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
OLevel Elementary Math
PSLE marks the graduation of Primary school students and their entry into Secondary schools as teenagers. Discuss all issues about Secondary schooling here.

wei lun  KiasuNewbie
 Posts: 3
 Joined: Tue Jan 17, 2012 10:27 am
 Total Likes: 0
Re: sec 3 math qn
by verykiasu2010 » 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 ?
 verykiasu2010
 KiasuGrandMaster
 Posts: 11696
 Joined: Mon Jan 04, 2010 11:06 pm
 Total Likes: 2
Re: sec 3 math qn
by wei lun » Tue Jan 17, 2012 12:27 pm
thks for your reply but need to use equation to solve this problem.

wei lun  KiasuNewbie
 Posts: 3
 Joined: Tue Jan 17, 2012 10:27 am
 Total Likes: 0
Re: sec 3 math qn
by VisualTextEducation » 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 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! =)

VisualTextEducation  OrangeBelt
 Posts: 70
 Joined: Thu Apr 28, 2011 1:01 am
 Total Likes: 0
Re: sec 3 math qn
by savoury sweet » 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
Find the value of X
2 logx 4  3 log4 X = 1

savoury sweet  GreenBelt
 Posts: 114
 Joined: Mon Feb 27, 2012 12:50 pm
 Total Likes: 0
Re: sec 3 math qn
by mum_sugoku » 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
(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)

mum_sugoku  BlueBelt
 Posts: 269
 Joined: Thu Dec 29, 2011 11:21 am
 Total Likes: 10
Re: sec 3 math qn
by savoury sweet » Fri Mar 02, 2012 12:56 pm
Thank you so much mum_sugoku. Will show my gal today

savoury sweet  GreenBelt
 Posts: 114
 Joined: Mon Feb 27, 2012 12:50 pm
 Total Likes: 0
OLevel Elementary Math
by anneshirleygilbert » Mon Nov 24, 2014 5:54 pm
I need help for this question. Can somebody pls help.
TIA
Anne
TIA
Anne

anneshirleygilbert  BlueBelt
 Posts: 238
 Joined: Mon Jan 31, 2011 11:27 am
 Total Likes: 0
Re: OLevel Elementary Math
by Rejoin » 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 = (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.

Rejoin  OrangeBelt
 Posts: 56
 Joined: Tue Oct 09, 2012 9:59 am
 Total Likes: 1
Re: OLevel Elementary Math
by iamastudent » Fri Jan 02, 2015 8:26 pm
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!!

iamastudent  OrangeBelt
 Posts: 31
 Joined: Tue Aug 14, 2012 8:35 pm
 Total Likes: 0
 VIEW:
 active •
 unanswered •
 your posts
 The team • Delete all board cookies • All times are UTC + 8 hours
Powered by phpBB® Forum Software © phpBB Group
phpBB Metro Theme by PixelGoose Studio