2012 Year 9 &10 Mathematics Marathon

**TerenceM** · 2 Jul 2012, 12:32 PM

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

**russ3l** · 2 Jul 2012, 12:53 PM

Originally Posted by Carrotsticks

Eliminate theta from the following expressions:

$x=a \sin \theta $ and $ y= b \cos \theta$

$a^2y^2+b^2x^2=a^2b^2$ ?

**ganji** · 2 Jul 2012, 12:59 PM

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

**twinklegal19** · 2 Jul 2012, 1:09 PM

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

**twinklegal19** · 2 Jul 2012, 1:13 PM

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

**bleakarcher** · 2 Jul 2012, 1:14 PM

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

**RivalryofTroll** · 2 Jul 2012, 1:17 PM

Originally Posted by ymcaec

$\frac{1}{9}$?$$

Correct. 3^-2 = 1/9

**russ3l** · 2 Jul 2012, 1:22 PM

simplify

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

**RivalryofTroll** · 2 Jul 2012, 1:26 PM

Here's another one for you smarties:

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

Solve for x.

**ymcaec** · 2 Jul 2012, 1:31 PM

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$

**Carrotsticks** · 2 Jul 2012, 1:32 PM

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.

**twinklegal19** · 2 Jul 2012, 1:36 PM

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.

**ymcaec** · 2 Jul 2012, 1:38 PM

Originally Posted by RivalryofTroll

Here's another one for you smarties:

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

Solve for x.

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

**twinklegal19** · 2 Jul 2012, 1:39 PM

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)

**ymcaec** · 2 Jul 2012, 1:55 PM

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.$$

**twinklegal19** · 2 Jul 2012, 2:06 PM

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

**ymcaec** · 2 Jul 2012, 2:08 PM

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

thanks for the advice

**twinklegal19** · 2 Jul 2012, 2:15 PM

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$$

**Peeik** · 2 Jul 2012, 2:16 PM

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 $can be written as the difference of two squares.$

**twinklegal19** · 2 Jul 2012, 2:17 PM

Originally Posted by mwong

simplify

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

Also, don't forget about this question

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

**russ3l** · 2 Jul 2012, 2:18 PM

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

**ymcaec** · 2 Jul 2012, 2:37 PM

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 $can be written as the difference of two squares.$

$(x+y\sqrt{d})^2 (x-y\sqrt{d})^2\\= (x^2+2xy\sqrt{d}+y^2d)(x^2 - 2xy\sqrt{d}+y^2d)\\ = [(x^2+y^2d)+2xy\sqrt{d}][(x^2+y^2d)-2xy\sqrt{d}]\\={(x^2+y^2d})}^2 - {(2xy\sqrt{d})}^2$

**Peeik** · 2 Jul 2012, 2:54 PM

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!

**ymcaec** · 2 Jul 2012, 2:54 PM

Originally Posted by mwong

simplify

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

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

**Carrotsticks** · 2 Jul 2012, 2:57 PM

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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