2012 Year 9 &10 Mathematics Marathon (1 Viewer)

twinklegal19 · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kev-kun said:
Well tried to stay to the question, guess I was wrong ==''

BTW just curious, are we allowed to submit questions from our textbooks?

go ahead

kazemagic · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
Sorry that I didn't get to your Q earlier but here it is now:

$\frac{6^n+3^n}{2^{n+1}+2}$

$\frac{3^n(2^n+1)}{2(2^n+1)}$

$=\frac{3^n}{2}$

Correct

iBibah · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kev-kun said:
Well tried to stay to the question, guess I was wrong ==''

BTW just curious, are we allowed to submit questions from our textbooks?

haha its ok, its just that if

f(x)=*insert quadratic function*

then x=*insert quadratic equation you just proved*

Your not trying to find another way of expressing it, but rather a general formula to find the roots, hence x=.

Just trying to help you understand what your doing a little more

And yes, you can post questions from texts.

ymcaec · Aug 18, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
Here's another one lol

Pythagoras

First we know BE = ED (diagonals of parallelogram bisect each other)

Then using Pythagoras:
${AB}^2+{BE}^2={AE}^2\\{AB}^2+{ED}^2={AE}^2\\{AB}^2={AE}^2-{ED}^2...(1)\\\\{AB}^2+{BD}^2={AD}^2\\{AB}^2+{(2ED)}^2={AD}^2...(2)\\\\$Substitute (1) into (2)$\\{AE}^2-{ED}^2+4{ED}^2={AD}^2\\{AD}^2={AE}^2+3{ED}^2$

twinklegal19 · Aug 18, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Have fun with this guys

Use a suitable substitution to solve

$\left ( x^2+\frac{1}{x^2} \right )^2-9\left ( x^2+\frac{1}{x^2} \right )+20=0$

kev-kun · Aug 18, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Have fun with this guys

Use a suitable substitution to solve

$\left ( x^2+\frac{1}{x^2} \right )^2-9\left ( x^2+\frac{1}{x^2} \right )+20=0$

$\\(x^2+\frac{1}{x^2})^2-9(x^2+\frac{1}{x^2})+20=0 \\Let\ y=(x^2+\frac{1}{x^2}) \\y^2-9y+20=0 \\(y-5)(y-4)=0 \\Sub\ y=(x^2+\frac{1}{x^2}) \\(x^2+\frac{1}{x^2}-5)(x^2+\frac{1}{x^2}-4)=0 \\That's\ all\ I \ got\ =="$

twinklegal19 · Aug 18, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kev-kun said:
$\\(x^2+\frac{1}{x^2})^2-9(x^2+\frac{1}{x^2})+20=0 \\Let\y=(x^2+\frac{1}{x^2}) \\y^2-9y+20=0 \\(y-5)(y-4)=0 \\Sub\ y=(x^2+\frac{1}{x^2}) \\(x^2+\frac{1}{x^2}-5)(x^2+\frac{1}{x^2}-4)=0 \\That's\ all\ I \ got\ =="$

finish solving your quadratic equation

i.e. y=5,4 and from there you can replace y with what you originally substituted

kazemagic · Aug 19, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
Pythagoras

First we know BE = ED (diagonals of parallelogram bisect each other)

Then using Pythagoras:
${AB}^2+{BE}^2={AE}^2\\{AB}^2+{ED}^2={AE}^2\\{AB}^2={AE}^2-{ED}^2...(1)\\\\{AB}^2+{BD}^2={AD}^2\\{AB}^2+{(2ED)}^2={AD}^2...(2)\\\\$Substitute (1) into (2)$\\{AE}^2-{ED}^2+4{ED}^2={AD}^2\\{AD}^2={AE}^2+3{ED}^2$

Yep good job lol

@Twinkle happy birthday m8 *insert twinkle twinkle little stars song here*

Carrotsticks · Aug 19, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
finish solving your quadratic equation i.e. y=5,4 and from there you can replace y with what you originally substituted

A quadratic within a quadratic.

kev-kun · Aug 19, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Carrotsticks said:
A quadratic within a quadratic.

