2012 Year 9 &10 Mathematics Marathon (1 Viewer)

RealiseNothing · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento, solution for the triangle question:

The area of a triangle is half times base times height. The longest height will have the shortest base, and the shortest height will have the longest base.

So let the longest height be 'H' and the shortest height be 'h'.

The area of the triangle using the longest height is:

$\frac{1}{2} 40H$

The area of the triangle using the shortest height is:

$\frac{1}{2} 80h$

Since these are just the areas of the triangle, they are equal, that is:

$\frac{1}{2} 40H = \frac{1}{2} 80h$

Simplifying gives:

$H = 2h$

Hence the longest height is twice that of the shortest height, and so k = 1/2.

twinklegal19 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Just curious, what maths topics have you guys done so far this year? And whether you are 5.1/5.2/5.3 or accelerated. Just want to make sure.

kazemagic · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

RealiseNothing said:
Demento, solution for the triangle question:

The area of a triangle is half times base times height. The longest height will have the shortest base, and the shortest height will have the longest base.

So let the longest height be 'H' and the shortest height be 'h'.

The area of the triangle using the longest height is:

$\frac{1}{2} 40H$

The area of the triangle using the shortest height is:

$\frac{1}{2} 80h$

Since these are just the areas of the triangle, they are equal, that is:

$\frac{1}{2} 40H = \frac{1}{2} 80h$

Simplifying gives:

$H = 2h$

Hence the longest height is twice that of the shortest height, and so k = 1/2.

Wow... you make it sound so simple

russ3l · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Just curious, what maths topics have you guys done so far this year? And whether you are 5.1/5.2/5.3 or accelerated. Just want to make sure.

To be honest, I have no fucking clue. I'm one of the top math students in my grade (for 5.3) and I still don't know how to do these questions. Ymceac is accelerated so he doesn't count

I feel like some of these questions are based more off 'general knowledge' other than "oh! ive done this topic at school, I know exactly what to do with a question like this!"...you know what I mean? :/

kazemagic · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
To be honest, I have no fucking clue. I'm one of the top math students in my grade (for 5.3) and I still don't know how to do these questions. Ymceac is accelerated so he doesn't count
I feel like some of these questions are based more off 'general knowledge' other than "oh! ive done this topic at school, I know exactly what to do with a question like this!"...you know what I mean? :/

So tru

Will anyone type up a worked solution for this question?

BTW heres a question.
A father in his will left all his money to his children in the following manner: $1k to first born and 1/10 of what then remains, then$2k to the second born and 1/10 of what then remains, then $3k to the third born and 1/10 of what then remains, and so on. When this was done each child had the same amount. How many children were there?

I forgot the answer to this lol

ymcaec · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
So tru

Will anyone type up a worked solution for this question?

We're trying to get rid of x.
$a = 3(x-x^{-1})\\\frac{a}{3} = x-x^{-1}\\\frac{a^2}{9} = x^2-2+x^{-2}\\\\\\b = 6(x+x^{-1})\\\frac{b}{6} = x+x^{-1}\\\frac{b^2}{36} = x^2+2+x^{-2}\\\\$Join them together (and also to get rid all x's)$\\\frac{a^2}{9}-\frac{b^2}{36}=x^2-2+x^{-2}-(x^2+2+x^{-2})\\\frac{a^2}{9}-\frac{b^2}{36}=-4\\4a^2-b^2=-144\\\therefore b^2=4a^2+144$

Demento1 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
So tru

Will anyone type up a worked solution for this question?

You're coming to meat tomorrow so perhaps I can explain it then. I will be arriving at 10am and you'll just have to find me.

twinklegal19 said:
Just curious, what maths topics have you guys done so far this year? And whether you are 5.1/5.2/5.3 or accelerated. Just want to make sure.

Advanced mathematics course in NSW (5.3 i'm guessing?). I don't actually accelerate, not really anyway as our school doesn't allow it.

RealiseNothing said:
Demento, solution for the triangle question:

The area of a triangle is half times base times height. The longest height will have the shortest base, and the shortest height will have the longest base.

So let the longest height be 'H' and the shortest height be 'h'.

The area of the triangle using the longest height is:

$\frac{1}{2} 40H$

The area of the triangle using the shortest height is:

$\frac{1}{2} 80h$

Since these are just the areas of the triangle, they are equal, that is:

$\frac{1}{2} 40H = \frac{1}{2} 80h$

Simplifying gives:

$H = 2h$

Hence the longest height is twice that of the shortest height, and so k = 1/2.

Appreciate the explanation, as it's probably a lot more orthodox then my working.

russ3l · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
So tru

Will anyone type up a worked solution for this question?

^Like this question for example. Is it even in the year 10 syllabus? I've never been taught how to find an independent of something. Maybe it's just me...

Demento1 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
^Like this question for example. Is it even in the year 10 syllabus? I've never been taught how to find an independent of something. Maybe it's just me...

The question is just worded a little differently to throw off some students. Really, the question can be interpreted as:

If $a=3(x-x^{-1})~and~b=6(x+x^{-1})$ find an equation which shows the relationship between a and b, without x terms.

kazemagic · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
We're trying to get rid of x.
$a = 3(x-x^{-1})\\\frac{a}{3} = x-x^{-1}\\\frac{a^2}{9} = x^2-2+x^{-2}\\\\\\b = 6(x+x^{-1})\\\frac{b}{6} = x+x^{-1}\\\frac{b^2}{36} = x^2+2+x^{-2}\\\\$Join them together (and also to get rid all x's)$\\\frac{a^2}{9}-\frac{b^2}{36}=x^2-2+x^{-2}-(x^2+2+x^{-2})\\\frac{a^2}{9}-\frac{b^2}{36}=-4\\4a^2-b^2=-144\\\therefore b^2=4a^2+144$

O I c. Thanks Ymcaec
Got a question for da genius lol. Have you seen these type of questions b4? Or you did it on the spot

RealiseNothing · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
Will anyone type up a worked solution for this question?

