# Thread: 2012 Year 9 &10 Mathematics Marathon

1. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by kev-
let the youngest sister be x
therefore oldest sister = x+2+2+2+2
2x=x+8
x+8
third youngest sister is 8+2+2 = 12 years old.

this question was in the Ngo and Sons Year 10 exit exam! O: where'd you find it?
Lol I did it myself. Failed it however cos I was sick and everything so barely studied and such LOL and silly mistakes TT

2. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by Carrotsticks
Eliminate theta from the following expressions:

$x=a \sin \theta and y= b \cos \theta$
$a^2y^2+b^2x^2=a^2b^2$ ?

3. ## Re: 2012 Year 9 &10 Mathematics Marathon

so you let x = that massive equation.

square both sides so..

x^2 = 7 - 6[insert that massive equation]

the massive equation = x

hence: x^2 = 7 - 6x
x^2 + 6x - 7 = 0
(x+7)(x-1)=0
x = -7 or 1
but because x>=0 because its a square root
therefore x=1

4. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by ymcaec
$i. \left ( x-3 \right )^{2}-6$
$ii. x = \ 3\pm \sqrt{y+6}$
$iii. All real y except y < -6$
Correct! Or alternately for iii, just $y\geq-6$ would probably be faster to write in an exam

5. ## Re: 2012 Year 9 &10 Mathematics Marathon

Here's a question from my year 10 yearly exam

$\\i. If k is any integer, write an expression for odd number n in terms of k.\\\\ii. If n is an odd integer, show that n^2+4n-1 is divisible by 4$

Also, don't forget to show full working for all questions in this thread

6. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by mwong
$a^2y^2+b^2x^2=a^2b^2$ ?
yep

simplifying leads to the general equation of an ellipse x^2/a^2 + y^2/b^2=1

7. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by ymcaec
$\frac{1}{9}?$
Correct. 3^-2 = 1/9

8. ## Re: 2012 Year 9 &10 Mathematics Marathon

simplify

$\frac{1}{(a-b)(a-c)} + \frac{1}{(b-a)(b-c)} + \frac{1}{(c-a)(c-b)}$

9. ## Re: 2012 Year 9 &10 Mathematics Marathon

Here's another one for you smarties:

16^x - 18(4^x) + 32 = 0

Solve for x.

10. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by twinklegal19
Here's a question from my year 10 yearly exam

$\\i. If k is any integer, write an expression for odd number n in terms of k.\\\\ii. If n is an odd integer, show that n^2+4n-1 is divisible by 4$

Also, don't forget to show full working for all questions in this thread
$i. n = 2k + 1 or n = 2k - 1\\\\ii. Substituting n = 2k + 1 into the equation n^2+4n-1 \\(2k + 1)^2 + 4 (2k+1) - 1\\= 4k^2 + 4k + 1 + 8k + 4 - 1\\= 4k^2 + 12k + 4\\Since everything are multiples of 4\\\therefore n^2 + 4n - 1 is divisible by 4 if n is an odd integer$

11. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by mwong
$a^2y^2+b^2x^2=a^2b^2$ ?
Correct! Divide both sides by a^2 b^2 to make it look nicer =)

You just found the equation of an ellipse.

12. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by ymcaec
$i. n = 2k + 1 or n = 2k - 1\\\\ii. Substituting n = 2k + 1 into the equation n^2+4n-1 \\(2k + 1)^2 + 4 (2k+1) - 1\\= 4k^2 + 4k + 1 + 8k + 4 - 1\\= 4k^2 + 4k + 8k + 4\\Since everything are multiples of 4\\\therefore n^2 + 4n - 1 is divisible by 4 if n is an odd integer$
Correct! Although it's best it you simplify and factorise your final line to make it more clear, like

$=4(k^2+3k+1)$

therefore it is a multiple of 4 etc.

13. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by RivalryofTroll
Here's another one for you smarties:

16^x - 18(4^x) + 32 = 0

Solve for x.
$16^x - 18(4^x) + 32 = 0\\ 4^{2x} - 18 \times 4^x + 32 = 0\\ 4^x (4^2 - 18) + 32 = 0\\ 4^x \times -2 + 32 = 0\\4^x\times -2=-32\\4^x = 16\\4^x = 4^2\\ \therefore x = 2$

14. ## Re: 2012 Year 9 &10 Mathematics Marathon

Another question from my Yr 10 yearly

$Prove that a^5-a^4b\geq ab^4-b^5 for all positive values of a and b$

(hint: start by putting everything into LHS and prove that expression to be greater than or equal to 0)

15. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by twinklegal19
Another question from my Yr 10 yearly

