O-Level Additional Math

[Acsian]

Herbie:

When to use HCF and LCM to solve problem qn? Can someone advise?

LCM can be used for questions like \" Bus A leaves the terminal every 10 minutes. Bus B leaves the terminal every 15 minutes. If the two buses leave the terminal at 08 00 when will they next leave together?\"

HCF can be used for questions like \" There are 48 studens from Mexico 96 students from Canada and 72 students from Canada attending a education conference. What is the greatest number of groups that can be formed so that students from each country can be distributed equally among all groups?\"

[jieheng]

Acsian:

when


10x^2+40xy-24x-96y is factorised the answer is (2x+8y)(5x-12) *not fully factorised*


But when 2ax-4ay+3bx-6by is factorised, why is the answer is
(2ab-3b)(x-2y)and not

(2ab+3b)(x-2y)

please help[/b]

2ax-4ay+3bx-6by

=2ax+3bx-4ay-6by

=x(2a+3b)-2y(2a+3b)

=(2a+3b)(x-2y)

[k1ndan]

Please help to solve these 2 Sec 2 Maths problem.


1) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.



2) Given that m^2 + 8m = v^2 + 8v and that m is not equal to v, find the value of 5(m + v).



Thanks.

[Sun_2010]

Hi k1ndan,


Let me give it a try and see if my rusty grey sells can work
1 ) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.
3a+e=7
(3a+e)^2 = 7^2
9a^2 + e^2 + 6ae= 49 -------(i)

4ae/3=2 so 6ae=9 -------------(ii)

substituting in (i)
9a^2 + e^2 +9= 49
9a^2 + e^2 =40 -------------(iii)

(e-3a)^2
= 9a^2 + e^2 - 6ae
= 40-9
=31

hence (e-3a) = square root of 31

gtg, will be back later for 2nd question

[Sun_2010]

2) Given that m^2 + 8m = v^2 + 8v and that m is not equal to v, find the value of 5(m + v).


m^2 + 8m = v^2 + 8v
m^2 - v^2 + 8v -8M = 0
(m+v)(m-v) + 8(m-v) = 0
(m-v) (m+v+8)=0
since mis not equal to v, m+v= -8
5(m+v) = -40

[CoffeeCat]

Sun_2010:

Hi k1ndan,


Let me give it a try and see if my rusty grey sells can work
1 ) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.
3a+e=7
(3a+e)^2 = 7^2
9a^2 + e^2 + 6ae= 49 -------(i)

4ae/3=2 so 6ae=9 -------------(ii)

substituting in (i)
9a^2 + e^2 +9= 49
9a^2 + e^2 =40 -------------(iii)

(e-3a)^2
= 9a^2 + e^2 - 6ae
= 40-9
=31

hence (e-3a) = square root of 31

gtg, will be back later for 2nd question

by the way, without any other restrictions on the values it is possible that the answer can be -sqrt(31) as well.

[JadeDry]

I currently use "New Syllabus Mathematics" (8th Grade) books, and would appreciate if you could recommend additional good quality publications.


Thanks in advance.

[PiggyLalala]

What kind of assessment books are you looking at? Topical or paper by paper? Are you looking for more challenging questions? Or are you looking for one that focus more on drill and practice.

[hometutors.018849sg]

JadeDry:

I currently use \"New Syllabus Mathematics\" (8th Grade) books, and would appreciate if you could recommend additional good quality publications.


Thanks in advance.

A good tutor / mentor would be able to select what is good for the student.

[elkniwt]

Hi,


For compound interest. If the principal amt is invested for a non-integral no. of the compounding period, how is it computed?

Amt = $1000
Interest = 1% per annum compounded yearly
Invested for 2yr and 3mths
Final amt = (1.01)^2 x 1000
OR
Final amt = (1.01)^2.25 x 1000 ??

Thanks.