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### sec 3 math qn

Posted: Tue Jan 17, 2012 10:51 am
can someone help me solve this sec 3 math problem

There is a 2 digit no. such that the sum of its digit is 6 while the product of the digits is 1/3 the original no. Find this no. (Hint: Let x be one of the digits)

thks

### Re: sec 3 math qn

Posted: Tue Jan 17, 2012 12:00 pm
wei lun wrote:can someone help me solve this sec 3 math problem

There is a 2 digit no. such that the sum of its digit is 6 while the product of the digits is 1/3 the original no. Find this no. (Hint: Let x be one of the digits)

thks
sum = 6 means only the following number pairs are possible :

1 5
2 4
3 3
4 2
5 1

and 1/3 of the original numbers are : 5, 8, 11, 14, 17
product of the digits are : 5, 8, 9, 8, 5

the answer is obvious, this is NMOS question ?

### Re: sec 3 math qn

Posted: Tue Jan 17, 2012 12:27 pm
thks for your reply but need to use equation to solve this problem.

### Re: sec 3 math qn

Posted: Tue Jan 17, 2012 1:13 pm
wei lun wrote:thks for your reply but need to use equation to solve this problem.
Hi wel lun,

Here you go:

Let the first digit be x,

Since the sum of the digits is 6, the second digit would be 6-x,

Therefore, the products of the digits would be = x(6-x)
= 6x-x^2

If the first digit is x, and it is a 2 digit number, x must be in the tens place, hence the NUMERICAL value of x would be = 10 x (x)
= 10x

Since the second digit is in the ones place, the NUMERICAL value of 6-x would be
= 1 x (6-x)
= 6-x

Hence, the NUMERICAL value of the 2 digit number would be = 10x + (6-x)
= 10x + 6 - x
= 9x +6

Since the products of the digits IS EQUAL to the 1/3 of the original number, we can construct the equation as follows,

6x-x^2 = (1/3) x (9x+6)

Rearranging the equation, we will have,

6x-x^2 = 3x+2
6x = x^2+3x+2
0 = x^2+3x-6x+2
0 = x^2-3x+2
x^2-3x+2 = 0
Factorising using the Cross technique you learnt in secondary 2,

You will get,

(x-2)(x-1)=0
x=2 or x=1 (first digit)
6-x=4 or x=5 (second digit)

Hence, the original number may be 24 or 15.

SOLVED! =D  You may call me at 92220737 if you require any further assistance in these maths questions, I will be glad to help! =)

### Re: sec 3 math qn

Posted: Thu Mar 01, 2012 10:08 pm
Can someone pls enlighten me the following maths qn as I am really clueless when come to my gal's Secondary maths qn. Thanks

Find the value of X
2 logx 4 - 3 log4 X = 1

### Re: sec 3 math qn

Posted: Thu Mar 01, 2012 10:36 pm
savoury sweet wrote:Can someone pls enlighten me the following maths qn as I am really clueless when come to my gal's Secondary maths qn. Thanks

Find the value of X
2 logx 4 - 3 log4 X = 1
first change log base x to log base 4, ie
logx 4=(log4 4) /(log4 x) = 1/(log4 x)

let y= log4 x

hence eqn becomes 2[1/y] - 3y =1
or 2 - 3y^2 = y
or 3y^2 + y -2 =0
(3y-2)(y+1)=0

hence y=2/3, or y=-1

log4 x= 2/3 --> x=4^(2/3)

or log4 x =-1 --> x=4^(-1)

### Re: sec 3 math qn

Posted: Fri Mar 02, 2012 12:56 pm
Thank you so much mum_sugoku. Will show my gal today ### O-Level Elementary Math

Posted: Mon Nov 24, 2014 5:54 pm I need help for this question. Can somebody pls help.
TIA
Anne

### Re: O-Level Elementary Math

Posted: Wed Dec 17, 2014 5:25 pm
Angle BOC = 64 x 2 = 128 degrees (because angle at the centre is 2x angle at the circumference)

Triangle BOC is isoceles. So that means angle OCB = angle OBC = (180 - 128)/2 = 26 degrees.

Angle AOC = 180 - 16 - 26 = 138 degrees.
Angle DOC = 180 - 138 = 42 degrees

Triangle OCD is isoceles. So that means angle OCD = angle DOC = (180-42)/2 = 69 degrees.

Angle ADB = 69-64 = 5 degrees.

Angle ABD = 180 - 5 - 16 = 159 degrees.

Angle DBC = 180 - 159 = 21 degrees.

### Re: O-Level Elementary Math

Posted: Fri Jan 02, 2015 8:26 pm
Hi,
I need help for the following questions
1) part d. 2) 3) 4) 5) Part iv   