O-Level 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.
Post Reply
wei lun
KiasuNewbie
KiasuNewbie
Posts: 3
Joined: Tue Jan 17,

sec 3 math qn

Post by wei lun » 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

verykiasu2010
KiasuGrandMaster
KiasuGrandMaster
Posts: 11696
Joined: Mon Jan 04,
Total Likes:2

Re: sec 3 math qn

Post 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 ?

wei lun
KiasuNewbie
KiasuNewbie
Posts: 3
Joined: Tue Jan 17,

Re: sec 3 math qn

Post by wei lun » Tue Jan 17, 2012 12:27 pm

thks for your reply but need to use equation to solve this problem.

VisualTextEducation
OrangeBelt
OrangeBelt
Posts: 70
Joined: Thu Apr 28,

Re: sec 3 math qn

Post 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 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 :imcool: :dancing:

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
GreenBelt
Posts: 114
Joined: Mon Feb 27,

Re: sec 3 math qn

Post 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


mum_sugoku
BrownBelt
BrownBelt
Posts: 606
Joined: Thu Dec 29,
Total Likes:15

Re: sec 3 math qn

Post 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
(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)

savoury sweet
GreenBelt
GreenBelt
Posts: 114
Joined: Mon Feb 27,

Re: sec 3 math qn

Post by savoury sweet » Fri Mar 02, 2012 12:56 pm

Thank you so much mum_sugoku. Will show my gal today :smile:

anneshirleygilbert
BlueBelt
BlueBelt
Posts: 238
Joined: Mon Jan 31,

O-Level Elementary Math

Post by anneshirleygilbert » Mon Nov 24, 2014 5:54 pm

Image I need help for this question. Can somebody pls help.
TIA
Anne

Rejoin
OrangeBelt
OrangeBelt
Posts: 57
Joined: Tue Oct 09,
Total Likes:1

Re: O-Level Elementary Math

Post 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 = (180-42)/2 = 69 degrees.

Angle ADB = 69-64 = 5 degrees.

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

Angle DBC = 180 - 159 = 21 degrees.

iamastudent
OrangeBelt
OrangeBelt
Posts: 54
Joined: Tue Aug 14,

Re: O-Level Elementary Math

Post by iamastudent » Fri Jan 02, 2015 8:26 pm

Hi,
I need help for the following questions
1)
Image
part d.

My answers for part a,b,c:
Image

2)
Image

3)
Image

4)
Image

5)
Image

Part iv

My answers for i,ii,iii:
Image
Image

6)
Image

Thank you!!

Post Reply