Here's my way:

Double the first expression to get:

$2a = 6(x - x^{-1})$

Now if we expand the brackets we will get:

$2a = 6x - \frac{6}{x} ~$and$~ b = 6x + \frac{6}{x}$

Add the two expressions together, and also subtract the two expressions, we will get:

$2a + b = 12x ~$and$~ 2a - b = -\frac{\12}{x}$

Now we just multiply them together:

$(2a + b)(2a - b) = \frac{-12 \times 12x}{x}$

Now notice the LHS is a difference of two squares and on the RHS the x's will cancel out:

$4a^2 - b^2 = -144$

Re-arranging gives the answer of:

$b^2 = 144 + 4a^2$

kazemagic · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

RealiseNothing said:
Here's my way:

Double the first expression to get:

$2a = 6(x - x^{-1})$

Now if we expand the brackets we will get:

$2a = 6x - \frac{6}{x} ~$and$~ b = 6x + \frac{6}{x}$

Add the two expressions together, and also subtract the two expressions, we will get:

$2a + b = 12x ~$and$~ 2a - b = -\frac{\12}{x}$

Now we just multiply them together:

$(2a + b)(2a - b) = \frac{-12 \times 12x}{x}$

Now notice the LHS is a difference of two squares and on the RHS the x's will cancel out:

$4a^2 - b^2 = -144$

Re-arranging gives the answer of:

$b^2 = 144 + 4a^2$

Wow

You're coming to meat tomorrow so perhaps I can explain it then. I will be arriving at 10am and you'll just have to find me.

Looking forward to meet you and Ealdoon

ymcaec · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
O I c. Thanks Ymcaec
Got a question for da genius lol. Have you seen these type of questions b4? Or you did it on the spot

i have done some "connecting" questions before

ymcaec · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

RealiseNothing said:
Here's my way:

Double the first expression to get:

$2a = 6(x - x^{-1})$

Now if we expand the brackets we will get:

$2a = 6x - \frac{6}{x} ~$and$~ b = 6x + \frac{6}{x}$

Add the two expressions together, and also subtract the two expressions, we will get:

$2a + b = 12x ~$and$~ 2a - b = -\frac{\12}{x}$

Now we just multiply them together:

$(2a + b)(2a - b) = \frac{-12 \times 12x}{x}$

Now notice the LHS is a difference of two squares and on the RHS the x's will cancel out:

$4a^2 - b^2 = -144$

Re-arranging gives the answer of:

$b^2 = 144 + 4a^2$

i think this way is better !

twinklegal19 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
^Like this question for example. Is it even in the year 10 syllabus? I've never been taught how to find an independent of something. Maybe it's just me...

Yes, that was in one of my school's past year 10 yearly papers, although admittedly that was one of the last questions.

Demento1 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
Wow

Looking forward to meet you and Ealdoon

We're the only year 10s, but meh, who cares. To be honest, I reckon russ3l and ymcaec should come tomorrow. It would be interesting.

twinklegal19 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

How strong is everyone at probability? I want to post another one hehe.

It'll have more algebra and less getting around the wording of the question

Demento1 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
How strong is everyone at probability? I want to post another one hehe.

It'll have more algebra and less getting around the wording of the question

Do we let something be x? If yes, I will absolutely pour my heart to answer it.

twinklegal19 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
Do we let something be x? If yes, I will absolutely pour my heart to answer it.

Yep! Here it is, have fun

Again, tree diagrams can help (although some pro people here don't seem to need them haha), just let p be the probably of winning and 1-p be losing. For part iii it's just basically a similar (but easier) version of the inequality question I posted before by putting everything on one side and proving it for be more than 0

Demento1 · Jul 3, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Well for i) Probability of winning first dart = p

Probability of winning with second dart given first dart did not hit $= 1 – (1-p)(1-p)$

If you expand that you should get: $2p -p^2$

For ii) To win, she needs to hit the target twice, there are 4 possibilities:

There are 3 possibilities of hitting 2 darts with one missing (i.e. 1st and 2nd, 2nd and 3rd, 1st and 3rd).

The probability of this is: $p*p*(1-p)$

1 permutation of hitting with all 3 darts: $p*p*p$

So the total probability $= 3p^2(1-p) + p^3 = 3p^2 – 2p^3$

For iii) I hope I didn't type anything wrong...

$p(winning game 1): data from (i)$
$p(winning game 2): data from (ii)$

If Darcy is to be more likely to win game 1 than game 2, then from above:

$2p – p^2 > 3p^2 – 2p^3$

$= p (2 – p) > p (3p – 2p^2)$

Since $0 < p < 1,~2 – p > 3p – 2p^2$

$2p^2 – 4p + 2 > 0$

$p^2 – 2p + 1 > 0$

$(p – 1)^2 > 0$

Therefore for 0 < p < 1, this statement is true and the induction holds.

For iv), algebra can really finally be used to some extent.

$2p – p^2 = 2(3p^2 – 2p^3)$

Divide by p, get it into the format of $ap^2 + bp + c = 0$ and advance:

$2p – p^2 = 6p^2 – 4p^3$
$2 – p = 6p – 4p^2$
$4p^2 – 7p + 2 = 0$

Thus, you have a quadratic equation to solve and the quadratic formula is used which I cannot bother with typing unfortunately...

Eventually you get p = 1.39, 0.36. Only one of these values is between our limit 0 < p < 1, so the answer would be when p = 0.36 (2 dp).

2012 Year 9 &10 Mathematics Marathon (1 Viewer)

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