OLevel Elementary Math
sec 3 math qn
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
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

 KiasuGrandMaster
 Posts: 11679
 Joined: Mon Jan 04,
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Re: sec 3 math qn
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 ?
 VisualTextEducation
 OrangeBelt
 Posts: 61
 Joined: Thu Apr 28,
Re: sec 3 math qn
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 6x,
Therefore, the products of the digits would be = x(6x)
= 6xx^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 6x would be
= 1 x (6x)
= 6x
Hence, the NUMERICAL value of the 2 digit number would be = 10x + (6x)
= 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,
6xx^2 = (1/3) x (9x+6)
Rearranging the equation, we will have,
6xx^2 = 3x+2
6x = x^2+3x+2
0 = x^2+3x6x+2
0 = x^23x+2
x^23x+2 = 0
Factorising using the Cross technique you learnt in secondary 2,
You will get,
(x2)(x1)=0
x=2 or x=1 (first digit)
6x=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! =)
 savoury sweet
 GreenBelt
 Posts: 114
 Joined: Mon Feb 27,
Re: sec 3 math qn
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
Find the value of X
2 logx 4  3 log4 X = 1
 mum_sugoku
 BrownBelt
 Posts: 606
 Joined: Thu Dec 29,
 Total Likes:15
Re: sec 3 math qn
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
(3y2)(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)
 savoury sweet
 GreenBelt
 Posts: 114
 Joined: Mon Feb 27,
 anneshirleygilbert
 BlueBelt
 Posts: 238
 Joined: Mon Jan 31,
Re: OLevel Elementary Math
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 = (18042)/2 = 69 degrees.
Angle ADB = 6964 = 5 degrees.
Angle ABD = 180  5  16 = 159 degrees.
Angle DBC = 180  159 = 21 degrees.
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 = (18042)/2 = 69 degrees.
Angle ADB = 6964 = 5 degrees.
Angle ABD = 180  5  16 = 159 degrees.
Angle DBC = 180  159 = 21 degrees.
 iamastudent
 OrangeBelt
 Posts: 43
 Joined: Tue Aug 14,
Re: OLevel Elementary Math
Hi,
I need help for the following questions
1)
part d.
My answers for part a,b,c:
2)
3)
4)
5)
Part iv
My answers for i,ii,iii:
6)
Thank you!!
I need help for the following questions
1)
part d.
My answers for part a,b,c:
2)
3)
4)
5)
Part iv
My answers for i,ii,iii:
6)
Thank you!!