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
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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
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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.

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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! =)

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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

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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)

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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

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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

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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.

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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!!

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