O-Level Additional Math
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PiggyLalala:
:thankyou: very much for the solution. Was using the wrong approach- change of base n made a big mess... haha..
Dear Sir/Mdm,mathtuition88:
[quote=\"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
Please see my answer at http://mathtuition88.com/2013/05/12/o-level-logarithm-question-challenging/
Best wishes.
I did not think of letting y = the log expression n change it into its exponential form.
Thank you so much.[/quote]You are welcome, glad to help.
That question is quite creative, enjoyed solving it. -
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? " -
Qwertymum:
Father's money [-1000-][1U][-------------------9U-------------------]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][--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 -
jieheng:
Father's money [-1000-][1U][-------------------9U-------------------]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? \"
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. -
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/2013/04/21/amaths-solve-cubic-equation-by-synthetic-division-or-comparing-coefficients-video/
Cheers! -
积少成多: How can doing at least one Maths question per day help you improve!
We all know the saying “an apple a day keeps the doctor away“. Many essential activities, like eating, exercising, sleeping, needs to be done on a daily basis.
Mathematics is no different!
Here is a surprising fact of how much students can achieve if they do at least one Maths question per day. (the question must be substantial and worth at least 5 marks)
This study plan is based on the concept of 积少成多, or “Many little things add up“. Also, this method prevents students from getting rusty in older topics, or totally forgetting the earlier topics. Also, this method makes use of the fact that the human brain learns during sleep, so if you do mathematics everyday, you are letting your brain learn during sleep everyday.
Let’s take the example of Additional Mathematics.
Exam is on 24/25 October 2013.
Let’s say the student starts the “One Question per day” Strategy on 20 May 2013
Days till exam: 157 days (22 weeks or 5 months, 4 days)
So, 157 days = 157 questions (or more!)
Read the full article at:
http://mathtuition88.com/2013/05/18/how-can-doing-at-least-one-maths-question-per-day-help-you-improve-maths-tuition-stud/ -
Hi, ps help to solve sec one math
1) the sum of first n terms of a series is (n^2 + 2n) for all values of n. find the first three terms of the series.
2. Find the sum of the even numbers from 50 to 100 inclusive.
3. Write down the nth term and the sum of the first n terms of an arithmetic progression whose first term is a and whose common difference is d. Use these formulae to find the sum of all the numbers between 100 and 200 that are divisible by 7.
4. In the arithmetic progression whose first term is -27,the tenth term is equal to the sum of the first nine terms. Calculate the common difference.
5. The sum of the first 6 terms of an arithmetical progression is 55.5 and the sum of the next 6 terms is 145.5. Find the common difference and first term.
Thanks . Ps show me the working. -
lost boy:
1)Hi, ps help to solve sec one math
1) the sum of first n terms of a series is (n^2 + 2n) for all values of n. find the first three terms of the series.
Sn (Sum of first n terms) = n^2 + 2n
T1 (1st term) = S1 (Sum of first term) = 1 + 2 = 3
S2 (Sum of first 2 terms) = 2^2 + 2*2 = 4 + 4 = 8
T2 (2nd term) = S2 - S1 = 8 -3 = 5
S3 = 3^2 + 2*3 = 9 + 6 = 15
T3 = S3 - S2 = 15 - 8 = 7 -
lost boy:
3)Hi, ps help to solve sec one math
3. Write down the nth term and the sum of the first n terms of an arithmetic progression whose first term is a and whose common difference is d. Use these formulae to find the sum of all the numbers between 100 and 200 that are divisible by 7.
a = first term , d = common difference
Tn (nth term) = a + (n-1)*d
Sn (sum of first n terms) = (n/2)*[ 2a + (n-1)*d ]
1st no that is divisible by 7 is 105
Last no that is divisible by 7 is 196
105 , 112 , ...................................... , 196
a = 105 , d = 7
Tn = a + (n-1)*d
196 = 105 + (n-1)*7
(n-1) = 13
n = 14
Sn = (n/2)*[ 2a + (n-1)*d ]
S14 = (14/2)*[2*105 + (14-1)*7] = 2107 -
lost boy:
Hi, ps help to solve sec one math
2. Find the sum of the even numbers from 50 to 100 inclusive.
a = first term , d = common difference
Tn (nth term) = a + (n-1)*d
Sn (sum of first n terms) = (n/2)*[ 2a + (n-1)*d ]
a = 50 , d = 2
Tn = a + (n-1)*d
100 = 50 + (n-1)*2
n = 26
Sn = (n/2)*[ 2a + (n-1)*d ]
S26 = (26/2)*[2*50 + (26-1)*2] = 1950
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