___________________________________
$\text{At this point, what I've effectively done is calculated the number of ways}\\ \text{in which we can ensure that }A\text{ and }B\textbf{ will not}\text{ face off in round 1}$
The reason I do this is because facing off in round 1 is different: (A,C) must form a pair. For ANY OTHER ROUND, (A,C) must NOT form a pair.
$\text{Now, I consider the probability of}\\ \textbf{the earliest round they can face off in.}$
$\text{If }A\text{ and }C\text{ are to face off in round 2}\\ \text{then we must've grouped the pairs like:}\\ (A,?)\, (C,?) \| (?,?), (?,?)\\ \text{There can only be 1 way of this happening.}$
$\text{If }A\text{ and }C\text{ are to face off in round 3}\\ \text{then we must've grouped the pairs in one of the following ways:}\\ (A,?)\, (?,?) \| (C,?), (?,?)\text{ or }(A,?)\, (C,?) \| (?,?), (C,?)\\ \text{There are 2 ways of this happening}$
$\text{So the probability }A\text{ and }C\text{ get the opportunity to face off in round 2 is }\frac13\\ \text{and thus the number of ways this can happen is }90\times \frac13 = \boxed{30}$
$\text{And the probability }A\text{ and }C\text{ get the opportunity to face off round 3 is }\frac23\\ \text{and thus the number of ways this can happen is }90\times \frac23 = \boxed{60}$
$\textit{From here, we can now successfully incorporate the winning probabilities}\\ \textit{and work our way towards the final answer}\\ \textit{i.e. the probability they ACTUALLY face off}$

aryak · « **Reply #1519 on:** October 22, 2017, 09:36:32 pm »

Hey for 2015, question 12 (d) I am very confused as to whether to use x or (3-x) for volumes by cylindrical shells. How do we know which one to use?
Please Help!
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-2.pdf

RuiAce · « **Reply #1520 on:** October 22, 2017, 09:44:02 pm »

Quote from: aryak on October 22, 2017, 09:36:32 pm

Hey for 2015, question 12 (d) I am very confused as to whether to use x or (3-x) for volumes by cylindrical shells. How do we know which one to use?
Please Help!
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-2.pdf

$\text{Since the orange line reflects the radius}\\ \text{the radius is }3-x$

aryak · « **Reply #1521 on:** October 22, 2017, 09:48:59 pm »

is it ok if you could explain this a bit more? I am still a bit confused

RuiAce · « **Reply #1522 on:** October 22, 2017, 09:49:59 pm »

Quote from: aryak on October 22, 2017, 09:48:59 pm

is it ok if you could explain this a bit more? I am still a bit confused

What part is the confusing part?

I essentially wrote down the two relevant lengths, and figured out which one was the radius of the cylinder.
$\text{Edit: One thing that you do need to recall is what the}\\ x\text{-coordinate actually is.}\\ \text{The }x\text{-coordinate is literally how far to the right/left of the }y\text{-axis you are.}$
$\text{That is the key thing to deducing the lengths}\\ \text{and is something every 4U student needs to know for the volumes topic}$

Dragomistress · « **Reply #1523 on:** October 23, 2017, 07:55:47 pm »

Would someone please do this. Solving quadratic equations with complex coefficients: finding the complex number equations.

4x^2-4(1+2i)x-(3-4i)=0

justwannawish · « **Reply #1524 on:** October 23, 2017, 08:36:46 pm »

Hi, guys, do you use De Moivre's Thereom to solve:
z^5= -32
Thank you

beau77bro · « **Reply #1525 on:** October 23, 2017, 08:54:14 pm »

What were the roots of the complex quadratic in the exam?

beau77bro · « **Reply #1526 on:** October 23, 2017, 09:33:33 pm »

And what was 16ciii? I got 2(3)^5 -3

pikachu975 · « **Reply #1527 on:** October 23, 2017, 10:02:17 pm »

Quote from: beau77bro on October 23, 2017, 09:33:33 pm

And what was 16ciii? I got 2(3)^5 -3

Answer was 480

beau77bro · « **Reply #1528 on:** October 23, 2017, 10:15:50 pm »

I think I can see why, the arrangements of only 2 colours can be happen in two ways.

As in like changing which way they alternate - 3 different pairs x2 because flip alternating. Damn was cheering

paigek3 · « **Reply #1529 on:** October 24, 2017, 10:49:08 am »

Yewww I can say I posted in 4u maths

Helloooo, this is such a dumb question that has no relevance to any mathematical problems whatsoever, but I am so confused about the structure of 4u maths so I thought I might as well ask here since the exam is over

Sooo I know you guys sit the exam the same time as Mathematics, meaning you don't do that exam, but do you do the Mathematics course still? Im 10/10 confused and always wanted to know the answer to this

- from your fellow 4 unit (English) student

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