O-Level Additional Math

[FrekiWang]

Xiao Hu:

Hi FrekiWang,


Need help on this question, I think it's 2010 O level Maths paper 1 question.
Part2 looks innocuously easy, but it trips me up. Difficult to put my reasonings in writings.

There are 2 parts to the question.
#1. Factorise 90. This is easy, 90=2x3x3x5

#2
LCM of 6,15 and x is 90. What are possible values of x if x is odd?

From #1, since 90=6x15, how can this be used to work out possible values of x?

The answers given for x are 9 and 45.

TIA,
Xiao Hu.

90=2*3^2*5
6=2*3
15=3*5
Therefore, x=2^a * 3^2 * 5^b
where a=0 or 1(reject as odd) and b=0 or 1
when a=0,b=0: x=3^2=9
when a=0,b=1: x=3^2*5^1=45

[Xiao Hu]

FrekiWang:

Xiao Hu:

Hi FrekiWang,


Need help on this question, I think it's 2010 O level Maths paper 1 question.
Part2 looks innocuously easy, but it trips me up. Difficult to put my reasonings in writings.

There are 2 parts to the question.
#1. Factorise 90. This is easy, 90=2x3x3x5

#2
LCM of 6,15 and x is 90. What are possible values of x if x is odd?

From #1, since 90=6x15, how can this be used to work out possible values of x?

The answers given for x are 9 and 45.

TIA,
Xiao Hu.

90=2*3^2*5
6=2*3
15=3*5
Therefore, x=2^a * 3^2 * 5^b
where a=0 or 1(reject as odd) and b=0 or 1
when a=0,b=0: x=3^2=9
when a=0,b=1: x=3^2*5^1=45

Hi FrekiWang,
Good one!! I could't have expessed the solution this way, I would have them in English.
This is a bit tough for O level Maths, isn't it? It's only 1 mark. Crazy.

Thanks very much, appreciate it.
Xiao Hu.

[Bestmark]

Hi


Any recommendation for maths tuition centre? which is better mavis, mind stretcher or smartlab or ms loi?

[JadeDry]

Hello,


Can you please help me with the following question?

Make "t" the subject of the formula:

T= 2(pi) * (the square root of(( (t^2) + (k^2)) / 2gt))

I have checked the answer which is:
t = (+ or -) (1/ (2pi)) (The square root of((2gt(T2)) - ((4(pi squared))(k squared))

I do not understand how they have used the lower case "t" in the formula.

Is it correct?

Thanks in advance.

[Belle2011]

Hello,

May I ask where can I get books/resources on math olympiad questions suitable for lower sec?
Thank-you.

cheers,
Belle.

[mum_sugoku]

JadeDry:

Hello,


Can you please help me with the following question?

Make \"t\" the subject of the formula:

T= 2(pi) * (the square root of(( (t^2) + (k^2)) / 2gt))

I have checked the answer which is:
t = (+ or -) (1/ (2*pi)) *(The square root of((2gt(T*2)) - ((4*(pi squared))(k squared))

I do not understand how they have used the lower case \"t\" in the formula.

Is it correct?

Thanks in advance.

Ya by right \"t\" shouldn't be in there if it is the subject.. So I too think that answer is incorrect.

Anyway this is a seemingly tedious sum (I mean the working). You'll need to apply the standard formula x=[-b +- sq rt (b^2 -4ac) ]/2a to solve it.. Here's what I've got after working it out, not sure if it's correct though:

t ={ (T^2)g +- {sq rt[(T^4)(g^2) - 16(pi^4)(k^2)]} } / (4pi^2)

[Xiao Hu]

Hi FrekiWang, other maths guru here,


Need help on this Q20, taken from Cedar Girl's 2011 Prelim2 papers. I just couldn't figure it out. Answer in the answer key is 90 degrees.

http://i40.tinypic.com/t9wj9c.jpg\">

Thanks in advance,
Xiao Hu

[njcstudent]

Hi, I am a sec 1 student in an IP school. Can anyone please recommend any good IP assessment books? Thanks in advance.

[mum_sugoku]

Xiao Hu:

Hi FrekiWang, other maths guru here,


Need help on this Q20, taken from Cedar Girl's 2011 Prelim2 papers. I just couldn't figure it out. Answer in the answer key is 90 degrees.

http://i40.tinypic.com/t9wj9c.jpg\">

Thanks in advance,
Xiao Hu

rhombus -> all 4 sides are equal, and AB//DC, AD//BC

a) DL is common to both triangles -->side
AD=CD -->side
tri ABD=tri CBD (SSS), therefore <ADL=<LDC -->angle

therefore tri ALD= tri CLD (SSA)

b) <ADL=<LBX (corr <), and
<DLX= <LBX+90 (ext < of tri)
but <DLX also = <DAL + <ADL (ext < of tri), and <ADL=<LBX
==> <DAL =90
==><LCD=90 (since tri ALD=tri CLD)

[Xiao Hu]

mum_sugoku:

Xiao Hu:

Hi FrekiWang, other maths guru here,


Need help on this Q20, taken from Cedar Girl's 2011 Prelim2 papers. I just couldn't figure it out. Answer in the answer key is 90 degrees.

http://i40.tinypic.com/t9wj9c.jpg\">

Thanks in advance,
Xiao Hu

rhombus -> all 4 sides are equal, and AB//DC, AD//BC

a) DL is common to both triangles -->side
AD=CD -->side
tri ABD=tri CBD (SSS), therefore <ADL=<LDC -->angle

therefore tri ALD= tri CLD (SSA)

b) <ADL=<LBX (corr <), and
<DLX= <LBX+90 (ext < of tri)
but <DLX also = <DAL + <ADL (ext < of tri), and <ADL=<LBX
==> <DAL =90
==><LCD=90 (since tri ALD=tri CLD)

Hi mum_sugoku,
Thanks so much! Yes, on hindsight, now I know I should have proceeded by using the congruent triangle results from part a.

I really appreciate your help,
Xiao Hu.