$Prove that a^5-a^4b\geq ab^4-b^5 for all positive values of a and b$

(hint: start by putting everything into LHS and prove that expression to be greater than or equal to 0)
$a^5-a^4b\geq ab^4-b^5\\a^5-a^4b - ab^4+b^5\geq 0 \\ a^4 (a-b) - b^4 (a-b) \geq 0 \\ (a^4-b^4)(a-b)\geq 0\\ (a^2+b^2)(a^2-b^2)(a-b)\geq 0 \\(a^2+b^2)(a+b)(a-b)(a-b) \geq 0 \\ (a^2+b^2)(a+b)(a-b)^2 \geq 0 \\\\ Since a and b are positive, and all squares are positive (or equal to 0)\\ so a^2 + b^2, a+b, (a-b)^2 are positive (or equal to 0) \\ \therefore a^5-a^4b - ab^4+b^5\geq 0 and a^5-a^4b\geq ab^4-b^5 is true.$

16. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by ymcaec
$a^5-a^4b\geq ab^4-b^5\\a^5-a^4b - ab^4+b^5\geq 0 \\ a^4 (a-b) - b^4 (a-b) \geq 0 \\ (a^4-b^4)(a-b)\geq 0\\ (a^2+b^2)(a^2-b^2)(a-b)\geq 0 (a^2+b^2)(a+b)(a-b)(a-b) \geq 0 \\ (a^2+b^2)(a+b)(a-b)^2 \geq 0 \\ Since a and b are positive, and all squares are positive (or equal to 0)\\ so a^2 + b^2, a+b, (a-b)^2 are positive (or equal to 0) \\ \therefore a^5-a^4b - ab^4+b^5\geq 0 and a^5-a^4b\geq ab^4-b^5 is true.$
Yep! Good work

It might be a good idea to mention somewhere along the lines that $given~a,b\geq 0, (a+b)\geq 0$

I know it's pretty obvious, but it's best to acknowledge the criteria in the question that a and b have to be positive themselves

17. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by twinklegal19
Yep! Good work

It might be a good idea to mention that $given~a,b\geq 0, (a+b)\geq 0$

I know it's pretty obvious, but its good to acknowledge the question's criteria for a and b somewhere

18. ## Re: 2012 Year 9 &10 Mathematics Marathon

Another maths yearly question!

$\\\\i. Show algebraically that x^2+y^2=18 and y=x-6 intersect at precisely one point.\\\\ii. Draw a diagram illustrating both graphs and their point of intersection.\\\\iv. Use the symmetry of your diagram to find the values of k for which the graph of y=x+k will intersect x^2+y^2=18 in precisely two points$

19. ## Re: 2012 Year 9 &10 Mathematics Marathon

Here's an interesting intimidating (yet very easy) question on algebraic factorisation.
$Show that the following algebraic expression (x+y\sqrt{d})^2 (x-y\sqrt{d})^2 can be written as the difference of two squares.$

20. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by mwong
simplify

$\frac{1}{(a-b)(a-c)} + \frac{1}{(b-a)(b-c)} + \frac{1}{(c-a)(c-b)}$

edit: just realised that this is an extension question in chapter 1 of Yr 11 Cambridge

21. ## Re: 2012 Year 9 &10 Mathematics Marathon

does anyone have any good "simplifying expressions" questions? they're my favourite

22. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by Peeik
Here's an interesting intimidating (yet very easy) question on algebraic factorisation.
$Show that the following algebraic expression (x+y\sqrt{d})^2 (x-y\sqrt{d})^2 can be written as the difference of two squares.$
$(x+y\sqrt{d})^2 (x-y\sqrt{d})^2\\= (x^2+2xy\sqrt{d}+y^2d)(x^2 - 2xy\sqrt{d}+y^2d)\\ = [(x^2+y^2d)+2xy\sqrt{d}][(x^2+y^2d)-2xy\sqrt{d}]\\={(x^2+y^2d})}^2 - {(2xy\sqrt{d})}^2$

23. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by ymcaec
$(x+y\sqrt{d})^2 (x-y\sqrt{d})^2\\= (x^2+2xy\sqrt{d}+y^2d)(x^2 - 2xy\sqrt{d}+y^2d)\\ = [(x^2+y^2d)+2xy\sqrt{d}][(x^2+y^2d)-2xy\sqrt{d}]\\={(x^2+y^2d})}^2 - {(2xy\sqrt{d})}^2$
Correct!

24. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by mwong
simplify

$\frac{1}{(a-b)(a-c)} + \frac{1}{(b-a)(b-c)} + \frac{1}{(c-a)(c-b)}$
$\frac{1}{(a-b)(a-c)} + \frac{1}{(b-a)(b-c)} + \frac{1}{(c-a)(c-b)}\\ = \frac{1}{(a-b)(a-c)} + \frac{1}{-(a-b)(b-c)} + \frac{1}{(a-c)(b-c)}\\ = \frac{(b-c)-(a-c)+(a-b)}{(a-b)(a-c)(b-c)}\\ = \frac{0}{(a-b)(a-c)(b-c)}\\ = 0$

25. ## Re: 2012 Year 9 &10 Mathematics Marathon

Originally Posted by mwong
does anyone have any good "simplifying expressions" questions? they're my favourite
$\\ Simplify \frac{x+1}{xy+x+1} + \frac{y+1}{yz+y+1} + \frac{z+1}{xz+z+1} where x,y,z \geq 0 and xyz=1.$

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