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Re: Secondary School Mathematics

Posted: Mon May 06, 2013 5:08 pm
by mathtuition88
the above method is very nice. :)

Expansion and Factorisation, HCI IP Sec 2 Revision Pack

Posted: Tue May 07, 2013 1:21 pm
by mathtuition88
Does your child need practice on Expansion and Factorisation, especially the more challenging questions? It is a key technique that needs to be honed for the O Levels.

Practice Hwa Chong Institution Expansion and Factorisation (Sec 2, IP) here.

http://mathtuition88.wordpress.com/2013 ... orisation/

Re: Secondary School Mathematics

Posted: Sat May 11, 2013 8:32 am
by PiggyLalala
Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA

Re: Secondary School Mathematics

Posted: Sun May 12, 2013 12:42 am
by mathtuition88
PiggyLalala wrote:Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA
Dear Sir/Mdm,

Please see my answer at http://mathtuition88.com/2013/05/12/o-l ... allenging/

Best wishes.

Re: Secondary School Mathematics

Posted: Sun May 12, 2013 8:40 am
by PiggyLalala
mathtuition88 wrote:
PiggyLalala wrote:Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA
Dear Sir/Mdm,

Please see my answer at http://mathtuition88.com/2013/05/12/o-l ... allenging/

Best wishes.
:thankyou: very much for the solution. Was using the wrong approach- change of base n made a big mess... haha..

I did not think of letting y = the log expression n change it into its exponential form.

Thank you so much.

Re: Secondary School Mathematics

Posted: Sun May 12, 2013 2:37 pm
by mathtuition88
PiggyLalala wrote:
mathtuition88 wrote:
PiggyLalala wrote:Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA
Dear Sir/Mdm,

Please see my answer at http://mathtuition88.com/2013/05/12/o-l ... allenging/

Best wishes.
:thankyou: very much for the solution. Was using the wrong approach- change of base n made a big mess... haha..

I did not think of letting y = the log expression n change it into its exponential form.

Thank you so much.
You are welcome, glad to help. :)
That question is quite creative, enjoyed solving it.

Re: Secondary School Mathematics

Posted: Thu May 16, 2013 8:41 am
by Qwertymum
My son asks me a Math question from his MY Revision Paper which I need help. Find it rather challenging! Or if there is something not right? Could anyone out there kindly help to solve, thanks in advance.

" A father in his will left all his money to his children as such: $1000 and 1/10 of what then remains to the first born; then $2000 and 1/10 of what then remains to the second; then $3000 and 1/10 of what then remains to the third born; and so no. When this was done each child had the same amount. How many children were there? "

Re: Secondary School Mathematics

Posted: Thu May 16, 2013 10:11 am
by jieheng
Qwertymum wrote:My son asks me a Math question from his MY Revision Paper which I need help. Find it rather challenging! Or if there is something not right? Could anyone out there kindly help to solve, thanks in advance.

" A father in his will left all his money to his children as such: $1000 and 1/10 of what then remains to the first born; then $2000 and 1/10 of what then remains to the second; then $3000 and 1/10 of what then remains to the third born; and so no. When this was done each child had the same amount. How many children were there? "
Father's money [-1000-][1U][-------------------9U-------------------]
Father's money [-1000-][1U][--2000--][1P][-----------9P-----------]

Father's money = 1000 + 10U

the money that 1st child will receive = 1000 + 1U

the money that 2nd child will receive = 2000 + 1P

Each child will receive the same amount of money
1000 + 1U = 2000 + 1P
1U = 1000 + 1P ------------(1)

9U = 2000 + 10P ----------(2)

(1)*9 , 9U = 9000 + 9P ---(3)

(2) = (3) ,

2000 + 10P = 9000 + 9P
1P = 7000

From (1)
1U = 1000 + 1P = 1000 + 7000 = 8000

Father's money = 1000 + 10U = 1000 + 10*8 = 81000

the amount of money that each child will receive = 1000 + 1U = 9000

No of children = 81000 / 9000 = 9

Re: Secondary School Mathematics

Posted: Thu May 16, 2013 11:26 am
by Qwertymum
jieheng wrote:
Qwertymum wrote:My son asks me a Math question from his MY Revision Paper which I need help. Find it rather challenging! Or if there is something not right? Could anyone out there kindly help to solve, thanks in advance.

" A father in his will left all his money to his children as such: $1000 and 1/10 of what then remains to the first born; then $2000 and 1/10 of what then remains to the second; then $3000 and 1/10 of what then remains to the third born; and so no. When this was done each child had the same amount. How many children were there? "
Father's money [-1000-][1U][-------------------9U-------------------]
Father's money [-1000-][1U][--2000--][1P][-----------9P-----------]

Father's money = 1000 + 10U

the money that 1st child will receive = 1000 + 1U

the money that 2nd child will receive = 2000 + 1P

Each child will receive the same amount of money
1000 + 1U = 2000 + 1P
1U = 1000 + 1P ------------(1)

9U = 2000 + 10P ----------(2)

(1)*9 , 9U = 9000 + 9P ---(3)

(2) = (3) ,

2000 + 10P = 9000 + 9P
1P = 7000

From (1)
1U = 1000 + 1P = 1000 + 7000 = 8000

Father's money = 1000 + 10U = 1000 + 10*8 = 81000

the amount of money that each child will receive = 1000 + 1U = 9000

No of children = 81000 / 9000 = 9

:thankyou: Wow! That's fast. Thanks a lot.

Re: Cubic Equations

Posted: Thu May 16, 2013 1:28 pm
by alwaysLovely
Solving cubic equations is taught in Secondary 3 Additional Maths.

There are three methods:
1) Synthetic division
2) Comparing coefficients
3) Long division

I've two videos on demonstrating how to solve a cubic equation using the first two methods.

I believe this post will be useful for you.
http://www.singaporeolevelmaths.com/201 ... nts-video/

Cheers!