QUADception

enoilgam · Aug 21, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

This isnt really a syllabus problem, but more of a logic problem - nonetheless though I think the skills needed are pretty important in maths. Some people may recognise it from somewhere, but I wont say where...

A person has a five litre jug and a three litre jug and they need to fill the five litre jug with exactly four litres of water. The person cannot guess the amount and they have no measurement aids other then the two jugs. Assuming they have an unlimited supply of water, how would they complete this task?

Fawun · Aug 21, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

enoilgam said:
This isnt really a syllabus problem, but more of a logic problem - nonetheless though I think the skills needed are pretty important in maths. Some people may recognise it from somewhere, but I wont say where...

A person has a five litre jug and a three litre jug and they need to fill the five litre jug with exactly four litres of water. The person cannot guess the amount and they have no measurement aids other then the two jugs. Assuming they have an unlimited supply of water, how would they complete this task?

To do it, you fill the 5 litre jug completely, which means you have 5 litres, then you pour the contents into the 3 litre jug until it is completely full. Which means that you now have 2 litres left in the 5 litre jug. You do it twice which means you will ahve 4 litres.

enoilgam · Aug 21, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Fawun said:
To do it, you fill the 5 litre jug completely, which means you have 5 litres, then you pour the contents into the 3 litre jug until it is completely full. Which means that you now have 2 litres left in the 5 litre jug. You do it twice which means you will ahve 4 litres.

You skipped a step at the end...elaborate.

Fawun · Aug 21, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Well since the question stated that you can have unlimited amount of water, after pouring the contents into the three litre jug Now you have 2 litres left in the 5 litre jug. You then pour the 2 litre jug in the other jug, then empty the three litre jug and start again to get another 2 which gives you four.

nightweaver066 · Aug 22, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Fawun said:
Well since the question stated that you can have unlimited amount of water, after pouring the contents into the three litre jug Now you have 2 litres left in the 5 litre jug. You then pour the 2 litre jug in the other jug, then empty the three litre jug and start again to get another 2 which gives you four.

i think you only have one 5L jug and one 3L jug. are you taking it to be 3 jugs?

@enoilgam, i've been playing a game exactly like this on my phone and it's so much fun haha

enoilgam · Aug 22, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

nightweaver066 said:
i think you only have one 5L jug and one 3L jug. are you taking it to be 3 jugs?

@enoilgam, i've been playing a game exactly like this on my phone and it's so much fun haha

There are only two jugs fawun. You are close, but the end is where you are going wrong.

kazemagic · Aug 22, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

enoilgam said:
This isnt really a syllabus problem, but more of a logic problem - nonetheless though I think the skills needed are pretty important in maths. Some people may recognise it from somewhere, but I wont say where...

A person has a five litre jug and a three litre jug and they need to fill the five litre jug with exactly four litres of water. The person cannot guess the amount and they have no measurement aids other then the two jugs. Assuming they have an unlimited supply of water, how would they complete this task?

Let x= 5 litre jug
y=3 litre jug

Fill up y, pour into x so x has 3 litre. Fill up y, then pour it to x so x has 5 litre and y has 1 litre. Pour x into sink, so nothing remains in it. Pour the 1 litre from y to x. Fill up y then pour it to x, then u get der 4 litres in 5 litre jug

enoilgam · Aug 22, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
Let x= 5 litre jug
y=3 litre jug

Fill up y, pour into x so x has 3 litre. Fill up y, then pour it to x so x has 5 litre and y has 1 litre. Pour x into sink, so nothing remains in it. Pour the 1 litre from y to x. Fill up y then pour it to x, then u get der 4 litres in 5 litre jug

Nice - that wasnt the method I had in mind but it works nonetheless.

kazemagic · Aug 22, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

enoilgam said:
Nice - that wasnt the method I had in mind but it works nonetheless.

lol are the jugs the same width tho?

enoilgam · Aug 22, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
lol are the jugs the same width tho?

The only information that is provided is their capacity. There is another way to do it, can anyone think of it.

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